# 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-

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 where is a homogeneous function of degree zero.

• First order linear differential equation-  A differential equation of the form where P and Q are constants or functions of x only.

## 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

## CBSE NCERT solutions for class 12 maths chapter 9 differential equations-Exercise: 9.1

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

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.

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

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

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

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

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

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

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

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

(A)    3

(B)    2

(C)    1

(D)    not defined

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

Question:12 The order of the differential equation  is

(A)    2

(B)    1

(C)    0

(D)    Not Defined

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

Solutions of NCERT for class 12 maths chapter 9 differential equations-Exercise: 9.2

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' =  ex - ex = 0 =  RHS.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

(D) 4

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

(D) 0

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

## NCERT solutions for class 12 maths chapter 9 differential equations-Exercise: 9.3

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

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

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

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

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 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

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

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

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 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

(A)

(B)

(C)

(D)

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

(A)

(B)

(C)

(D)

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)

## CBSE NCERT solutions for class 12 maths chapter 9 differential equations-Exercise: 9.4

Question:1 Find the general solution:

Given,

Question:2 Find the general solution:

Given, in the question

The required general solution:

Question:3 Find the general solution:

Given, in the question

The required general equation

Question:4 Find the general solution:

Given,

Now, let tany = t and tanx = u

Question:5 Find the general solution:

Given, in the question

Let,

This is the general solution

Question:6 Find the general solution:

Given, in the question

Question:7 Find the general solution:

Given,

let logy = t

=> 1/ydy = dt

This is the general solution

Question:8 Find the general solution:

Given, in the question

This is the required general equation.

Question:9 Find the general solution:

Given, in the question

Now,

Here, u = and v = 1

Question:10 Find the general solution

Given,

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: 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,

Given,

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

Given,

Now, y=1 when x =0

1 = ksec0

k = 1

Putting the vlue of k:

y = sec x

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:

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:

According to the question,

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

x =0 and y = -2

Putting the value of c, we get

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

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:

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%

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

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 solutions for class 12 maths chapter 9 differential equations-Exercise: 9.5

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

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
Differentiating 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

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.

we can write it 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

.............[]
This is the required solution.

............(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

.................................(i)

henxe 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,

Substitute the value of v=y/x , we get

Required solution

Question:7 Solve.

......................(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.

...............................(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.

..................(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.

.......................................(i)

Hence it is a homogeneous equation.

Now, to solve substitute x= yv

Differentiating 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.

..........................(i)

We can clearly say that 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

......................(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.

...............................(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.

..................(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.

....................................(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.

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)

(A)

(B)

(C)

(D)

for solving this type of equation put x/y = v
x = vy

option C is correct

(A)

(B)

(C)

(D)

Option D is the right answer.

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:

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:

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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

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

(A)

(B)

(C)

(D)

Given equation is

we can rewrite it as

Now,
It is      type of equation  where
Now,

Therefore, the correct answer is (C)

(A)

(B)

(C)

(D)

Given equation is

we can rewrite it as

It is        type of equation where
Now,

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)

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.

(ii)

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.

(iii)

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

(i)