NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

 

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations: In class 11th you have already learnt 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 differential equations. Important topics that are going to be discussed in this chapter are some basic concepts related to the differential equation, particular and general solutions of a differential equation. In CBSE NCERT solutions for class 12 maths chapter 9 differential equations article, questions from all these topics are covered. The equation of function and its one or more derivative is called as a differential equation. You will also learn some methods to solve the differential equations, the formation of differential equations, and discuss some applications of differential equations in different areas in this chapter. Questions related to these topics also covered in the solutions of NCERT for class 12 maths chapter 9 differential equations article. Check all NCERT solutions from class 6 to 12 at a single place, which is helpful to learn CBSE maths and science.

This chapter has 5 marks weightage in 12th board final examination. Generally, one question is asked from this chapter in the 12th board final exam. You can score these 5 marks very easily with the help of these CBSE NCERT solutions for class 12 maths chapter 9 differential equations. This is a very important chapter for the students from 12th board exam point of view. This chapter holds good weightage in the 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 CBSE NCERT solutions for class 12 maths chapter 9 differential equations article. 

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 the power of the highest order derivative in the differential equation.

  • Homogeneous differential equation- A differential equation which 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. 

Topics of NCERT class 12 maths chapter 9 Differential Equations

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 solutions for class 12 maths chapter 9 Differential Equations- Solved exercise questions

CBSE NCERT solutions for class 12 maths chapter 9 differential equations-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)

Solutions of NCERT for class 12 maths chapter 9 differential equations-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' =  ex - ex = 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.

NCERT solutions for class 12 maths chapter 9 differential equations-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:


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:


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)

CBSE NCERT solutions for class 12 maths chapter 9 differential equations-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,

Now comparing the coefficients

 A + B = 2;  B + C = 1;  A + C = 0

Solving these:

Putting the values of A,B,C:

Therefore,

Now, y= 1 when x = 0

c = 1

Putting the value of c, we get:

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,

Now comparing the values of A,B,C

A + B + C = 0;  B-C = 0;  A = -1

Solving these:

 

Now putting the values of A,B,C

Given, y =0 when x =2

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 (loge 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 (e0.5 = 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 solutions for class 12 maths chapter 9 differential equations-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
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{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
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{ 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

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+\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

Differentiating 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

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-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 

Solutions of NCERT for class 12 maths chapter 9 differential equations-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)

CBSE NCERT solutions for class 12 maths chapter 9 differential equations-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,