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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
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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.
Also read :
>> 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
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 solutions - Exercise: 9.1
Question:1 Determine order and degree (if defined) of differential equation
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation 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
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation 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
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation 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.
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation 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.
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:6 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 3 Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 2
Therefore, it's degree is 2
Question:7 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 3
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:8 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:9 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:10 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:11 The degree of the differential equation is
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation 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 is
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Therefore, answer is (A)
NCERT differential equations class 12 solutions - Exercise: 9.2
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. 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.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y’ in the given differential equations,
.
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y’ in the given differential equations,
.
Therefore, the given function is not the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y in RHS,
.
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
.
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
Substituting the values of y in RHS.
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
y' + siny.y' = 1
y'(1 + siny) = 1
Substituting the values of y and y' in LHS,
= (x + cosy) = y = RHS
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y and y' in LHS,
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
(D) 4
The number of constants in the general solution of a differential equation of order n is equal to its order.
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
Answer:
Given equation is
Differentiate both the sides w.r.t x
Now, again differentiate it w.r.t x
Therefore, the required differential equation is or
Answer:
Given equation is
Differentiate both the sides w.r.t x
-(i)
Now, again differentiate it w.r.t x
-(ii)
Now, divide equation (i) and (ii)
Therefore, the required differential equation is
Answer:
Given equation is
-(i)
Differentiate both the sides w.r.t x
-(ii)
Now, again differentiate w.r.t. x
-(iii)
Now, multiply equation (i) with 2 and add equation (ii)
-(iv)
Now, multiply equation (i) with 3 and subtract from equation (ii)
-(v)
Now, put values from (iv) and (v) in equation (iii)
Therefore, the required differential equation is
Answer:
Given equation is
-(i)
Now, differentiate w.r.t x
-(ii)
Now, again differentiate w.r.t x
-(iii)
Now, multiply equation (ii) with 2 and subtract from equation (iii)
-(iv)
Now,put the value in equation (iii)
Therefore, the required equation is
Answer:
Given equation is
-(i)
Now, differentiate w.r.t x
-(ii)
Now, again differentiate w.r.t 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
Therefore, the required equation is
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
-(i)
Now, differentiate w.r.t x
-(ii)
Put equation (ii) in equation (i)
Therefore, the required equation is
Answer:
Equation of perabola having vertex at origin and axis along positive y-axis is
(i)
Now, differentiate w.r.t. c
-(ii)
Put value from equation (ii) in (i)
Therefore, the required equation is
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
-
Now, differentiate w..r.t. x
-(i)
Now, again differentiate w.r.t. x
-(ii)
Put value from equation (ii) in (i)
Our equation becomes
Therefore, the required equation is
Answer:
Equation of hyperbolas having foci on x-axis and centre at the origin
Now, differentiate w..r.t. x
-(i)
Now, again differentiate w.r.t. x
-(ii)
Put value from equation (ii) in (i)
Our equation becomes
Therefore, the required equation is
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
Now, differentiate w.r.t x
we get,
Put value fro equation (ii) in (i)
Therefore, the required differential equation is
Question:11 Which of the following differential equations has as the general solution?
Answer:
Given general solution is
Differentiate it w.r.t x
we will get
Again, Differentiate it w.r.t x
Therefore, (B) is the correct answer
Question:12 Which of the following differential equations has as one of its particular solution?
Answer:
Given equation is
Now, on differentiating it w.r.t x
we get,
and again on differentiating it w.r.t x
we get,
Now, on substituting the values of in all the options we will find that only option c which is satisfies
Therefore, the correct answer is (C)
NCERT class 12 maths chapter 9 question answer - Exercise: 9.4
Question:1 Find the general solution:
Answer:
Given,
Question:5 Find the general solution:
Answer:
Given, in the question
Let,
This is the general solution
Question:7 Find the general solution:
Answer:
Given,
let logy = t
=> 1/ydy = dt
This is the general solution
Question:8 Find the general solution:
Answer:
Given, in the question
This is the required general equation.
Question:11 Find a particular solution satisfying the given condition:
Answer:
Given, in the question
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:
Answer:
Given, in the question
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,
Question:13 Find a particular solution satisfying the given condition:
Answer:
Given,
Now, y =1 when x =0
1 = 0 + c
Therefore, c = 1
Putting the value of c:
Question:14 Find a particular solution satisfying the given condition:
Answer:
Given,
Now, y=1 when x =0
1 = ksec0
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 .
Answer:
We first find the general solution of the given differential equation
Given,
Now, Since the curve passes through (0,0)
y = 0 when x =0
Putting the value of c, we get:
Question:16 For the differential equation , find the solution curve passing through the point (1, –1).
Answer:
We first find the general solution of the given differential equation
Given,
Now, Since the curve passes through (1,-1)
y = -1 when x = 1
Putting the value of C:
Answer:
According to the question,
Now, Since the curve passes through (0,-2).
x =0 and y = -2
Putting the value of c, we get
Answer:
Slope m of line joining (x,y) and (-4,-3) is
According to the question,
Now, Since the curve passes through (-2,1)
x = -2 , y =1
Putting the value of k, we get
Answer:
Volume of a sphere,
Given, Rate of change is constant.
Now, at t=0, r=3 and at t=3 , r =6
Putting these value:
Also,
Putting the value of c and k:
Answer:
Let p be the principal amount and t be the time.
According to question,
Now, at t =0 , p = 100
and at t =10, p = 200
Putting these values,
Also,
,
So value of r = 6.93%
Answer:
Let p be the principal amount and t be the time.
According to question,
Now, at t =0 , p = 1000
Putting these values,
Also, At t=10
,
After 10 years, the total amount would be Rs.1648
Answer:
Let n be the number of bacteria at any time t.
According to question,
Now, at t=0, n = 100000
Again, at t=2, n= 110000
Using these values, for n= 200000
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.
Answer:
The given diffrential eq can be written as
Let
Now,
Hence, it is a homogeneous equation.
To solve it put y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both side, we get;
Again substitute the value ,we get;
This is the required solution of given diff. equation
Question:2 Show that the given differential equation is homogeneousand solve each of them.
Answer:
the above differential eq can be written as,
............................(i)
Now,
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both sides, we get; (and substitute the value of )
this is the required solution
Question:3 Show that the given differential equation is homogeneous and solve each of them.
Answer:
The given differential eq can be written as;
....................................(i)
Hence it is a homogeneous equation.
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both sides, we get;
again substitute the value of
This is the required solution.
Question:4 Show that the given differential equation is homogeneous and solve each of them.
Answer:
we can write it as;
...................................(i)
Hence it is a homogeneous equation
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get
.............[ ]
This is the required solution.
Question:5 Show that the given differential equation is homogeneous and solve it.
Answer:
............(i)
Hence it is a homogeneous eq
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides, we get;
after substituting the value of
This is the required solution
Question:6 Show that the given differential equation is homogeneous and solve it.
Answer:
.................................(i)
henxe it is a homogeneous equation
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides,
Substitute the value of v=y/x , we get
Required solution
Question:7 Solve.
Answer:
......................(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
Substitute this value in equation (i)
integrating on both sides, we get
substitute the value of v= y/x , we get
Required solution
Question:8 Solve.
Answer:
...............................(i)
it is a homogeneous equation
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides we get;
Required solution
Question:9 Solve.
Answer:
..................(i)
hence it is a homogeneous eq
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get; ( substituting v =y/x)
This is the required solution of the given differential eq
Question:10 Solve.
Answer:
.......................................(i)
Hence it is a homogeneous equation.
Now, to solve substitute x = yv
Diff erentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both sides, we get;
This is the required solution of the diff equation.
Question:11 Solve for particular solution.
Answer:
..........................(i)
We can clearly say that it is a homogeneous equation.
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides
......................(ii)
Now, y=1 and x= 1
After substituting the value of 2k in eq. (ii)
This is the required solution.
Question:12 Solve for particular solution.
Answer:
...............................(i)
Hence it is a homogeneous equation
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i), we get
Integrating on both sides, we get;
replace the value of v=y/x
.............................(ii)
Now y =1 and x = 1
therefore,
Required solution
Question:13 Solve for particular solution.
Answer:
..................(i)
Hence it is a homogeneous eq
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
on integrating both sides, we get;
On substituting v =y/x
............................(ii)
Now,
put this value of C in eq (ii)
Required solution.
Question:14 Solve for particular solution.
Answer:
....................................(i)
the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
on integrating both sides, we get;
.................................(ii)
now y = 0 and x =1 , we get
put the value of C in eq 2
Question:15 Solve for particular solution.
Answer:
The above eq can be written as;
By looking, we can say that it is a homogeneous equation.
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get;
.............................(ii)
Now, y = 2 and x =1, we get
C =-1
put this value in equation(ii)
Question:16 A homogeneous differential equation of the from can be solved by making the substitution.
Answer:
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?
Answer:
Option D is the right answer.
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:
Answer:
Given equation is
This is type where p = 2 and Q = sin x
Now,
Now, the solution of given differential equation is given by relation
Let
Put the value of I in our equation
Now, our equation become
Therefore, the general solution is
Question:2 Solve for general solution:
Answer:
Given equation is
This is type where p = 3 and
Now,
Now, the solution of given differential equation is given by the relation
Therefore, the general solution is
Question:3 Find the general solution
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:4 Solve for General Solution.
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:5 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is where and
Now,
Now, the solution of given differential equation is given by relation
take
Now put again
Put this value in our equation
Therefore, the general solution is
Question:6 Solve for General Solution.
Answer:
Given equation is
Wr can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Let
Put this value in our equation
Therefore, the general solution is
Question:7 Solve for general solutions.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
take
Put this value in our equation
Therefore, the general solution is
Question:8 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of the given differential equation is given by the relation
Therefore, the general solution is
Question:9 Solve for general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of the given differential equation is given by the relation
Lets take
Put this value in our equation
Therefore, the general solution is
Question:10 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Lets take
Put this value in our equation
Therefore, the general solution is
Question:11 Solve for general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:12 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:13 Solve for particular solution.
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when
at
Now,
Therefore, the particular solution is
Question:14 Solve for particular solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
Now,
Therefore, the particular solution is
Question:15 Find the particular solution.
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when
at
Now,
Therefore, the particular solution is
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
Now, it is given that
It is type of equation where
Now,
Now,
Now, Let
Put this value in our equation
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)
Our final equation becomes
Therefore, the required equation of the curve is
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
Now, it is given that
It is type of equation where
Now,
Now,
Now, Let
Put this value in our equation
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
Our final equation becomes
Therefore, the required equation of curve is
Question:18 The Integrating Factor of the differential equation is
Answer:
Given equation is
we can rewrite it as
Now,
It is type of equation where
Now,
Therefore, the correct answer is (C)
Question:19 The Integrating Factor of the differential equation is
Answer:
Given equation is
we can rewrite it as
It is type of equation where
Now,
Therefore, the correct answer is (D)
Class 12 Maths Chapter 9 NCERT solutions - Miscellaneous Exercise
Question:1 Indicate Order and Degree.
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation 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.
Answer:
Given function is
We can rewrite it as
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.
Answer:
Given function is
We can rewrite it as
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
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Putting values in LHS
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given equation is
we can rewrite it as
-(i)
Differentiate both the sides w.r.t x
-(ii)
Put value from equation (ii) in (i)
Therefore, the required differential equation is
Question:4 Prove that is the general solution of differential equation , where c is a parameter.
Answer:
Given,
Now, let y = vx
Substituting the values of y and y' in the equation,
Integrating both sides we get,
Now,
Let
Now,
Let v 2 = p
Now, substituting the values of I 1 and I 2 in the above equation, we get,
Thus,
Answer:
Now, equation of the circle with center at (x,y) and radius r is
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
-(i)
Differentiate it w.r.t x
we will get
-(ii)
Put value from equation (ii) in equation (i)
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is
Question:6 Find the general solution of the differential equation
Answer:
Given equation is
we can rewrite it as
Now, integrate on both the sides
Therefore, the general solution of the differential equation is
Question:7 Show that the general solution of the differential equation is given by , where A is parameter.
Answer:
Given,
Integrating both sides,
Let
Let A = ,
Hence proved.
Question:8 Find the equation of the curve passing through the point whose differential equation is
Answer:
Given equation is
we can rewrite it as
Integrate both the sides
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point
So,
Now,
Therefore, the equation of the curve passing through the point whose differential equation is is
Question:9 Find the particular solution of the differential equation , given that when .
Answer:
Given equation is
we can rewrite it as
Now, integrate both the sides
Put
Put again
Put this in our equation
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
Now, put the value of C
Therefore, the particular solution of the differential equation is
Question:10 Solve the differential equation
Answer:
Given,
Let
Differentiating it w.r.t. y, we get,
Thus from these two equations,we get,
Question:11 Find a particular solution of the differential equation , given that , when . (Hint: put )
Answer:
Given equation is
Now, integrate both the sides
Put
Now, given equation become
Now, integrate both the sides
Put again
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
Now, put the value of C
Therefore, the particular solution of the differential equation is
Question:12 Solve the differential equation .
Answer:
Given,
This is equation is in the form of
p = and Q =
Now, I.F. =
We know that the solution of the given differential equation is:
Question:13 Find a particular solution of the differential equation , given that .
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when
at
Now, put the value of C
Therefore, the particular solution is
Question:14 Find a particular solution of the differential equation , given that when
Answer:
Given equation is
we can rewrite it as
Integrate both the sides
Put
put again
Put this in our equation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
Now, put the value of C
Therefore, the particular solution is
Answer:
Let n be the population of the village at any time t.
According to question,
Now, at t=0, n = 20000 (Year 1999)
Again, at t=5, n= 25000 (Year 2004)
Using these values, at t =10 (Year 2009)
Therefore, the population of the village in 2009 will be 31250.
Question:16 The general solution of the differential equation is
Answer:
Given equation is
we can rewrite it as
Integrate both the sides
we will get
Therefore, answer is (C)
Question:17 The general solution of a differential equation of the type is
Answer:
Given equation is
and we know that the general equation of such type of differential equation is
Therefore, the correct answer is (C)
Question:18 The general solution of the differential equation is
Answer:
Given equation is
we can rewrite it as
It is type of equation where
Now,
Now, the general solution is
Therefore, (C) is the correct answer
If you want to get command on concepts then differential equations solutions of NCERT exercise are listed below
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.
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
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-
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-
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:
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.
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.
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.
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.
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 maths chapter 9 differential equations are very helpful for the preparation of this chapter. Here are some tips to get command on it.
Happy learning !!!
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.
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.
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.
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.
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.
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.
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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.
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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.
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.
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.
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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.
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.
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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.
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.
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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.
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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.
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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.
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).
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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.
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