Careers360 Logo
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

Edited By Ramraj Saini | Updated on Sep 14, 2023 10:22 PM IST | #CBSE Class 12th

NCERT Differential Equations Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations are provided here. In class 11th, you have already learned how to differentiate a given function (f) with respect to an independent variable. In this article, you will get NCERT solutions for class 12 maths chapter 9 for all major topics of NCERT Class 12 maths syllabus. The equation of function and its one or more derivatives is called a differential equation.

In this differential equations class 12 questions and answers, some basic concepts related to the differential equations solutions, particular solutions, and general solutions of differential equations class 12 will be comprehensively discussed. In NCERT solutions for chapter 9 class 12 maths, questions from all these topics are covered in this article. If you are interested in other subjects then you can refer to NCERT solutions for class 12

Apply to Aakash iACST Scholarship Test 2024

Applications for Admissions are open.

You will also learn some methods to find the differential equations solutions, the formation of differential equations class 12, and applications of differential equations in different areas in this NCERT class 12 maths chapter 9 question answer are also explained in details. Questions related to these topics are also covered in the NCERT solutions for class 12 maths ch 9 differential equations article. You can refer to NCERT solutions from classes 6 to 12 to learn CBSE maths and science.

ALLEN JEE Exam Prep

Start your JEE preparation with ALLEN

Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

Also read :

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

NCERT Differential Equations Class 12 Questions And Answers PDF Free Download

Download PDF

NCERT Differential Equations Class 12 Questions And Answers - Important Formulae

>> Ordinary Differential Equations (ODEs): Ordinary Differential Equations involve derivatives of a function concerning a single independent variable. They are commonly used to model dynamic systems and phenomena.

>> Partial Differential Equations (PDEs): Partial Differential Equations involve derivatives of a function concerning multiple independent variables. They are frequently used in physics to describe phenomena like heat diffusion, wave propagation, and fluid dynamics.

>> Types of Differential Equations: Differential equations can be categorised based on their order, linearity, and specific properties. Common types include:

  • First-Order Differential Equations

  • Second-Order Differential Equations

  • Linear Differential Equations

  • Nonlinear Differential Equations

  • Homogeneous Differential Equations

  • Non-Homogeneous Differential Equations

JEE Main Important Physics formulas

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

>> Methods for Solving Differential Equations: Various techniques can be employed to solve differential equations, including:

  • Separation of Variables

  • Integrating Factors

  • Exact Differential Equations

  • Linear Differential Equations with Constant Coefficients

  • Method of Undetermined Coefficients

  • Variation of Parameters

  • Laplace Transforms

>> Applications of Differential Equations: Differential equations have widespread applications in science and engineering. Some examples include modelling population growth, describing electrical circuits, predicting radioactive decay, and simulating fluid flow.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations for CBSE Exam.

NCERT Differential Equations Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT differential equations class 12 solutions - Exercise: 9.1

Question:1 Determine order and degree (if defined) of differential equation \frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0

Answer:

Given function is
\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0
We can rewrite it as
y^{''''}+\sin(y''') =0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''''}

Therefore, the order of the given differential equation \frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0 is 4
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:2 Determine order and degree (if defined) of differential equation y' + 5y = 0

Answer:

Given function is
y' + 5y = 0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{'}
Therefore, the order of the given differential equation y' + 5y = 0 is 1
Now, the given differential equation is a polynomial equation in its derivatives and its highest power raised to y ' is 1
Therefore, it's a degree is 1.

Question:3 Determine order and degree (if defined) of differential equation \left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0

Answer:

Given function is
\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0
We can rewrite it as
(s^{'})^4+3s.s^{''} =0
Now, it is clear from the above that, the highest order derivative present in differential equation is s^{''}

Therefore, the order of the given differential equation \left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0 is 2
Now, the given differential equation is a polynomial equation in its derivatives and power raised to s '' is 1
Therefore, it's a degree is 1

Question:4 Determine order and degree (if defined) of differential equation.

\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0

Answer:

Given function is
\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0
We can rewrite it as
(y^{''})^2+\cos y^{''} =0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''}

Therefore, the order of the given differential equation \left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0 is 2
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:5 Determine order and degree (if defined) of differential equation.

\frac{d^2y}{dx^2} = \cos 3x + \sin 3x

Answer:

Given function is
\frac{d^2y}{dx^2} = \cos 3x + \sin 3x
\Rightarrow \frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0

Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''}\left ( \frac{d^2y}{dx^2} \right )

Therefore, order of given differential equation \frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0 is 2
Now, the given differential equation is a polynomial equation in it's dervatives \frac{d^2y}{dx^2} and power raised to \frac{d^2y}{dx^2} is 1
Therefore, it's degree is 1

Question:6 Determine order and degree (if defined) of differential equation (y''')^2 + (y'')^3 + (y')^4 + y^5= 0

Answer:

Given function is
(y''')^2 + (y'')^3 + (y')^4 + y^5= 0 Now, it is clear from the above that, the highest order derivative present in differential equation is y^{'''}

Therefore, order of given differential equation (y''')^2 + (y'')^3 + (y')^4 + y^5= 0 is 3 Now, the given differential equation is a polynomial equation in it's dervatives y^{'''} , y^{''} \ and \ y^{'} and power raised to y^{'''} is 2
Therefore, it's degree is 2

Question:7 Determine order and degree (if defined) of differential equation

y''' + 2y'' + y' =0

Answer:

Given function is
y''' + 2y'' + y' =0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{'''}

Therefore, order of given differential equation y''' + 2y'' + y' =0 is 3
Now, the given differential equation is a polynomial equation in it's dervatives y^{'''} , y^{''} \ and \ y^{'} and power raised to y^{'''} is 1
Therefore, it's degree is 1

Question:8 Determine order and degree (if defined) of differential equation

y' + y = e^x

Answer:

Given function is
y' + y = e^x
\Rightarrow y^{'}+y-e^x=0

Now, it is clear from the above that, the highest order derivative present in differential equation is y^{'}

Therefore, order of given differential equation y^{'}+y-e^x=0 is 1
Now, the given differential equation is a polynomial equation in it's dervatives y^{'} and power raised to y^{'} is 1
Therefore, it's degree is 1

Question:9 Determine order and degree (if defined) of differential equation

y'' + (y')^2 + 2y = 0

Answer:

Given function is
y'' + (y')^2 + 2y = 0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''}

Therefore, order of given differential equation y'' + (y')^2 + 2y = 0 is 2
Now, the given differential equation is a polynomial equation in it's dervatives y^{''} \ and \ y^{'} and power raised to y^{''} is 1
Therefore, it's degree is 1

Question:10 Determine order and degree (if defined) of differential equation

y'' + 2y' + \sin y = 0

Answer:

Given function is
y'' + 2y' + \sin y = 0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''}

Therefore, order of given differential equation y'' + 2y' + \sin y = 0 is 2
Now, the given differential equation is a polynomial equation in it's dervatives y^{''} \ and \ y^{'} and power raised to y^{''} is 1
Therefore, it's degree is 1

Question:11 The degree of the differential equation \left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0 is

(A) 3

(B) 2

(C) 1

(D) not defined

Answer:

Given function is
\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0
We can rewrite it as
(y^{''})^3+(y^{'})^2+\sin y^{'}+1=0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''}

Therefore, order of given differential equation \left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0 is 2
Now, the given differential equation is a not polynomial equation in it's dervatives
Therefore, it's degree is not defined

Therefore, answer is (D)

Question:12 The order of the differential equation 2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0 is

(A) 2

(B) 1

(C) 0

(D) Not Defined

Answer:

Given function is
2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0
We can rewrite it as
2x.y^{''}-3y^{'}+y=0
Now, it is clear from the above that, the highest order derivative present in differential equation is y^{''}

Therefore, order of given differential equation 2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0 is 2

Therefore, answer is (A)


NCERT differential equations class 12 solutions - Exercise: 9.2

Question:1 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = e^x + 1 \qquad :\ y'' -y'=0

Answer:

Given,

y = e^x + 1

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x

Again, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y' }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x

\implies y'' = e^x

Substituting the values of y’ and y'' in the given differential equations,

y'' - y' = e x - e x = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = x^2 + 2x + C\qquad:\ y' -2x - 2 =0

Answer:

Given,

y = x^2 + 2x + C

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 2x + C) = 2x + 2

Substituting the values of y’ in the given differential equations,

y' -2x - 2 =2x + 2 - 2x - 2 = 0= RHS .

Therefore, the given function is the solution of the corresponding differential equation.

Question:3. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = \cos x + C\qquad :\ y' + \sin x = 0

Answer:

Given,

y = \cos x + C

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(cosx + C) = -sinx

Substituting the values of y’ in the given differential equations,

y' - \sin x = -sinx -sinx = -2sinx \neq RHS .

Therefore, the given function is not the solution of the corresponding differential equation.

Question:4. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}

Answer:

Given,

y = \sqrt{1 + x^2}

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{1 + x^2}) = \frac{2x}{2\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}}

Substituting the values of y in RHS,

\frac{x\sqrt{1+x^2}}{1 + x^2} = \frac{x}{\sqrt{1+x^2}} = LHS .

Therefore, the given function is a solution of the corresponding differential equation.

Question:5 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax\qquad :\ xy' = y\;(x\neq 0)

Answer:

Given,

y = Ax

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(Ax) = A

Substituting the values of y' in LHS,

xy' = x(A) = Ax = y = RHS .

Therefore, the given function is a solution of the corresponding differential equation.

Question:6. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)

Answer:

Given,

y = x\sin x

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(xsinx) = sinx + xcosx

Substituting the values of y' in LHS,

xy' = x(sinx + xcosx)

Substituting the values of y in RHS.

\\xsinx + x\sqrt{x^2 - x^2sin^2x} = xsinx + x^2\sqrt{1-sinx^2} = x(sinx+xcosx) = LHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:7 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)

Answer:

Given,

xy = \log y + C

Now, differentiating both sides w.r.t. x,

\\ y + x\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(logy) = \frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}\\ \\ \implies y^2 + xyy' = y' \\ \\ \implies y^2 = y'(1-xy) \\ \\ \implies y' = \frac{y^2}{1-xy}

Substituting the values of y' in LHS,

y' = \frac{y^2}{1-xy} = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:8 In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y

Answer:

Given,

y - cos y = x

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} +siny\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x) = 1

\implies y' + siny.y' = 1

\implies y'(1 + siny) = 1

\implies y' = \frac{1}{1+siny}

Substituting the values of y and y' in LHS,

(\ (x+cosy)\sin y + \cos y + x) (\frac{1}{1+siny})

= [x(1+siny) + cosy(1+siny)]\frac{1}{1+siny}

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:9 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0

Answer:

Given,

x + y = \tan^{-1}y

Now, differentiating both sides w.r.t. x,

\\ 1 + \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 + y^2}\frac{\mathrm{d} y}{\mathrm{d} x}\\ \\ \implies1+y^2 = y'(1-(1+y^2)) = -y^2y' \\ \implies y' = -\frac{1+y^2}{y^2}

Substituting the values of y' in LHS,

y^2(-\frac{1+y^2}{y^2}) + y^2 + 1 = -1- y^2+ y^2 +1 = 0 = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:10 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)

Answer:

Given,

y = \sqrt{a^2 - x^2}

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{a^2 - x^2}) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}

Substituting the values of y and y' in LHS,

x + y \frac{dy}{dx} = x + (\sqrt{a^2 - x^2})(\frac{-x}{\sqrt{a^2 - x^2}}) = 0 = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:11 The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4

Answer:

(D) 4

The number of constants in the general solution of a differential equation of order n is equal to its order.

Question:12 The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0

Answer:

(D) 0

In a particular solution of a differential equation, there is no arbitrary constant.


Differential Equations Class 12 NCERT Solutions - Exercise: 9.3

Question:1 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

\frac{x}{a} + \frac{y}{b} = 1

Answer:

Given equation is

\frac{x}{a} + \frac{y}{b} = 1
Differentiate both the sides w.r.t x
\frac{d\left ( \frac{x}{a}+\frac{y}{b} \right )}{dx}=\frac{d(1)}{dx}
\frac{1}{a}+\frac{1}{b}.\frac{dy}{dx} = 0\\ \frac{dy}{dx} = -\frac{b}{a}
Now, again differentiate it w.r.t x
\frac{d^2y}{dx^2} =0
Therefore, the required differential equation is \frac{d^2y}{dx^2} =0 or y^{''} =0

Question:2 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y^2 = a(b^2 - x^2)

Answer:

Given equation is
y^2 = a(b^2 - x^2)
Differentiate both the sides w.r.t x
\frac{d\left ( y^2 \right )}{dx}=\frac{d(a(b^2-x^2))}{dx}
2y\frac{dy}{dx}= -2ax\\ \\ y.\frac{dy}{dx}= -ax\\ \\ y.y^{'}=-ax -(i)
Now, again differentiate it w.r.t x
y^{'}.y^{'}+y.y^{''}= -a\\ (y^{'})^2+y.y^{''}=-a -(ii)
Now, divide equation (i) and (ii)
\frac{(y^{'})^2+y.y^{''}}{y.y^{'}}= \frac{-a}{-ax}\\ \\ x(y^{'})^2+x.y.y^{''}=y.y^{'}\\ \\ x(y^{'})^2+x.y.y^{''}-y.y^{'}=0
Therefore, the required differential equation is x(y^{'})^2+x.y.y^{''}-y.y^{'}=0

Question:3 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y = ae^{3x} + b e^{-2x}

Answer:

Given equation is
y = ae^{3x} + b e^{-2x} -(i)
Differentiate both the sides w.r.t x
\frac{d\left ( y \right )}{dx}=\frac{d(ae^{3x}+be^{-2x})}{dx}
y^{'}=\frac{dy}{dx}= 3ae^{3x}-2be^{-2x}\\ \\ -(ii)
Now, again differentiate w.r.t. x
y^{''}= \frac{d^2y}{dx^2} = 9ae^{3x}+4be^{-2x} -(iii)
Now, multiply equation (i) with 2 and add equation (ii)
2(ae^{3x}+be^{-2x})+(3a-2be^{-x}) = 2y+y^{'}\\ 5ae^{3x} = 2y+y^{'}\\ ae^{3x}= \frac{2y+y^{'}}{5} -(iv)
Now, multiply equation (i) with 3 and subtract from equation (ii)
3(ae^{3x}+be^{-2x})-(3a-2be^{-x}) = 3y-y^{'}\\ 5be^{-2x} = 3y-y^{'}\\ be^{-2x}= \frac{3y-y^{'}}{5} -(v)
Now, put values from (iv) and (v) in equation (iii)
y^{''}= 9.\frac{2y+y^{'}}{5}+4.\frac{3y-y^{'}}{5}\\ \\ y^{''}= \frac{18y+9y^{'}+12y-4y^{'}}{5}\\ \\ y^{''}= \frac{5(6y-y^{'})}{5}=6y-y^{'}\\ \\ y^{''}+y^{'}-6y=0

Therefore, the required differential equation is y^{''}+y^{'}-6y=0

Question:4 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y = e^{2x}(a+bx)

Answer:

Given equation is
y = e^{2x}(a+bx) -(i)
Now, differentiate w.r.t x
\frac{dy}{dx}= \frac{d(e^{2x}(a+bx))}{dx}= 2e^{2x}(a+bx)+e^{2x}.b -(ii)
Now, again differentiate w.r.t x
y^{''}= \frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx} = 4e^{2x}(a+bx)+2be^{2x}+2be^{2x}= 4e^{2x}(a+bx)+4be^{2x} -(iii)
Now, multiply equation (ii) with 2 and subtract from equation (iii)
4e^{2x}(a+bx)+4be^{2x}-2\left ( 2e^{2x}(a+bx)+be^{2x} \right )=y^{''}-2y^{'}\\ \\ 2be^{2x} = y^{''}-2y^{'}\\ \\ be^{2x}= \frac{y^{''}-2y^{'}}{2} -(iv)
Now,put the value in equation (iii)
y^{''}=4y+4.\frac{y^{''}-2y^{'}}{2}\\ \\ y^{''}= 4y+2y^{''}-4y^{'}\\ \\ y^{''}-4y^{'}+4y=0
Therefore, the required equation is y^{''}-4y^{'}+4y=0

Question:5 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y=e^x(a\cos x + b\sin x)

Answer:

Given equation is
y=e^x(a\cos x + b\sin x) -(i)
Now, differentiate w.r.t x
\frac{dy}{dx}= \frac{d(e^{x}(a\cos x+b\sin x))}{dx}= e^{x}(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x ) -(ii)
Now, again differentiate w.r.t x
y^{''}= \frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx} =e^{x}(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x ) +e^x(-a\sin x+b\cos x )+e^x(-a\cos x-b\sin x)
=2e^x(-a\sin x+b\cos x ) -(iii)
Now, multiply equation (i) with 2 and multiply equation (ii) with 2 and add and subtract from equation (iii) respectively
we will get

y^{''}-2y^{'}+2y = 0
Therefore, the required equation is y^{''}-2y^{'}+2y = 0

Question:6 Form the differential equation of the family of circles touching the y-axis at origin.

Answer:

1628484808301 If the circle touches y-axis at the origin then the centre of the circle lies at the x-axis
Let r be the radius of the circle
Then, the equation of a circle with centre at (r,0) is
(x-r)^2+(y-0)^2 = r^2
x^2+r^2-2xr+y^2=r^2\\ x^2+y^2-2xr=0 -(i)
Now, differentiate w.r.t x
2x+2y\frac{dy}{dx}-2r=0\\ y\frac{dy}{dx}\Rightarrow yy^{'}+x-r=0
yy^{'}+x=r -(ii)
Put equation (ii) in equation (i)
x^2+y^2=2x(yy^{'}+x)\\ y^2=2xyy^{'}+x^2
Therefore, the required equation is y^2=2xyy^{'}+x^2

Question:7 Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

Equation of perabola having vertex at origin and axis along positive y-axis is
x^2= 4ay (i)
Now, differentiate w.r.t. c
2x= 4a\frac{dy}{dx}\\ \\ \frac{dy}{dx} =y^{'}= \frac{x}{2a}
a=\frac{x}{2y^{'}} -(ii)
Put value from equation (ii) in (i)
x^2= 4y.\frac{x}{2y^{'}}\\ xy^{'}-2y = 0
Therefore, the required equation is xy^{'}-2y = 0

Question:8 Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Answer:

Equation of ellipses having foci on y-axis and centre at origin is
\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 -
Now, differentiate w..r.t. x
\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\ -(i)
Now, again differentiate w.r.t. x
\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right ) -(ii)
Put value from equation (ii) in (i)
Our equation becomes
\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0
Therefore, the required equation is xyy^{''}-x(y^{'})^2+yy^{'}= 0

Question:9 Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Answer:

Equation of hyperbolas having foci on x-axis and centre at the origin
\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1
Now, differentiate w..r.t. x
\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\ -(i)
Now, again differentiate w.r.t. x
\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right ) -(ii)
Put value from equation (ii) in (i)
Our equation becomes
\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0
Therefore, the required equation is xyy^{''}-x(y^{'})^2+yy^{'}= 0

Question:10 Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer:

1628484856150 Equation of the family of circles having centre on y-axis and radius 3 units
Let suppose centre is at (0,b)
Now, equation of circle with center (0,b) an radius = 3 units
(x-0)^2+(y-b)^2=3^2 \ \ \ \ \ \ \ \ \ \ \ -(i)\\ x^2+y^2+b^2-2yb = 9
Now, differentiate w.r.t x
we get,
2x+2yy^{'}-2by^{'}= 0\\ 2x+2y(y-b)= 0\\ (y-b)=\frac{-x}{y^{'}} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
Put value fro equation (ii) in (i)
(x-0)^2+(\frac{-x}{y^{'}})^2=3^2 \\ x^2+\frac{x^2}{(y^{'})^2}=9\\ x^2(y^{'})^2+x^2=9(y^{'})^2\\ \\ (x^2-9)(y^{'})^2+x^2 = 0
Therefore, the required differential equation is (x^2-9)(y^{'})^2+x^2 = 0

Question:11 Which of the following differential equations has y = c_1e^x + c_2e^{-x} as the general solution?

(A) \frac{d^2y}{dx^2} + y = 0

(B) \frac{d^2y}{dx^2} - y = 0

(C) \frac{d^2y}{dx^2} +1 = 0

(D) \frac{d^2y}{dx^2} -1 = 0

Answer:

Given general solution is
y = c_1e^x + c_2e^{-x}
Differentiate it w.r.t x
we will get
\frac{dy}{dx} = c_1e^x-c_2e^{-x}
Again, Differentiate it w.r.t x
\frac{d^2y}{dx^2} = c_1e^x+c_2e^{x}=y\\ \frac{d^2y}{dx^2} - y = 0
Therefore, (B) is the correct answer

Question:12 Which of the following differential equations has y = x as one of its particular solution?

(A) \frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =x

(B) \frac{d^2y}{dx^2} + x\frac{dy}{dx} + xy =x

(C) \frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =0

(D) \frac{d^2y}{dx^2} + x\frac{dy}{dx} + xy =0

Answer:

Given equation is
y = x
Now, on differentiating it w.r.t x
we get,
\frac{dy}{dx} = 1
and again on differentiating it w.r.t x
we get,

\frac{d^2y}{dx^2} = 0
Now, on substituting the values of \frac{d^2y}{dx^2} , \frac{dy}{dx} \ and \ y in all the options we will find that only option c which is \frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =0 satisfies
Therefore, the correct answer is (C)


NCERT class 12 maths chapter 9 question answer - Exercise: 9.4

Question:1 Find the general solution: \frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}

Answer:

Given,

\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}

\\ \implies\frac{dy}{dx} = \frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}} = tan^2\frac{x}{2} \\ \implies dy = (sec^2\frac{x}{2} - 1)dx

\\ \implies \int dy = \int sec^2\frac{x}{2}dx - \int dx \\ \implies y = 2tan^{-1}\frac{x}{2} - x + C

Question:2 Find the general solution: \frac{dy}{dx} = \sqrt{4-y^2}\ (-2 < y < 2)

Answer:

Given, in the question

\frac{dy}{dx} = \sqrt{4-y^2}

\\ \implies \frac{dy}{\sqrt{4-y^2}} = dx \\ \implies \int \frac{dy}{\sqrt{4-y^2}} = \int dx

\\ (\int \frac{dy}{\sqrt{a^2-y^2}} = sin^{-1}\frac{y}{a})\\

The required general solution:

\\ \implies sin^{-1}\frac{y}{2} = x + C

Question:3 Find the general solution: \frac{dy}{dx} + y = 1 (y\neq 1)

Answer:

Given, in the question

\frac{dy}{dx} + y = 1

\\ \implies \frac{dy}{dx} = 1- y \\ \implies \int\frac{dy}{1-y} = \int dx

(\int\frac{dx}{x} = lnx)

\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\ \implies log k(1-y) = -x \\ \implies 1- y = \frac{1}{k}e^{-x} \\

The required general equation

\implies y = 1 -\frac{1}{k}e^{-x}

Question:4 Find the general solution: \sec^2 x \tan y dx + \sec^2 y \tan x dy = 0

Answer:

Given,

\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0

\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx \\ \implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx

Now, let tany = t and tanx = u

sec^2 y dy = dt\ and\ sec^2 x dx = du

\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} \\ \implies log t = -log u +logk \\ \implies t = \frac{1}{ku} \\ \implies tany = \frac{1}{ktanx}

Question:5 Find the general solution:

(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0

Answer:

Given, in the question

(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0

\\ \implies dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})}dx

Let,

\\ (e^x + e^{-x}) = t \\ \implies (e^x - e^{-x})dx = dt

\\ \implies \int dy = \int \frac{dt}{t} \\ \implies y = log t + C \\ \implies y = log(e^x + e^{-x}) + C

This is the general solution

Question:6 Find the general solution: \frac{dy}{dx} = (1+x^2)(1+y^2)

Answer:

Given, in the question

\frac{dy}{dx} = (1+x^2)(1+y^2)

\\ \implies \int \frac{dy}{(1+y^2)} = \int (1+x^2)dx

(\int \frac{dx}{(1+x^2)} =tan^{-1}x +c)

\\ \implies tan^{-1}y = x+\frac{x^3}{3} + C

Question:7 Find the general solution: y\log y dx - x dy = 0

Answer:

Given,

y\log y dx - x dy = 0

\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx

let logy = t

=> 1/ydy = dt

\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx \\ \implies \log t = \log x + \log k \\ \implies t = kx \\ \implies \log y = kx

This is the general solution

Question:8 Find the general solution: x^5\frac{dy}{dx} = - y^5

Answer:

Given, in the question

x^5\frac{dy}{dx} = - y^5

\\ \implies \int \frac{dy}{y^5} = - \int \frac{dx}{x^5} \\ \implies \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C \\ \implies \frac{1}{y^4} + \frac{1}{x^4} = C

This is the required general equation.

Question:9 Find the general solution: \frac{dy}{dx} = \sin^{-1}x

Answer:

Given, in the question

\frac{dy}{dx} = \sin^{-1}x

\implies \int dy = \int \sin^{-1}xdx

Now,

\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx

Here, u = \sin^{-1}x and v = 1

\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx

\\ Let\ 1- x^2 = t \\ \implies -2xdx = dt \implies xdx = -dt/2

\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) \\ \implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C \\ \implies y = x\sin^{-1}x + \sqrt{1-x^2} + C

Question:10 Find the general solution e^x\tan y dx + (1-e^x)\sec^2 y dy = 0

Answer:

Given,

e^x\tan y dx + (1-e^x)\sec^2 y dy = 0

\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy \\ \implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx

\\ let\ tany = t \ and \ 1-e^x = u \\ \implies \sec^2 ydy = dt\ and \ -e^xdx = du

\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} \\ \implies \log t = \log u + \log k \\ \implies t = ku \\ \implies \tan y= k (1-e^x)

Question:11 Find a particular solution satisfying the given condition:

(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0

Answer:

Given, in the question

(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x

\\ \implies \int dy = \int\frac{2x^2 + x}{(x^3 + x^2 + x + 1)}dx

(x^3 + x^2 + x + 1) = (x +1)(x^2+1)

Now,

1517900463071792

1517900463878580

1517900464704155

1517900465489721

Now comparing the coefficients

A + B = 2; B + C = 1; A + C = 0

Solving these:

1517900467056911

Putting the values of A,B,C:

1517900467842360

Therefore,

1517900468626376

1517900469359944

1517900470082760

1517900470844580

151790047163329

151790047240963

1517900473174127

1517900473936125

1517900474702185

1517900475486768

Now, y= 1 when x = 0

1517900477940526

c = 1

Putting the value of c, we get:

1517900478708257

Question:12 Find a particular solution satisfying the given condition:

x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2

Answer:

Given, in the question

x(x^2 -1)\frac{dy}{dx} =1

\\ \implies \int dy=\int \frac{dx}{x(x^2 -1)} \\ \implies \int dy=\int \frac{dx}{x(x -1)(x+1)}

Let,

1628485409903

1517900483388423

1517900484190962

Now comparing the values of A,B,C

A + B + C = 0; B-C = 0; A = -1

Solving these:

1517900484965476

Now putting the values of A,B,C

15179004857815100

1517900486545318

1517900487390967

1517900488152880

Given, y =0 when x =2

1517900489745379

1517900490509113

1517900492095338

Therefore,

\\ \implies y = \frac{1}{2}\log[\frac{4(x-1)(x+1)}{3x^2}]

\\ \implies y = \frac{1}{2}\log[\frac{4(x^2-1)}{3x^2}]

Question:13 Find a particular solution satisfying the given condition:

\cos\left(\frac{dy}{dx} \right ) = a\ (a\in R);\ y = 1\ \textup{when}\ x = 0

Answer:

Given,

\cos\left(\frac{dy}{dx} \right ) = a

\\ \implies \frac{dy}{dx} = \cos^{-1}a \\ \implies \int dy = \int\cos^{-1}a\ dx \\ \implies y = x\cos^{-1}a + c

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

\implies y = x\cos^{-1}a + 1

Question:14 Find a particular solution satisfying the given condition:

\frac{dy}{dx} = y\tan x; \ y =1\ \textup{when}\ x = 0

Answer:

Given,

\frac{dy}{dx} = y\tan x

\\ \implies \int \frac{dy}{y} = \int \tan x\ dx \\ \implies \log y = \log \sec x + \log k \\ \implies y = k\sec x

Now, y=1 when x =0

1 = ksec0

\implies k = 1

Putting the vlue of k:

y = sec x

Question:15 Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x\sin x .

Answer:

We first find the general solution of the given differential equation

Given,

y' = e^x\sin x

\\ \implies \int dy = \int e^x\sin xdx

\\ Let I = \int e^x\sin xdx \\ \implies I = \sin x.e^x - \int(\cos x. e^x)dx \\ \implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] \\ \implies 2I = e^x(\sin x - \cos x) \\ \implies I = \frac{1}{2}e^x(\sin x - \cos x)

\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + c

Now, Since the curve passes through (0,0)

y = 0 when x =0

\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\ \implies c = \frac{1}{2}

Putting the value of c, we get:

\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\ \implies 2y -1 = e^x(\sin x - \cos x)

Question:16 For the differential equation xy\frac{dy}{dx} = (x+2)(y+2) , find the solution curve passing through the point (1, –1).

Answer:

We first find the general solution of the given differential equation

Given,

xy\frac{dy}{dx} = (x+2)(y+2)

\\ \implies \int \frac{y}{y+2}dy = \int \frac{x+2}{x}dx \\ \implies \int \frac{(y+2)-2}{y+2}dy = \int (1 + \frac{2}{x})dx \\ \implies \int (1 - \frac{2}{y+2})dy = \int (1 + \frac{2}{x})dx \\ \implies y - 2\log (y+2) = x + 2\log x + C

Now, Since the curve passes through (1,-1)

y = -1 when x = 1

\\ \therefore -1 - 2\log (-1+2) = 1 + 2\log 1 + C \\ \implies -1 -0 = 1 + 0 +C \\ \implies C = -2

Putting the value of C:

\\ y - 2\log (y+2) = x + 2\log x + -2 \\ \implies y -x + 2 = 2\log x(y+2)

Question:17 Find the equation of a curve passing through the point ( 0 ,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

According to the question,

y\frac{dy}{dx} =x

\\ \implies \int ydy =\int xdx \\ \implies \frac{y^2}{2} = \frac{x^2}{2} + c

Now, Since the curve passes through (0,-2).

x =0 and y = -2

\\ \implies \frac{(-2)^2}{2} = \frac{0^2}{2} + c \\ \implies c = 2

Putting the value of c, we get

\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4

Question:18 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is \frac{y+3}{x+4}

According to the question,

\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\ \implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \\ \implies \log (y+3) = 2\log (x+4) + \log k \\ \implies (y+3) = k(x+4)^2

Now, Since the curve passes through (-2,1)

x = -2 , y =1

\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1

Putting the value of k, we get

\\ \implies y+3 = (x+4)^2

Question:19 The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Volume of a sphere, V = \frac{4}{3}\pi r ^3

Given, Rate of change is constant.

\\ \therefore \frac{dV}{dt} = c \\ \implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c \\ \implies \int d(\frac{4}{3}\pi r ^3) = c\int dt \\ \implies \frac{4}{3}\pi r ^3 = ct + k

Now, at t=0, r=3 and at t=3 , r =6

Putting these value:

\frac{4}{3}\pi (3) ^3 = c(0) + k \\ \implies k = 36\pi

Also,

\frac{4}{3}\pi (6) ^3 = c(3) + 36\pi \\ \implies 3c = 252\pi \\ \implies c = 84\pi

Putting the value of c and k:

\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\ \implies r ^3 = (21 t + 9)(3) = 62t + 27 \\ \implies r = \sqrt[3]{62t + 27}

Question:20 In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to question,

\frac{dp}{dt} = (\frac{r}{100})p

\\ \implies \int\frac{dp}{p} = \int (\frac{r}{100})dt \\ \implies \log p = \frac{r}{100}t + C

\\ \implies p = e^{\frac{rt}{100} + C}

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

\\ \implies 100 = e^{\frac{r(0)}{100} + C} = e^C

Also,

, \\ \implies 200 = e^{\frac{r(10)}{100} + C} = e^{\frac{r}{10}}.e^C = e^{\frac{r}{10}}.100 \\ \implies e^{\frac{r}{10}} = 2 \\ \implies \frac{r}{10} = \ln 2 = 0.6931 \\ \implies r = 6.93

So value of r = 6.93%

Question:21 In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e 0.5 = 1.648).

Answer:

Let p be the principal amount and t be the time.

According to question,

\frac{dp}{dt} = (\frac{5}{100})p

\\ \implies \int\frac{dp}{p} = \int (\frac{1}{20})dt \\ \implies \log p = \frac{1}{20}t + C

\\ \implies p = e^{\frac{t}{20} + C}

Now, at t =0 , p = 1000

Putting these values,

\\ \implies 1000 = e^{\frac{(0)}{20} + C} = e^C

Also, At t=10

, \\ \implies p = e^{\frac{(10)}{20} + C} = e^{\frac{1}{2}}.e^C = e^{\frac{1}{2}}.1000 \\ \implies p =(1.648)(1000) = 1648

After 10 years, the total amount would be Rs.1648

Question:22 In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let n be the number of bacteria at any time t.

According to question,

\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)

\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C

Now, at t=0, n = 100000

\\ \implies \log (100000) = k(0) + C \\ \implies C = 5

Again, at t=2, n= 110000

\\ \implies \log (110000) = k(2) + 5 \\ \implies \log 11 + 4 = 2k + 5 \\ \implies 2k = \log 11 -1 =\log \frac{11}{10} \\ \implies k = \frac{1}{2}\log \frac{11}{10}

Using these values, for n= 200000

\\ \implies \log (200000) = kt + C \\ \implies \log 2 +5 = kt + 5 \\ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 \\ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}


NCERT class 12 maths chapter 9 question answer - Exercise: 9.5

Question:1 Show that the given differential equation is homogeneous and solve each of them. (x^2 + xy)dy = (x^2 + y^2)dx

Answer:

The given diffrential eq can be written as
\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}
Let F(x,y)=\frac{x^{2}+y^{2}}{x^{2}+xy}
Now, F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}
=\frac{x^{2}+y^{2}}{x^{2}+xy} = \lambda ^{0}F(x,y) Hence, it is a homogeneous equation.

To solve it put y = vx
Diff
erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v +x\frac{dv}{dx} = \frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ v +x\frac{dv}{dx} = \frac{1+v^{2}}{1+v}

x\frac{dv}{dx} = \frac{(1+v^{2})-v(1+v)}{1+v} = \frac{1-v}{1+v}

( \frac{1+v}{1-v})dv = \frac{dx}{x}

( \frac{2}{1-v}-1)dv = \frac{dx}{x}
Integrating on both side, we get;
\\-2\log(1-v)-v=\log x -\log k\\ v= -2\log (1-v)-\log x+\log k\\ v= \log\frac{k}{x(1-v)^{2}}\\
Again substitute the value y = \frac{v}{x} ,we get;

\\\frac{y}{x}= \log\frac{kx}{(x-y)^{2}}\\ \frac{kx}{(x-y)^{2}}=e^{y/x}\\ (x-y)^{2}=kxe^{-y/x}
This is the required solution of given diff. equation

Question:2 Show that the given differential equation is homogeneousand solve each of them. y' = \frac{x+y}{x}

Answer:

the above differential eq can be written as,

\frac{dy}{dx} = F(x,y)=\frac{x+y}{x} ............................(i)

Now, F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x} = \lambda ^{0}F(x,y)
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx}= \frac{x+vx}{x} = 1+v
\\x\frac{dv}{dx}= 1\\ dv = \frac{dx}{x}
Integrating on both sides, we get; (and substitute the value of v =\frac{y}{x} )

\\v =\log x+C\\ \frac{y}{x}=\log x+C\\ y = x\log x +Cx
this is the required solution

Question:3 Show that the given differential equation is homogeneous and solve each of them.

(x-y)dy - (x+y)dx = 0

Answer:

The given differential eq can be written as;

\frac{dy}{dx}=\frac{x+y}{x-y} = F(x,y)(let\ say) ....................................(i)

F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}= \lambda ^{0}F(x,y)
Hence it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v+x\frac{dv}{dx}= \frac{1+v}{1-v}\\ x\frac{dv}{dx} = \frac{1+v}{1-v}-v =\frac{1+v^{2}}{1-v}

\frac{1-v}{1+v^{2}}dv = (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}
Integrating on both sides, we get;

\tan^{-1}v-1/2 \log(1+v^{2})=\log x+C
again substitute the value of v=y/x
\\\tan^{-1}(y/x)-1/2 \log(1+(y/x)^{2})=\log x+C\\ \tan^{-1}(y/x)-1/2 [\log(x^{2}+y^{2})-\log x^{2}]=\log x+C\\ tan^{-1}(y/x) = 1/2[\log (x^{2}+y^{2})]+C

This is the required solution.

Question:4 Show that the given differential equation is homogeneous and solve each of them.

(x^2 - y^2)dx + 2xydy = 0

Answer:

we can write it as;

\frac{dy}{dx}= -\frac{(x^{2}-y^{2})}{2xy} = F(x,y)\ (let\ say) ...................................(i)

F(\lambda x,\lambda y) = \frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)} = \lambda ^{0}.F(x,y)
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx} = \frac{ x^{2}-(vx)^{2}}{2x(vx)} =\frac{v^{2}-1}{2v}
\\x\frac{dv}{dx} =\frac{v^{2}+1}{2v}\\ \frac{2v}{1+v^{2}}dv=\frac{dx}{x}
integrating on both sides, we get

\log (1+v^{2})= -\log x +\log C = \log C/x
\\= 1+v^{2} = C/x\\ = x^2+y^{2}=Cx .............[ v =y/x ]
This is the required solution.

Question:5 Show that the given differential equation is homogeneous and solve it.

x^2\frac{dy}{dx} = x^2 - 2y^2 +xy

Answer:

\frac{dy}{dx}= \frac{x^{2}-2y^{2}+xy}{x^{2}} = F(x,y)\ (let\ say)

F(\lambda x,\lambda y)= \frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda .\lambda )xy}{(\lambda x)^{2}} = \lambda ^{0}.F(x,y) ............(i)
Hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v+x\frac{dv}{dx}= 1-2v^{2}+v\\ x\frac{dv}{dx} = 1-2v^{2}\\ \frac{dv}{1-2v^{2}}=\frac{dx}{x}

1/2[\frac{dv}{(1/\sqrt{2})^{2}-v^{2}}] = \frac{dx}{x}

On integrating both sides, we get;

\frac{1}{2\sqrt{2}}\log (\frac{1/\sqrt{2}+v}{1/\sqrt{2}-v}) = \log x +C
after substituting the value of v= y/x

\frac{1}{2\sqrt{2}}\log (\frac{x+\sqrt{2}y}{x-\sqrt{2}y}) = \log \left | x \right | +C

This is the required solution

Question:6 Show that the given differential equation is homogeneous and solve it.

xdy - ydx = \sqrt{x^2 + y^2}dx

Answer:

\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} = F(x,y) .................................(i)

F(\mu x,\mu y)=\frac{\mu y+\sqrt{(\mu x)^{2}+(\mu y)^{2}}}{\mu x} =\mu^{0}.F(x,y)
henxe it is a homogeneous equation

Now, to solve substitute y = vx

Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx}= v+\sqrt{1+v^{2}}=\sqrt{1+v^{2}}

=\frac{dv}{\sqrt{1+v^{2}}} =\frac{dx}{x}

On integrating both sides,

\Rightarrow \log \left | v+\sqrt{1+v^{2}} \right | = \log \left | x \right |+\log C
Substitute the value of v=y/x , we get

\\\Rightarrow \log \left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right | = \log \left | Cx \right |\\ y+\sqrt{x^{2}+y^{2}} = Cx^{2}

Required solution

Question:7 Solve.

\left\{x\cos\left(\frac{y}{x} \right ) + y\sin\left(\frac{y}{x} \right ) \right \}ydx = \left\{y\sin\left(\frac{y}{x} \right ) - x\cos\left(\frac{y}{x} \right ) \right \}xdy

Answer:

\frac{dy}{dx} =\frac{x \cos(y/x)+y\sin(y/x)}{y\sin(y/x)-x\cos(y/x)}.\frac{y}{x} = F(x,y) ......................(i)
By looking at the equation we can directly say that it is a homogenous equation.

Now, to solve substitute y = vx

Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx} =\frac{v \cos v+v^{2}\sin v}{v\sin v-\cos v}\\ =x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v-\cos v}\\ =(\tan v-1/v)dv = \frac{2dx}{x}

integrating on both sides, we get

\\=\log(\frac{\sec v}{v})= \log (Cx^{2})\\=\sec v/v =Cx^{2}
substitute the value of v= y/x , we get

\\\sec(y/x) =Cxy \\ xy \cos (y/x) = k

Required solution

Question:8 Solve.

x\frac{dy}{dx} - y + x\sin\left(\frac{y}{x}\right ) = 0

Answer:

\frac{dy}{dx}=\frac{y-x \sin(y/x)}{x} = F(x,y) ...............................(i)

F(\mu x, \mu y)=\frac{\mu y-\mu x \sin(\mu y/\mu x)}{\mu x} = \mu^{0}.F(x,y)
it is a homogeneous equation

Now, to solve substitute y = vx

Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx}= v- \sin v = -\sin v
\Rightarrow -\frac{dv}{\sin v} = -(cosec\ v)dv=\frac{dx}{x}

On integrating both sides we get;

\\\Rightarrow \log \left | cosec\ v-\cot v \right |=-\log x+ \log C\\ \Rightarrow cosec (y/x) - \cot (y/x) = C/x

= x[1-\cos (y/x)] = C \sin (y/x) Required solution

Question:9 Solve.

ydx + x\log\left(\frac{y}{x} \right ) -2xdy = 0

Answer:

\frac{dy}{dx}= \frac{y}{2x-x \log(y/x)} = F(x,y) ..................(i)

\frac{\mu y}{2\mu x-\mu x \log(\mu y/\mu x)} = F(\mu x,\mu y) = \mu^{0}.F(x,y)

hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}= \frac{v}{2-\log v}\\ =x\frac{dv}{dx} = \frac{v\log v-v}{2-\log v}\\ =[\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}
integrating on both sides, we get; ( substituting v =y/x)

\\\Rightarrow \log[\log(y/x)-1]-\log(y/x)=\log(Cx)\\\Rightarrow \frac{x}{y}[\log(y/x)-1]=Cx\\ \Rightarrow \log (y/x)-1=Cy

This is the required solution of the given differential eq

Question:10 Solve.

\left(1 + e^{\frac{x}{y}} \right )dx + e^\frac{x}{y}\left(1-\frac{x}{y}\right )dy = 0

Answer:

\frac{dx}{dy}=\frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} = F(x,y) .......................................(i)

= F(\mu x,\mu y)=\frac{-e^{\mu x/\mu y}(1-\mu x/\mu y)}{1+e^{\mu x/\mu y}} =\mu^{0}.F(x,y)
Hence it is a homogeneous equation.

Now, to solve substitute x = yv

Diff erentiating on both sides wrt x
\frac{dx}{dy}= v +y\frac{dv}{dy}

Substitute this value in equation (i)

\\=v+y\frac{dv}{dy} = \frac{-e^{v}(1-v)}{1+e^{v}} \\ =y\frac{dv}{dy} = -\frac{v+e^{v}}{1+e^{v}}\\ =\frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}

Integrating on both sides, we get;

\dpi{100} \log(v+e^{v})=-\log y+ \log c =\log (c/y)\\ =[\frac{x}{y}+e^{x/y}]= \frac{c}{y}\\\Rightarrow x+ye^{x/y}=c
This is the required solution of the diff equation.

Question:11 Solve for particular solution.

(x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1

Answer:

\frac{dy}{dx}=\frac{-(x-y)}{x+y} =F(x,y) ..........................(i)

We can clearly say that it is a homogeneous equation.

Now, to solve substitute y = vx

Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v+x\frac{dv}{dx}=\frac{v-1}{v+1}\\ \Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}

\frac{1+v}{1+v^{2}}dv = [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv=-\frac{dx}{x}

On integrating both sides

\\=\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k\\ =\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k\\ =\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k\\ =\log(x^{2}+y^{2})+2\tan^{-1}(y/x) = 2k ......................(ii)

Now, y=1 and x= 1


\\=\log 2 +2\tan^{-1}1=2k\\ =\pi/2+\log 2 = 2k\\

After substituting the value of 2k in eq. (ii)

\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2

This is the required solution.

Question:12 Solve for particular solution.

x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1

Answer:

\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y) ...............................(i)

F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i), we get

\\=v+\frac{xdv}{dx}= -v- v^{2}\\ =\frac{xdv}{dx}=-v(v+2)\\ =\frac{dv}{v+2}=-\frac{dx}{x}\\ =1/2[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}

Integrating on both sides, we get;

\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\ =\frac{v}{v+2}=(C/x)^{2}

replace the value of v=y/x

\frac{x^{2}y}{y+2x}=C^{2} .............................(ii)

Now y =1 and x = 1

C = 1/\sqrt{3}
therefore,

\frac{x^{2}y}{y+2x}=1/3

Required solution

Question:13 Solve for particular solution.

\left [x\sin^2\left(\frac{y}{x} \right ) - y \right ]dx + xdy = 0;\ y =\frac{\pi}{4}\ when \ x = 1

Answer:

\frac{dy}{dx}=\frac{-[x\sin^{2}(y/x)-y]}{x} = F(x,y) ..................(i)

F(\mu x,\mu y)=\frac{-[\mu x\sin^{2}(\mu y/\mu x)-\mu y]}{\mu x}=\mu ^{0}.F(x,y)

Hence it is a homogeneous eq

Now, to solve substitute y = vx

Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

on integrating both sides, we get;

\\-\cot v =\log\left | x \right | -C\\ =\cot v = \log\left | x \right |+\log C

On substituting v =y/x

=\cot (y/x) = \log\left | Cx \right | ............................(ii)

Now, y = \pi/4\ @ x=1

\\\cot (\pi/4) = \log C \\ =C=e^{1}

put this value of C in eq (ii)

\cot (y/x)=\log\left | ex \right |

Required solution.

Question:14 Solve for particular solution.

\frac{dy}{dx} - \frac{y}{x} + \textup{cosec}\left (\frac{y}{x} \right ) = 0;\ y = 0 \ \textup{when}\ x = 1

Answer:

\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y) ....................................(i)

the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}=v- cosec\ v\\ =x\frac{dv}{dx} = -cosec\ v\\ =-\frac{dv}{cosec\ v}= \frac{dx}{x}\\ =-\sin v dv = \frac{dx}{x}

on integrating both sides, we get;

\\=cos\ v = \log x +\log C =\log Cx\\ =\cos(y/x)= \log Cx .................................(ii)

now y = 0 and x =1 , we get

C =e^{1}

put the value of C in eq 2

\cos(y/x)=\log \left | ex \right |

Question:15 Solve for particular solution.

2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1

Answer:

The above eq can be written as;

\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)
By looking, we can say that it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}= \frac{2v+v^{2}}{2}\\ =x\frac{dv}{dx} = v^{2}/2\\ = \frac{2dv}{v^{2}}=\frac{dx}{x}

integrating on both sides, we get;

\\=-2/v=\log \left | x \right |+C\\ =-\frac{2x}{y}=\log \left | x \right |+C .............................(ii)

Now, y = 2 and x =1, we get

C =-1
put this value in equation(ii)

\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}

Question:16 A homogeneous differential equation of the from \frac{dx}{dy}= h\left(\frac{x}{y} \right ) can be solved by making the substitution.

(A) y = vx

(B) v = yx

(C) x = vy

(D) x =v

Answer:

\frac{dx}{dy}= h\left(\frac{x}{y} \right )
for solving this type of equation put x/y = v
x = vy

option C is correct

Question:17 Which of the following is a homogeneous differential equation?

(A) (4x + 6x +5)dy - (3y + 2x +4)dx = 0

(B) (xy)dx - (x^3 + y^3)dy = 0

(C) (x^3 +2y^2)dx + 2xydy =0

(D) y^2dx + (x^2 -xy -y^2)dy = 0

Answer:

Option D is the right answer.

y^2dx + (x^2 -xy -y^2)dy = 0
\frac{dy}{dx}=\frac{y^{2}}{x^{2}-xy-y^{2}} = F(x,y)
we can take out lambda as a common factor and it can be cancelled out


NCERT class 12 maths chapter 9 question answer - Exercise: 9.6

Question:1 Find the general solution:

\frac{dy}{dx} + 2y = \sin x

Answer:

Given equation is
\frac{dy}{dx} + 2y = \sin x
This is \frac{dy}{dx} + py = Q type where p = 2 and Q = sin x
Now,
I.F. = e^{\int pdx}= e^{\int 2dx}= e^{2x}
Now, the solution of given differential equation is given by relation
Y(I.F.) =\int (Q\times I.F.)dx +C
Y(e^{\int 2x }) =\int (\sin x\times e^{\int 2x })dx +C
Let I =\int (\sin x\times e^{\int 2x })
I = \sin x \int e^{2x}dx- \int \left ( \frac{d(\sin x)}{dx}.\int e^{2x}dx \right )dx\\ \\ I = \sin x.\frac{e^{2x}}{2}- \int \left ( \cos x.\frac{e^{2x}}{2} \right )\\ \\ I = \sin x. \frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x\int e^{2x}dx- \left ( \frac{d(\cos x)}{dx}.\int e^{2x}dx \right ) \right )dx\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+ \int \left ( \sin x.\frac{e^{2x}}{2} \right ) \right )\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+\frac{I}{2} \right ) \ \ \ \ \ \ \ \ \ \ \ (\because I = \int \sin xe^{2x})\\ \\ \frac{5I}{4}= \frac{e^{2x}}{4}\left ( 2\sin x-\cos x \right )\\ \\ I = \frac{e^{2x}}{5}\left ( 2\sin x-\cos x \right )
Put the value of I in our equation
Now, our equation become
Y.e^{x^2 }= \frac{e^{2x}}{5}\left (2 \sin x-\cos x \right )+C
Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}
Therefore, the general solution is Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}

Question:2 Solve for general solution:

\frac{dy}{dx} + 3y = e^{-2x}

Answer:

Given equation is
\frac{dy}{dx} + 3y = e^{-2x}
This is \frac{dy}{dx} + py = Q type where p = 3 and Q = e^{-2x}
Now,
I.F. = e^{\int pdx}= e^{\int 3dx}= e^{3x}
Now, the solution of given differential equation is given by the relation
Y(I.F.) =\int (Q\times I.F.)dx +C
Y(e^{ 3x }) =\int (e^{-2x}\times e^{ 3x })dx +C
Y(e^{ 3x }) =\int (e^{x})dx +C\\ Y(e^{3x})= e^x+C\\ Y = e^{-2x}+Ce^{-3x}
Therefore, the general solution is Y = e^{-2x}+Ce^{-3x}

Question:3 Find the general solution

\frac{dy}{dx} + \frac{y}{x} = x^2

Answer:

Given equation is
\frac{dy}{dx} + \frac{y}{x} = x^2
This is \frac{dy}{dx} + py = Q type where p = \frac{1}{x} and Q = x^2
Now,
I.F. = e^{\int pdx}= e^{\int \frac{1}{x}dx}= e^{\log x}= x
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(x) =\int (x^2\times x)dx +C
y(x) =\int (x^3)dx +C\\ y.x= \frac{x^4}{4}+C\\
Therefore, the general solution is yx =\frac{x^4}{4}+C

Question:4 Solve for General Solution.

\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )

Answer:

Given equation is
\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )
This is \frac{dy}{dx} + py = Q type where p = \sec x and Q = \tan x
Now,
I.F. = e^{\int pdx}= e^{\int \sec xdx}= e^{\log |\sec x+ \tan x|}= \sec x+\tan x (\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sec x+\tan x) =\int ((\sec x+\tan x)\times \tan x)dx +C
y(\sec x+ \tan x) =\int (\sec x\tan x+\tan^2 x)dx +C\\y(\sec x+ \tan x) =\sec x+\int (\sec^2x-1)dx +C\\ y(\sec x+ \tan x) = \sec x +\tan x - x+C
Therefore, the general solution is y(\sec x+ \tan x) = \sec x +\tan x - x+C

Question:5 Find the general solution.

\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )

Answer:

Given equation is
\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )
we can rewrite it as
\frac{dy}{dx}+\sec^2x y= \sec^2x\tan x
This is \frac{dy}{dx} + py = Q where p = \sec ^2x and Q =\sec^2x \tan x
Now,
I.F. = e^{\int pdx}= e^{\int \sec^2 xdx}= e^{\tan x}
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(e^{\tan x}) =\int ((\sec^2 x\tan x)\times e^{\tan x})dx +C
ye^{\tan x} =\int \sec^2 x\tan xe^{\tan x}dx+C\\
take
e^{\tan x } = t\\ \Rightarrow \sec^2x.e^{\tan x}dx = dt
\int t.\log t dt = \log t.\int tdt-\int \left ( \frac{d(\log t)}{dt}.\int tdt \right )dt \\ \\ \int t.\log t dt = \log t . \frac{t^2}{2}- \int (\frac{1}{t}.\frac{t^2}{2})dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \int \frac{t}{2}dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \frac{t^2}{4}\\ \\ \int t.\log t dt = \frac{t^2}{4}(2\log t -1)
Now put again t = e^{\tan x}
\int \sec^2x\tan xe^{\tan x}dx = \frac{e^{2\tan x}}{4}(2\tan x-1)
Put this value in our equation

ye^{\tan x} =\frac{e^{2\tan x}}{4}(2\tan x-1)+C\\ \\
Therefore, the general solution is y =\frac{e^{\tan x}}{4}(2\tan x-1)+Ce^{-\tan x }\\

Question:6 Solve for General Solution.

x\frac{dy}{dx} + 2y = x^2\log x

Answer:

Given equation is
x\frac{dy}{dx} + 2y = x^2\log x
Wr can rewrite it as
\frac{dy}{dx} +2.\frac{y}{x}= x\log x
This is \frac{dy}{dx} + py = Q type where p = \frac{2}{x} and Q = x\log x
Now,
I.F. = e^{\int pdx}= e^{\int \frac{2}{x}dx}= e^{2\log x}=e^{\log x^2} = x^2 (\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(x^2) =\int (x\log x\times x^2)dx +C
x^2y = \int x^3\log x+ C
Let
I = \int x^3\log x\\ \\ I = \log x\int x^3dx-\int \left ( \frac{d(\log x)}{dx}.\int x^3dx \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{1}{x}.\frac{x^4}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{x^3}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}-\frac{x^4}{16}
Put this value in our equation
x^2y =\log x.\frac{x^4}{4}-\frac{x^4}{16}+ C\\ \\ y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}
Therefore, the general solution is y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}

Question:7 Solve for general solutions.

x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x

Answer:

Given equation is
x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x
we can rewrite it as
\frac{dy}{dx}+\frac{y}{x\log x}= \frac{2}{x^2}
This is \frac{dy}{dx} + py = Q type where p = \frac{1}{x\log x} and Q =\frac{2}{x^2}
Now,
I.F. = e^{\int pdx}= e^{\int \frac{1}{x\log x} dx}= e^{\log(\log x)} = \log x
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\log x) =\int ((\frac{2}{x^2})\times \log x)dx +C

take
I=\int ((\frac{2}{x^2})\times \log x)dx
I = \log x.\int \frac{2}{x^2}dx-\int \left ( \frac{d(\log x)}{dt}.\int \frac{x^2}{2}dx \right )dx \\ \\ I= -\log x . \frac{2}{x}+ \int (\frac{1}{x}.\frac{2}{x})dx\\ \\ I = -\log x.\frac{2}{x}+ \int \frac{2}{x^2}dx\\ \\I = -\log x.\frac{2}{x}- \frac{2}{x}\\ \\
Put this value in our equation

y\log x =-\frac{2}{x}(\log x+1)+C\\ \\
Therefore, the general solution is y\log x =-\frac{2}{x}(\log x+1)+C\\ \\

Question:8 Find the general solution.

(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)

Answer:

Given equation is
(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)
we can rewrite it as
\frac{dy}{dx}+\frac{2xy}{(1+x^2)}= \frac{\cot x}{1+x^2}
This is \frac{dy}{dx} + py = Q type where p = \frac{2x}{1+ x^2} and Q =\frac{\cot x}{1+x^2}
Now,
I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+ x^2} dx}= e^{\log(1+ x^2)} = 1+x^2
Now, the solution of the given differential equation is given by the relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(1+x^2) =\int ((\frac{\cot x}{1+x^2})\times (1+ x^2))dx +C
y(1+x^2) =\int \cot x dx+C\\ \\ y(1+x^2)= \log |\sin x|+ C\\ \\ y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}
Therefore, the general solution is y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}

Question:9 Solve for general solution.

x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)

Answer:

Given equation is
x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)
we can rewrite it as
\frac{dy}{dx}+y.\left ( \frac{1}{x}+\cot x \right )= 1
This is \frac{dy}{dx} + py = Q type where p =\left ( \frac{1}{x}+\cot x \right ) and Q =1
Now,
I.F. = e^{\int pdx}= e^{\int \left ( \frac{1}{x}+\cot x \right ) dx}= e^{\log x +\log |\sin x|} = x.\sin x
Now, the solution of the given differential equation is given by the relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(x.\sin x) =\int 1\times x\sin xdx +C
y(x.\sin x) =\int x\sin xdx +C
Lets take
I=\int x\sin xdx \\ \\ I = x .\int \sin xdx-\int \left ( \frac{d(x)}{dx}.\int \sin xdx \right )dx\\ \\ I =- x.\cos x+ \int (\cos x)dx\\ \\ I = -x\cos x+\sin x
Put this value in our equation
y(x.\sin x)= -x\cos x+\sin x + C\\ y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}
Therefore, the general solution is y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}

Question:10 Find the general solution.

(x+y)\frac{dy}{dx} = 1

Answer:

Given equation is
(x+y)\frac{dy}{dx} = 1
we can rewrite it as
\frac{dy}{dx} = \frac{1}{x+y}\\ \\ x+ y =\frac{dx}{dy}\\ \\ \frac{dx}{dy}-x=y
This is \frac{dx}{dy} + px = Q type where p =-1 and Q =y
Now,
I.F. = e^{\int pdy}= e^{\int -1 dy}= e^{-y}
Now, the solution of given differential equation is given by relation
x(I.F.) =\int (Q\times I.F.)dy +C
x(e^{-y}) =\int y\times e^{-y}dy +C
xe^{-y}= \int y.e^{-y}dy + C
Lets take
I=\int ye^{-y}dy \\ \\ I = y .\int e^{-y}dy-\int \left ( \frac{d(y)}{dy}.\int e^{-y}dy \right )dy\\ \\ I =- y.e^{-y}+ \int e^{-y}dy\\ \\ I = - ye^{-y}-e^{-y}
Put this value in our equation
x.e^{-y} = -e^{-y}(y+1)+C\\ x = -(y+1)+Ce^{y}\\ x+y+1=Ce^y
Therefore, the general solution is x+y+1=Ce^y

Question:11 Solve for general solution.

y dx + (x - y^2)dy = 0

Answer:

Given equation is
y dx + (x - y^2)dy = 0
we can rewrite it as
\frac{dx}{dy}+\frac{x}{y}=y
This is \frac{dx}{dy} + px = Q type where p =\frac{1}{y} and Q =y
Now,
I.F. = e^{\int pdy}= e^{\int \frac{1}{y} dy}= e^{\log y } = y
Now, the solution of given differential equation is given by relation
x(I.F.) =\int (Q\times I.F.)dy +C
x(y) =\int y\times ydy +C
xy= \int y^2dy + C
xy = \frac{y^3}{3}+C
x = \frac{y^2}{3}+\frac{C}{y}
Therefore, the general solution is x = \frac{y^2}{3}+\frac{C}{y}

Question:12 Find the general solution.

(x+3y^2)\frac{dy}{dx} = y\ (y > 0)

Answer:

Given equation is
(x+3y^2)\frac{dy}{dx} = y\ (y > 0)
we can rewrite it as
\frac{dx}{dy}-\frac{x}{y}= 3y
This is \frac{dx}{dy} + px = Q type where p =\frac{-1}{y} and Q =3y
Now,
I.F. = e^{\int pdy}= e^{\int \frac{-1}{y} dy}= e^{-\log y } =y^{-1}= \frac{1}{y}
Now, the solution of given differential equation is given by relation
x(I.F.) =\int (Q\times I.F.)dy +C
x(\frac{1}{y}) =\int 3y\times \frac{1}{y}dy +C
\frac{x}{y}= \int 3dy + C
\frac{x}{y}= 3y+ C
x = 3y^2+Cy
Therefore, the general solution is x = 3y^2+Cy

Question:13 Solve for particular solution.

\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}

Answer:

Given equation is
\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}
This is \frac{dy}{dx} + py = Q type where p = 2\tan x and Q = \sin x
Now,
I.F. = e^{\int pdx}= e^{\int 2\tan xdx}= e^{2\log |\sec x|}= \sec^2 x
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sec^2 x) =\int ((\sin x)\times \sec^2 x)dx +C
y(\sec^2 x) =\int (\sin \times \frac{1}{\cos x}\times \sec x)dx +C\\ \\ y(\sec^2 x) = \int \tan x\sec xdx+ C\\ \\ y.\sec^2 x= \sec x+C
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x= \frac{\pi}{3}
at x= \frac{\pi}{3}
0.\sec \frac{\pi}{3} = \sec \frac{\pi}{3}+C\\ \\ C = - 2
Now,

y.\sec^2 x= \sec x - 2\\ \frac{y}{\cos ^2x}= \frac{1}{\cos x}- 2\\ y = \cos x- 2\cos ^2 x
Therefore, the particular solution is y = \cos x- 2\cos ^2 x

Question:14 Solve for particular solution.

(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1

Answer:

Given equation is
(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1
we can rewrite it as
\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}
This is \frac{dy}{dx} + py = Q type where p =\frac{2x}{1+x^2} and Q = \frac{1}{(1+x^2)^2}
Now,
I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+x^2}dx}= e^{\log |1+x^2|}= 1+x^2
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(1+ x^2) =\int (\frac{1}{(1+x^2)^2}\times (1+x^2))dx +C
y(1+x^2) =\int \frac{1}{(1+x^2)}dx +C\\ \\ y(1+x^2) = \tan^{-1}x+ C\\ \\
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
0.(1+1^2) = \tan^{-1}1+ C\\ \\ C =- \frac{\pi}{4}
Now,
y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}
Therefore, the particular solution is y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}

Question:15 Find the particular solution.

\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}

Answer:

Given equation is
\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}
This is \frac{dy}{dx} + py = Q type where p =-3\cot x and Q =\sin 2x
Now,
I.F. = e^{\int pdx}= e^{-3\cot xdx}= e^{-3\log|\sin x|}= \sin ^{-3}x= \frac{1}{\sin^3x}
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\frac{1}{\sin^3 x}) =\int (\sin 2x\times\frac{1}{\sin^3 x})dx +C
\frac{y}{\sin^3 x} =\int (2\sin x\cos x\times\frac{1}{\sin^3 x})dx +C
\frac{y}{\sin^3 x} =\int (2\times \frac{\cos x}{\sin x}\times\frac{1}{\sin x})dx +C
\frac{y}{\sin^3 x} =\int (2\times\cot x\times cosec x)dx +C
\frac{y}{\sin^3 x} =-2cosec x +C
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when x= \frac{\pi}{2}
at x= \frac{\pi}{2}
\frac{2}{\sin^3\frac{\pi}{2}} = -2cosec \frac{\pi}{2}+C\\ \\ 2 = -2 +C\\ C = 4
Now,
y= 4\sin^3 x-2\sin^2x\\
Therefore, the particular solution is y= 4\sin^3 x-2\sin^2x\\

Question:16 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

Let f(x , y) is the curve passing through origin
Then, the slope of tangent to the curve at point (x , y) is given by \frac{dy}{dx}
Now, it is given that
\frac{dy}{dx} = y + x\\ \\ \frac{dy}{dx}-y=x
It is \frac{dy}{dx}+py=Q type of equation where p = -1 \ and \ Q = x
Now,
I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}
Now,
y(I.F.)= \int (Q \times I.F. )dx+ C
y(e^{-x})= \int (x \times e^{-x} )dx+ C
Now, Let
I= \int (x \times e^{-x} )dx \\ \\ I = x.\int e^{-x}dx-\int \left ( \frac{d(x)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -xe^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}\\ \\ I = -e^{-x}(x+1)
Put this value in our equation
ye^{-x}= -e^{-x}(x+1)+C
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
0.e^{-0}= -e^{-0}(0+1)+C\\ \\ C = 1
Our final equation becomes
ye^{-x}= -e^{-x}(x+1)+1\\ y+x+1=e^x
Therefore, the required equation of the curve is y+x+1=e^x

Question:17 Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let f(x , y) is the curve passing through point (0 , 2)
Then, the slope of tangent to the curve at point (x , y) is given by \frac{dy}{dx}
Now, it is given that
\frac{dy}{dx} +5= y + x \\ \\ \frac{dy}{dx}-y=x-5
It is \frac{dy}{dx}+py=Q type of equation where p = -1 \ and \ Q = x- 5
Now,
I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}
Now,
y(I.F.)= \int (Q \times I.F. )dx+ C
y(e^{-x})= \int ((x-5) \times e^{-x} )dx+ C
Now, Let
I= \int ((x-5) \times e^{-x} )dx \\ \\ I = (x-5).\int e^{-x}dx-\int \left ( \frac{d(x-5)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -(x-5)e^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}+5e^{-x}\\ \\ I = -e^{-x}(x-4)
Put this value in our equation
ye^{-x}= -e^{-x}(x-4)+C
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
2.e^{-0}= -e^{-0}(0-4)+C\\ \\ C = -2
Our final equation becomes
ye^{-x}= -e^{-x}(x-4)-2\\ y=4-x-2e^x
Therefore, the required equation of curve is y=4-x-2e^x

Question:18 The Integrating Factor of the differential equation x\frac{dy}{dx} - y = 2x^2 is

(A) e^{-x}

(B) e^{-y}

(C) \frac{1}{x}

(D) x

Answer:

Given equation is
x\frac{dy}{dx} - y = 2x^2
we can rewrite it as
\frac{dy}{dx}-\frac{y}{x}= 2x
Now,
It is \frac{dy}{dx}+py=Q type of equation where p = \frac{-1}{x} \ and \ Q = 2x
Now,
I.F. = e^{\int pdx} = e^{\int \frac{-1}{x}dx}= e^{\int -\log x }= x^{-1}= \frac{1}{x}
Therefore, the correct answer is (C)

Question:19 The Integrating Factor of the differential equation (1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1) is

(A) \frac{1}{{y^2 -1}}

(B) \frac{1}{\sqrt{y^2 -1}}

(C) \frac{1}{{1 - y^2 }}

(D) \frac{1}{\sqrt{1 - y^2 }}

Answer:

Given equation is
(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)
we can rewrite it as
\frac{dx}{dy}+\frac{yx}{1-y^2}= \frac{ay}{1-y^2}
It is \frac{dx}{dy}+px= Q type of equation where p = \frac{y}{1-y^2}\ and \ Q = \frac{ay}{1-y^2}
Now,
I.F. = e^{\int pdy}= e^{\int \frac{y}{1-y^2}dy}= e^{\frac{\log |1 - y^2|}{-2}}= (1-y^2)^{\frac{-1}{2}}= \frac{1}{\sqrt{1-y^2}}
Therefore, the correct answer is (D)


Class 12 Maths Chapter 9 NCERT solutions - Miscellaneous Exercise

Question:1 Indicate Order and Degree.

(i) \frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x

Answer:

Given function is
\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x
We can rewrite it as
y''+5x(y')^2-6y = \log x
Now, it is clear from the above that, the highest order derivative present in differential equation is y''

Therefore, the order of the given differential equation \frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x is 2
Now, the given differential equation is a polynomial equation in its derivative y '' and y 'and power raised to y '' is 1
Therefore, it's degree is 1

Question:1 Indicate Order and Degree.

(ii) \left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x

Answer:

Given function is
\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x
We can rewrite it as
(y')^3-4(y')^2+7y=\sin x
Now, it is clear from the above that, the highest order derivative present in differential equation is y'

Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives y 'and power raised to y ' is 3
Therefore, it's degree is 3

Question:1 Indicate Order and Degree.

(iii) \frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0

Answer:

Given function is
\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0
We can rewrite it as
y''''-\sin y''' = 0
Now, it is clear from the above that, the highest order derivative present in differential equation is y''''

Therefore, order of given differential equation is 4
Now, the given differential equation is not a polynomial equation in it's dervatives
Therefore, it's degree is not defined

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) xy = ae^x + be^{-x} + x^2\qquad :\ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy +x^2 -2 =0

Answer:

Given,

xy = ae^x + be^{-x} + x^2

Now, differentiating both sides w.r.t. x,

x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x

Again, differentiating both sides w.r.t. x,

\\ (x\frac{d^2y}{dx^2} + \frac{dy}{dx}) + \frac{dy}{dx} = ae^x + be^{-x} + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ae^x + be^{-x} + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy -x^2 + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 + 2

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(ii) y = e^x(a\cos x + b \sin x )\qquad : \ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

Answer:

Given,

y = e^x(a\cos x + b \sin x )

Now, differentiating both sides w.r.t. x,

\frac{dy}{dx} = e^x(-a\sin x + b \cos x ) + e^x(a\cos x + b \sin x ) =e^x(-a\sin x + b \cos x ) +y

Again, differentiating both sides w.r.t. x,

\\ \frac{d^2y}{dx^2} = e^x(-a\cos x - b \sin x ) + e^x(-a\sin x + b \cos x ) + \frac{dy}{dx} \\ = -y + (\frac{dy}{dx} -y) + \frac{dy}{dx} \\ \implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii) y= x\sin 3x \qquad : \ \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0

Answer:

Given,

y= x\sin 3x

Now, differentiating both sides w.r.t. x,

y= x\sin 3x \frac{dy}{dx} = x(3\cos 3x) + \sin 3x

Again, differentiating both sides w.r.t. x,

\\ \frac{d^2y}{dx^2} = 3x(-3\sin 3x) + 3\cos 3x + 3\cos 3x \\ = -9y + 6\cos 3x \\ \implies \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv) x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0

Answer:

Given,

x^2 = 2y^2\log y

Now, differentiating both sides w.r.t. x,

\\ 2x = (2y^2.\frac{1}{y} + 2(2y)\log y)\frac{dy}{dx} = 2(y + 2y\log y)\frac{dy}{dx} \\ \implies \frac{dy}{dx} = \frac{x}{y(1+ 2\log y)}

Putting \frac{dy}{dx}\ and \ x^2 values in LHS

\\ (2y^2\log y + y^2)\frac{dy}{dx} - xy = y^2(2\log y + 1)\frac{x}{y(1+ 2\log y)} -xy \\ = xy - xy = 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question:3 Form the differential equation representing the family of curves given by (x-a)^2 + 2y^2 = a^2 , where a is an arbitrary constant.

Answer:

Given equation is
(x-a)^2 + 2y^2 = a^2
we can rewrite it as
2y^2 = a^2-(x-a)^2 -(i)
Differentiate both the sides w.r.t x
\frac{d\left ( 2y^2 \right )}{dx}=\frac{d(a^2-(x-a)^2)}{dx}
4yy^{'}=4y\frac{dy}{dx}=-2(x-a)\\ \\
(x-a)= -2yy'\Rightarrow a = x+2yy' -(ii)
Put value from equation (ii) in (i)
(-2yy')^2+2y^2= (x+2yy')^2\\ 4y^2(y')^2+2y^2= x^2+4y^2(y')^2+4xyy'\\ y' = \frac{2y^2-x^2}{4xy}
Therefore, the required differential equation is y' = \frac{2y^2-x^2}{4xy}

Question:4 Prove that x^2 - y^2 = c (x^2 + y^2 )^2 is the general solution of differential equation (x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy , where c is a parameter.

Answer:

Given,

(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy

\implies \frac{ dy}{dx} = \frac{(x^3 - 3x y^2 )}{(y^3 - 3x^2 y)}

Now, let y = vx

\implies \frac{ dy}{dx} = \frac{ d(vx)}{dx} = v + x\frac{dv}{dx}

Substituting the values of y and y' in the equation,

v + x\frac{dv}{dx} = \frac{(x^3 - 3x (vx)^2 )}{((vx)^3 - 3x^2 (vx))}

\\\implies v + x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v}\\ \implies x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v} -v = \frac{1 - v^4 }{v^3 - 3v}

\implies (\frac{v^3 - 3v }{1 - v^4})dv = \frac{dx}{x}

Integrating both sides we get,

1517901119530459

Now, 1517901120316962

1517901121031268

Let 151790112181538

151790112256298

\implies 1517901123344265

\implies 1517901124127226

Now, 1517901124894198

1517901125674574

Let v 2 = p

151790112647622

1517901127259961

1517901128042674

1517901128803851

Now, substituting the values of I 1 and I 2 in the above equation, we get,

1517901129585614

Thus,

1517901130366853

1517901131127676

1517901131975586

1517901132764415

\\ (x^2 - y^2)^2 = C'^4(x^2 + y^2 )^4 \\ \implies (x^2 - y^2) = C'^2(x^2 + y^2 )^2 \\ \implies (x^2 - y^2) = K(x^2 + y^2 )^2, where\ K = C'^2

Question:5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

1646973511355 Now, equation of the circle with center at (x,y) and radius r is
(x-a)^2+(y-b)^2 = r^2
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
(x-a)^2+(y-a)^2 = a^2 -(i)
Differentiate it w.r.t x
we will get
2(x-a)+2(y-a)y'= 0\\ \\ 2x-2a+2yy'-2ay' = 0\\ a=\frac{x+yy'}{1+y'} -(ii)
Put value from equation (ii) in equation (i)
(x-\frac{x+yy'}{1+y'})^2+(y-\frac{x+yy'}{1+y'})^2=\left ( \frac{x+yy'}{1+y'} \right )^2\\ \\ (x+xy'-x-yy')^2+(y+yy'-x-yy')^2=(x+yy')^2\\ \\ (y')^2(x-y)^2+(x-y)^2=(x+yy')^2\\ \\ (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2

Question:6 Find the general solution of the differential equation \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0

Answer:

Given equation is
\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0
we can rewrite it as
\frac{dy}{dx } =- \sqrt{\frac{1-y^2}{1-x^2}}\\ \\ \frac{dy}{\sqrt{1-y^2}}= \frac{-dx}{\sqrt{1-x^2}}
Now, integrate on both the sides
\sin^{-1}y + C =- \sin ^{-1}x + C'\\ \\ \sin^{-1}y+\sin^{-1}x= C
Therefore, the general solution of the differential equation \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0 is \sin^{-1}y+\sin^{-1}x= C

Question:7 Show that the general solution of the differential equation \frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0 is given by (x + y + 1) = A (1 - x - y - 2xy) , where A is parameter.

Answer:

Given,

\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0

1517901142938688

1517901143721318

1517901144865249

Integrating both sides,

15179011455756

1517901146296446

1517901147071246

1517901147830720

1517901148634163

1517901149451944

1517901150235484

1517901151028875

1517901151812195

Let 1517901152593257

1517901153375926

Let A = 1517901154163100 ,

1517901154943810

Hence proved.

Question:8 Find the equation of the curve passing through the point \left(0,\frac{\pi}{4} \right ) whose differential equation is \sin x \cos y dx + \cos x \sin y dy = 0.

Answer:

Given equation is
\sin x \cos y dx + \cos x \sin y dy = 0.
we can rewrite it as
\frac{dy}{dx}= -\tan x\cot y\\ \\ \frac{dy}{\cot y}= -\tan xdx\\ \\ \tan y dy =- \tan x dx
Integrate both the sides
\log |\sec y|+C' = -\log|sec x|- C''\\ \\ \log|\sec y | +\log|\sec x| = C\\ \\ \sec y .\sec x = e^{C}
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point \left(0,\frac{\pi}{4} \right )
So,
\sec \frac{\pi}{4} .\sec 0 = e^{C}\\ \\ \sqrt2.1= e^C\\ \\ C = \log \sqrt2
Now,
\sec y.\sec x= e^{\log \sqrt 2}\\ \\ \frac{\sec x}{\cos y} = \sqrt 2\\ \\ \cos y = \frac{\sec x}{\sqrt 2}
Therefore, the equation of the curve passing through the point \left(0,\frac{\pi}{4} \right ) whose differential equation is \sin x \cos y dx + \cos x \sin y dy = 0. is \cos y = \frac{\sec x}{\sqrt 2}

Question:9 Find the particular solution of the differential equation (1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0 , given that y = 1 when x = 0 .

Answer:

Given equation is
(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0
we can rewrite it as
\frac{dy}{dx}= -\frac{(1+y^2)e^x}{(1+e^{2x})}\\ \\ \frac{dy}{1+y^2}= \frac{-e^xdx}{1+e^{2x}}
Now, integrate both the sides
\tan^{-1}y + C' =\int \frac{-e^{x}dx}{1+e^{2x}}
\int \frac{-e^{x}dx}{1+e^{2x}}\\
Put
e^x = t \\ e^xdx = dt
\int \frac{dt}{1+t^2}= \tan^{-1}t + C''
Put t = e^x again
\int \frac{-e^{x}dx}{1+e^{2x}} = -\tan ^{-1}e^x+C''
Put this in our equation
\tan^{-1}y = -\tan ^{-1}e^x+C\\ \tan^{-1}y +\tan ^{-1}e^x=C
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
\\ \tan^{-1}1 +\tan ^{-1}e^0=C\\ \\ \frac{\pi}{4}+\frac{\pi}{4}= C\\ C = \frac{\pi}{2}
Now, put the value of C

\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}
Therefore, the particular solution of the differential equation (1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0 is \tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}

Answer:

Given,

ye^\frac{x}{y}dx = (xe^\frac{x}{y} + y^2)dy

\\ ye^\frac{x}{y}\frac{dx}{dy} = xe^\frac{x}{y} + y^2 \\ \implies e^\frac{x}{y}[y\frac{dx}{dy} -x] = y^2 \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} = 1

Let \large e^\frac{x}{y} = t

Differentiating it w.r.t. y, we get,

\\ \frac{d}{dy}e^\frac{x}{y} = \frac{dt}{dy} \\ \implies e^\frac{x}{y}.\frac{d}{dy}(\frac{x}{y}) = \frac{dt}{dy} \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} =\frac{dt}{dy}

Thus from these two equations,we get,

\\ \frac{dt}{dy} = 1 \\ \implies \int dt = \int dy \\ \implies t = y + C

1517901178627857

Question:11 Find a particular solution of the differential equation (x - y) (dx + dy) = dx - dy, , given that y = -1 , when x = 0 . (Hint: put x - y = t )

Answer:

Given equation is
(x - y) (dx + dy) = dx - dy,
Now, integrate both the sides
Put
(x-y ) = t\\ dx - dy = dt
Now, given equation become
dx+dy= \frac{dt}{t}
Now, integrate both the sides
x+ y + C '= \log t + C''
Put t = x- y again
x+y = \log (x-y)+ C
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
0+(-1) = \log (0-(-1))+ C\\ C = -1
Now, put the value of C

x+y = \log |x-y|-1\\ \log|x-y|= x+y+1
Therefore, the particular solution of the differential equation (x - y) (dx + dy) = dx - dy, is \log|x-y|= x+y+1

Question:12 Solve the differential equation \left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0) .

Answer:

Given,

\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1

1517901190189935

1517901190951594

This is equation is in the form of 1517901191734385

p = 1517901192495126 and Q = 151790119325976

Now, I.F. = 1517901194040946

We know that the solution of the given differential equation is:

y(I.F.) = \int(Q\timesI.F.)dx + C

151790119561795

1517901196417819

151790119717945

Question:13 Find a particular solution of the differential equation \frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0) , given that y = 0 \ \textup{when}\ x = \frac{\pi}{2} .

Answer:

Given equation is
\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)
This is \frac{dy}{dx} + py = Q type where p =\cot x and Q = 4xcosec x Q = 4x \ cosec x
Now,
I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C
y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\ \\ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x= \frac{\pi}{2}
at x= \frac{\pi}{2}
0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\ \\ C = - \frac{\pi^2}{2}
Now, put the value of C
y\sin x= 2x^2-\frac{\pi^2}{2}
Therefore, the particular solution is y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)

Question:14 Find a particular solution of the differential equation (x+1)\frac{dy}{dx} = 2e^{-y} -1 , given that y = 0 when x = 0

Answer:

Given equation is
(x+1)\frac{dy}{dx} = 2e^{-y} -1
we can rewrite it as
\frac{e^ydy}{2-e^y}= \frac{dx}{x+1}\\
Integrate both the sides
\int \frac{e^ydy}{2-e^y}= \log |x+1|\\
\int \frac{e^ydy}{2-e^y}
Put
2-e^y = t\\ -e^y dy = dt
\int \frac{-dt}{t}=- \log |t|
put t = 2- e^y again
\int \frac{e^ydy}{2-e^y} =- \log |2-e^y|
Put this in our equation
\log |2-e^y| + C'= \log|1+x| + C''\\ \log (2-e^y)^{-1}= \log (1+x)+\log C\\ \frac{1}{2-e^y}= C(1+x)

Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
\frac{1}{2-e^0}= C(1+0)\\ C = \frac{1}{2}
Now, put the value of C
\frac{1}{2-e^y} = \frac{1}{2}(1+x)\\ \\ \frac{2}{1+x}= 2-e^y\\ \frac{2}{1+x}-2= -e^y\\ -\frac{2x-1}{1+x} = -e^y\\ y = \log \frac{2x-1}{1+x}
Therefore, the particular solution is y = \log \frac{2x-1}{1+x}, x\neq-1

Question:15 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let n be the population of the village at any time t.

According to question,

\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)

\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C

Now, at t=0, n = 20000 (Year 1999)

\\ \implies \log (20000) = k(0) + C \\ \implies C = \log2 + 4

Again, at t=5, n= 25000 (Year 2004)

\\ \implies \log (25000) = k(5) + \log2 + 4 \\ \implies \log 25 + 3 = 5k + \log2 +4 \\ \implies 5k = \log 25 - \log2 -1 =\log \frac{25}{20} \\ \implies k = \frac{1}{5}\log \frac{5}{4}

Using these values, at t =10 (Year 2009)

\\ \implies \log n = k(10)+ C \\ \implies \log n = \frac{1}{5}\log \frac{5}{4}(10) + \log2 + 4 \\ \implies \log n = \log(\frac{25.2.10000}{16}) = \log(31250) \\ \therefore n = 31250

Therefore, the population of the village in 2009 will be 31250.

Question:16 The general solution of the differential equation \frac{ydx - xdy}{y} = 0 is

(A) xy = C

(B) x = Cy^2

(C) y = Cx

(D) y = Cx^2

Answer:

Given equation is
\frac{ydx - xdy}{y} = 0
we can rewrite it as
dx = \frac{x}{y}dy\\ \frac{dy}{y}=\frac{dx}{x}
Integrate both the sides
we will get
\log |y| = \log |x| + C\\ \log \frac{y}{x} = C \\ \frac{y}{x} = e^C\\ \frac{y}{x} = C\\ y = Cx
Therefore, answer is (C)

Question:17 The general solution of a differential equation of the type \frac{dx}{dy} + P_1 x = Q_1 is

(A) ye^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C

(B) ye^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C

(C) xe^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C

(D) xe^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C

Answer:

Given equation is
\frac{dx}{dy} + P_1 x = Q_1
and we know that the general equation of such type of differential equation is

xe^{\int p_1dy} = \int (Q_1e^{\int p_1dy})dy+ C
Therefore, the correct answer is (C)

Question:18 The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is

(A) xe^y + x^2 = C

(B) xe^y + y^2 = C

(C) ye^x + x^2 = C

(D) ye^y + x^2 = C

Answer:

Given equation is
e^x dy + (y e^x + 2x) dx = 0
we can rewrite it as
\frac{dy}{dx}+y=-2xe^{-x}
It is \frac{dy}{dx}+py=Q type of equation where p = 1 \ and \ Q = -2xe^{-x}
Now,
I.F. = e^{\int p dx }= e^{\int 1dx}= e^x
Now, the general solution is
y(I.F.) = \int (Q\times I.F.)dx+C
y(e^x) = \int (-2xe^{-x}\times e^x)dx+C\\ ye^x= \int -2xdx + C\\ ye^x=- x^2 + C\\ ye^x+x^2 = C
Therefore, (C) is the correct answer

If you want to get command on concepts then differential equations solutions of NCERT exercise are listed below

More About NCERT Solutions for Class 12 Maths Chapter 9

This class 12 differential equations NCERT solutions has 5 marks weightage in 12th board final examination. Generally, one question is asked from this Chapter 9 Class 12 Maths that can be studied in detail from the NCERT Class 12 maths book in the 12th board final exam. You can score these 5 marks very easily with the help of these Ncert Solutions For Class 12 Maths Chapter 9 Differential Equations.

Class 12 Maths ch 9 is very important for the students aspiring for the 12th board exam. This NCERT Class 12 Maths Chapter 9 solutions holds good weightage in competitive exams like JEE Main, VITEEE, BITSAT. In this chapter, there are 6 exercises with 95 questions. All these questions are prepared and explained in this class 12 differential equations NCERT solutions article.

Differential Equations Class 12 - Topics

9.1 Introduction

9.2 Basic Concepts

9.2.1. Order of a differential equation

9.2.2 Degree of a differential equation

9.3. General and Particular Solutions of a Differential Equation

9.4 Formation of a Differential Equation whose General Solution is given

9.4.1 Procedure to form a differential equation that will represent a given family of curves

9.5. Methods of Solving First Order, First Degree Differential Equations

9.5.1 Differential equations with variables separable

9.5.2 Homogeneous differential equations

9.5.3 Linear differential equations

NCERT Exemplar Class 12 Solutions - Subject Wise

So, what is basically a differential equation? A differential equation is an equation in which derivatives of the dependent variable with respect to independent variables involved. Let's understand it with an example from NCERT chapter 9 differential equations-

\\x^2-3x+3=0\;\:\:\:\:\:\:\:\:\:\:\:\:.....(1)\\ Sin\:x+cos\:x=0\:\:\:\:\:\:\:\:\:\:\:\:....(2)\\x\:+\:y\:=0\;\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....(3)\\x\frac{dy}{dx}+y=0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....(4)

From the above equations, we notice that equations (1), (2) and (3) involve dependent variable(variables) and/or independent only but equation (4) involves variables as well as derivative of the dependent variable (y) with respect to the independent variable (x). That type of equation is known as the differential equation.

Important terms used in class 12 chapter 9 differential equations-

  • Order of a differential equation - It is the order of the highest order derivative present in the equation.
  • Degree of a differential equation - It is the power of the highest order derivative in the differential equation.
  • Homogeneous differential equation - A differential equation that can be expressed in the form \frac{dy}{dx}=f(x,\:y) where f(x,\:y) is a homogeneous function of degree zero.
  • First order linear differential equation - A differential equation of the form \dpi{80} \frac{dy}{dx}\:+\:Py=Q where P and Q are constants or functions of x only.

NCERT solutions for class 12 maths - Chapter Wise

Key Features of Class 12 Differential Equations NCERT Solutions – Differential Equations

Differential equations class 12 ncert solutions are designed to help students understand the various concepts and techniques involved in solving differential equations. Some of the key features of these solutions are:

  1. Comprehensive coverage: The class 12 maths ch 9 question answer cover all the topics included in the Class 12 Maths syllabus, ensuring that students are well-prepared for their exams.

  2. Simple language: The class 12 maths ch 9 question answer are written in simple language, making it easy for students to understand the concepts and techniques involved in solving differential equations.

  3. Step-by-step approach: The class 12 differential equations solutions follow a step-by-step approach, which helps students to understand the solution process in a structured way.

  4. Well-illustrated solutions: The maths chapter 9 class 12 solutions are accompanied by diagrams and illustrations, which help students to visualize the solution process and understand the concepts better.

  5. Conceptual clarity: The maths chapter 9 class 12 solutions aim to develop the conceptual clarity of students, rather than just providing them with the final answers. This helps students to build a strong foundation in the subject.

NCERT solutions for class 12 subject wise

NCERT Solutions class wise

Tips to use NCERT Class 12 Maths Chapter 9 Solutions

NCERT solutions for class 12 maths chapter 9 differential equations are very helpful for the preparation of this chapter. Here are some tips to get command on it.

  • Differential equations are the easiest part of the class 12 calculus. If your concepts of integration are clear then it won't take much effort to get command on this chapter.
  • First, solve all NCERT problems including examples and miscellaneous exercise on your own. If you are not able to solve you can take the help of NCERT solutions for ch 9 maths class 12 differential equations which are provided here
  • If you have solved NCERT problems, you can solve previous years paper. It gives you an idea about the type of questions and difficulty levels of questions that have been asked in previous years

NCERT Books and NCERT Syllabus

Happy learning !!!

Frequently Asked Question (FAQs)

1. How the NCERT solutions are helpful in the board exam ?

As CBSE board exam paper is designed entirely based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. Only knowing the answer it not enough to perform well in the exam. In the NCERT solutions students will get know how best to write answer in the board exam in order to get good marks.

2. What is the weightage of the chapter Differential Equations for CBSE board exam ?

Generally, one question of 5 marks is asked from this chapter in the 12th board final exam. you should refer NCER syllabus for it. NCERT textbook and NCERT Notes are recommended if you want to obtain meritious marks in the Board exam.

3. What are the important topics in chapter Differential Equations ?

Basic concepts of differential equation, order and degree of the differential equation, general and particular solutions of a differential equation, formation of a differential equation, methods of solving first order,first degree differential equations, homogeneous differential equations and linear differential equations are the important topics of this chapter.

4. Are the answers to all the textbook questions available in the differential equations ncert solutions maths class 12 chapter 9?

The NCERT class 12 maths differential equations are available in PDF format and have been created by subject experts in line with the textbook questions. These solutions for ch 9 maths class 12 adhere to the latest CBSE Syllabus for 2023 and encompass all the significant concepts for the board exam. The problems in the textbook are solved step by step in accordance with the marks weightage in the CBSE Board exams. Careers360 website offers both chapter-wise and exercise-wise PDF links that can be used by students to instantly clarify their doubts.

Articles

Explore Top Universities Across Globe

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary. 

4 Jobs Available
Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available
Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive. 

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available
Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction. 

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions. 

3 Jobs Available
Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
Back to top