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 12^{th} 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.
9.2 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
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)
Solutions of NCERT for class 12 maths chapter 9 differential equationsExercise: 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.
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 yaxis at origin.
Answer:
If the circle touches yaxis at the origin then the centre of the circle lies at the xaxis
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 yaxis 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 yaxis and centre at origin.
Answer:
Equation of ellipses having foci on yaxis 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 xaxis 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 yaxis and radius 3 units.
Answer:
Equation of the family of circles having centre on yaxis 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)
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; BC = 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
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
Differentiating 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
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
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
Differentiating 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
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.
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
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.
Answer:
..........................(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.
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
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)
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,