# 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 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 chapter 9 NCERT Class 12 Maths solutions, 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 solutions 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.

• Class 12 Maths Chapter 9 Differential Equations Notes

• NCERT Exemplar Solutions for Class 12 Maths Chapter 9 Differential Equations

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

Question:1 Determine order and degree (if defined) of differential equation

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

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

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.

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.

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

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

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

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

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

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

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## NCERT solutions for class 12 maths chapter 9 differential equations-Exercise: 9.2

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

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.

Question:2 Verify that the given functions (explicit or implicit) is a 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.

Question:3. Verify that the given functions (explicit or implicit) is a 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.

Question:4. Verify that the given functions (explicit or implicit) is a 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.

Question:5 Verify that the given functions (explicit or implicit) 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.

Question:6. Verify that the given functions (explicit or implicit) 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.

Question:7 Verify that the given functions (explicit or implicit) 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.

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

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.

Question:9 Verify that the given functions (explicit or implicit) 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.

Question:10 Verify that the given functions (explicit or implicit) 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.

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

(D) 4

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

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

(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

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

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

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

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

Question:3 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

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

Question:4 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

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

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

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.

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

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

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.

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

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

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.

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?

(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

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

(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)

## 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,

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

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:

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:

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:

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 .

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).

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:

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

According to the question,

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

x =0 and y = -2

Putting the value of c, we get

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

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

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

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:

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

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%

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

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

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

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:23 The general solution of the differential equation is

(A)

(B)

(C)

(D)

Given,

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

Question:1 Show that the given differential equation is homogeneous and solve each of them.

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.

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.

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.

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.

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

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

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

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.

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

...............................(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)

Question:16 A homogeneous differential equation of the from can be solved by making the substitution.

(A)

(B)

(C)

(D)

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

option C is correct

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

(A)

(B)

(C)

(D)

Option D is the right answer.

we can take out lambda as a common factor and it can be cancelled out

## NCERT solutions 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

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

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

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

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

(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)

Question:19 The Integrating Factor of the differential equation is

(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)

## 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

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

(i)

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.

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

(ii)

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.

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

(iii)

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.

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

(iv)

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.

Question:3 Form the differential equation representing the family of curves given by , where a is an arbitrary constant.

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.

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,

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

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

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.

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

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 .

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

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 )

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 .

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 .

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

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

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

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

(A)

(B)

(C)

(D)

Given equation is

we can rewrite it as

Integrate both the sides
we will get

Question:17 The general solution of a differential equation of the type is

(A)

(B)

(C)

(D)

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

(A)

(B)

(C)

(D)

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

• Differential Equations Class 12 Exercise 9.1
• Differential Equations Class 12 Exercise 9.2
• Differential Equations Class 12 Exercise 9.3
• Differential Equations Class 12 Exercise 9.4
• Differential Equations Class 12 Exercise 9.5
• Differential Equations Class 12 Exercise 9.6
• Differential Equations Class 12 Miscellaneous Exercise

## More About NCERT Solutions for Class 12 Maths Chapter 9

This Chapter Differential Equations Class 12 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 NCERT solutions for class 12 maths chapter 9 differential equations article.

Topics of NCERT class 12 maths chapter 9 Differential Equations

9.1 Introduction

9.2 Basic Concepts

9.2.1. Order of a differential equation

9.2.2 Degree of a differential equation

9.3. General and Particular Solutions of a Differential Equation

9.4 Formation of a Differential Equation whose General Solution is given

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

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

9.5.1 Differential equations with variables separable

9.5.2 Homogeneous differential equations

9.5.3 Linear differential equations

Also, check NCERT Exemplar Class 12 Solutions

• NCERT Exemplar Class 12th Maths Solutions

• NCERT Exemplar Class 12th Physics Solutions

• NCERT Exemplar Class 12th Chemistry Solutions

• NCERT Exemplar Class 12th Biology Solutions

• NCERT Exemplar Class 12 Maths Solutions Chapter 9

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 is the power of the highest order derivative in the differential equation.
• Homogeneous differential equation - A differential equation that can be expressed in the form 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.

NCERT solutions for class 12 maths - Chapter-wise

 chapter 1 NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions chapter 2 NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 NCERT solutions for class 12 maths chapter 4 Determinants chapter 5 NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals chapter 8 NCERT solutions for class 12 maths chapter 8 Application of Integrals chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra chapter 11 NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 NCERT solutions for class 12 maths chapter 13 Probability

NCERT solutions for class 12 subject wise

• NCERT solutions for class 12 mathematics
• NCERT solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Solutions class wise

• NCERT solutions for class 12
• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

### Tips to use NCERT Class 12 Maths Chapter 9 Solutions

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

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

Happy learning !!!