Class XII                           Chapter 9 – Differential Equations                                Maths

Exercise 9.1

Question 1:

Determine order and degree(if defined) of differential equation   Answer

The highest order derivative present in the differential equation is  . Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Question 2:

Determine order and degree(if defined) of differential equation  Answer

The given differential equation is:

The highest order derivative present in the differential equation is    . Therefore, its order is one.

It is a polynomial equation in     . The highest power raised to is 1. Hence, its degree is one.

Question 3:

Page 1 of 120

The highest order derivative present in the given differential equation is        . Therefore, its order is two.

It is a polynomial equation in  and  . The power raised to  is 1. Hence, its degree is one.

Question 4:

The highest order derivative present in the given differential equation is . Therefore, its order is 2.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Question 5:

The highest order derivative present in the differential equation is . Therefore, its order is two.

It is a polynomial equation in        and the power raised to        is 1. Hence, its degree is one.

Question 6:

Determine order and degree(if defined) of differential equation

Answer

The highest order derivative present in the differential equation is . Therefore, its order is three.

The given differential equation is a polynomial equation in . The highest power raised to is 2. Hence, its degree is 2.

Question 7:

Determine order and degree(if defined) of differential equation  Answer

The highest order derivative present in the differential equation is . Therefore, its order is three.

It is a polynomial equation in . The highest power raised to is 1. Hence, its degree is 1.

Question 8:

 

The highest order derivative present in the differential equation is . Therefore, its order is one.

The given differential equation is a polynomial equation in  and the highest power raised to is one. Hence, its degree is one.

Question 9:

The highest order derivative present in the differential equation is  . Therefore, its order is two.

The given differential equation is a polynomial equation in  and    and the highest

power raised to  is one. Hence, its degree is one.

Question 10:

The highest order derivative present in the differential equation is  . Therefore, its order is two.

This is a polynomial equation in  and and the highest power raised to  is one. Hence, its degree is one.

Question 11:

The degree of the differential equation

is

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

Question 12:

The order of the differential equation

is

The highest order derivative present in the given differential equation is  . Therefore, its order is two.

Hence, the correct answer is A.

Exercise 9.2

Question 1:

Answer

Differentiating both sides of this equation with respect to x, we get:

Now, differentiating equation (1) with respect to x, we get:

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

Question 2:

Answer

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of  in the given differential equation, we get:

L.H.S. =     = R.H.S.

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

Question 3:

Answer

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of    in the given differential equation, we get:

L.H.S. =     = R.H.S.

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

Question 4:

Answer

Differentiating both sides of the equation with respect to x, we get:

L.H.S. = R.H.S.

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

Question 5:

Answer

Differentiating both sides with respect to x, we get:

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

Question 6:

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of    in the given differential equation, we get:

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

Question 7:

Answer

Differentiating both sides of this equation with respect to x, we get:

L.H.S. = R.H.S.

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

Question 8:

Answer

Differentiating both sides of the equation with respect to x, we get:

Substituting the value of    in equation (1), we get:

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

Question 9:

Answer

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of    in the given differential equation, we get:

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

Question 10:

Answer

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of  in the given differential equation, we get:

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

Question 11:

The numbers of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Answer

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

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is D.

Question 12:

The numbers of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Answer

In a particular solution of a differential equation, there are no arbitrary constants. Hence, the correct answer is D.

Exercise 9.3

Question 1:

Answer

Differentiating both sides of the given equation with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Hence, the required differential equation of the given curve is

Question 2:

Answer

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Dividing equation (2) by equation (1), we get:

This is the required differential equation of the given curve.

Question 3:

Answer

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Multiplying equation (1) with (2) and then adding it to equation (2), we get:

Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we get:

Substituting the values of  in equation (3), we get:

This is the required differential equation of the given curve.

Question 4:

Answer

Differentiating both sides with respect to x, we get:

Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we get:

Differentiating both sides with respect to x, we get:

Dividing equation (4) by equation (3), we get:

This is the required differential equation of the given curve.

Question 5:

Answer

Differentiating both sides with respect to x, we get:

Again, differentiating with respect to x, we get:

Adding equations (1) and (3), we get:

This is the required differential equation of the given curve.

Question 6:

Form the differential equation of the family of circles touching the y-axis at the origin. Answer

The centre of the circle touching the y-axis at origin lies on the x-axis. Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

Differentiating equation (1) with respect to x, we get:

Now, on substituting the value of a in equation (1), we get:

This is the required differential equation.

Question 7:

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

Answer

The equation of the parabola having the vertex at origin and the axis along the positive

y-axis is:

Differentiating equation (1) with respect to x, we get:

Dividing equation (2) by equation (1), we get:

This is the required differential equation.

Question 8:

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Answer

Differentiating equation (1) with respect to x, we get:

Again, differentiating with respect to x, we get:

Substituting this value in equation (2), we get:

This is the required differential equation.

Question 9:

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

Answer

Differentiating both sides of equation (1) with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Substituting the value of      in equation (2), we get:

This is the required differential equation.

Question 10:

Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer

Let the centre of the circle on y-axis be (0, b).

Differentiating equation (1) with respect to x, we get:

Substituting the value of (y – b) in equation (1), we get:

This is the required differential equation.

Question 11:

Which of the following differential equations has as the general solution?

A.

B.

C.

D.

Answer

The given equation is:

Differentiating with respect to x, we get:

Again, differentiating with respect to x, we get:

This is the required differential equation of the given equation of curve. Hence, the correct answer is B.

Question 12:

Which of the following differential equation has          as one of its particular solution?

A.

B.

C.

D.

Answer

Again, differentiating with respect to x, we get:

Now, on substituting the values of y,  from equation (1) and (2) in each of

the given alternatives, we find that only the differential equation given in alternative C is correct.

Hence, the correct answer is C.

Exercise 9.4

Question 1:

Answer

The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Question 2:

Answer

The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Question 3:

Answer

The given differential equation is:

Now, integrating both sides, we get:

This is the required general solution of the given differential equation.

Question 4:

Answer

The given differential equation is:

Integrating both sides of this equation, we get:

Substituting these values in equation (1), we get:

This is the required general solution of the given differential equation.

Question 5:

Answer

The given differential equation is:

Integrating both sides of this equation, we get:

Let (e^x + e^–x) = t.

Differentiating both sides with respect to x, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Question 6:

Answer

The given differential equation is:

Integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Question 7:

Answer

The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Question 8:

Answer

The given differential equation is:

Integrating both sides, we get:

This is the required general solution of the given differential equation.

Question 9:

Answer

The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Question 10:

Answer

The given differential equation is:

Integrating both sides, we get:

Substituting the values of                                        in equation (1), we get:

This is the required general solution of the given differential equation.

Question 11:

Answer

The given differential equation is:

Integrating both sides, we get:

Comparing the coefficients of x² and x, we get:

A + B = 2 B + C = 1 A + C = 0

Solving these equations, we get:

Substituting the values of A, B, and C in equation (2), we get:

Therefore, equation (1) becomes:

Substituting C = 1 in equation (3), we get:

Question 12:

Answer

Integrating both sides, we get:

Comparing the coefficients of x², x, and constant, we get:

Solving these equations, we get         Substituting the values of A, B, and C in equation (2), we get:

Therefore, equation (1) becomes:

Substituting the value of k² in equation (3), we get:

Question 13:

Answer

Integrating both sides, we get:

Substituting C = 1 in equation (1), we get:

Question 14:

Answer

Integrating both sides, we get:

Substituting C = 1 in equation (1), we get:

y = sec x

Question 15:

Find the equation of a curve passing through the point (0, 0) and whose differential

equation is  .

Answer

The differential equation of the curve is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Now, the curve passes through point (0, 0).

Hence, the required equation of the curve is

Question 16:

For the differential equation  find the solution curve passing through the point (1, –1).

Answer

The differential equation of the given curve is:

Integrating both sides, we get:

Now, the curve passes through point (1, –1).

Substituting C = –2 in equation (1), we get:

This is the required solution of the given curve.

Question 17:

Find the equation of a curve passing through the point (0, –2) 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.

Answer

Let x and y be the x-coordinate and y-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the

According to the given information, we get:

Integrating both sides, we get:

Now, the curve passes through point (0, –2).

∴ (–2)² – 0² = 2C

⇒ 2C = 4

Substituting 2C = 4 in equation (1), we get:

y² – x² = 4

This is the required equation of the curve.

Question 18:

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

Answer

It is given that (x, y) is the point of contact of the curve and its tangent.

The slope (m₁) of the line segment joining (x, y) and (–4, –3) is

We know that the slope of the tangent to the curve is given by the relation,

According to the given information:

Integrating both sides, we get:

This is the general equation of the curve.

It is given that it passes through point (–2, 1).

Substituting C = 1 in equation (1), we get:

y + 3 = (x + 4)²

This is the required equation of the curve.

Question 19:

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

Answer

Let the rate of change of the volume of the balloon be k (where k is a constant).

Integrating both sides, we get:

⇒ 4π × 3³ = 3 (k × 0 + C)

⇒ 108π = 3C

⇒ C = 36π

At t = 3, r = 6:

⇒ 4π × 6³ = 3 (k × 3 + C)

⇒ 864π = 3 (3k + 36π)

⇒ 3k = –288π – 36π = 252π

⇒ k = 84π

Substituting the values of k and C in equation (1), we get:

Thus, the radius of the balloon after t seconds is .

Question 20:

In a bank, principal increases continuously at the rate of r% per year. Find the value of r

if Rs 100 doubles itself in 10 years (log_e 2 = 0.6931).

Answer

Integrating both sides, we get:

It is given that when t = 0, p = 100.

⇒ 100 = e^k … (2)

Hence, the value of r is 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 . Answer

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.

Integrating both sides, we get:

Now, when t = 0, p = 1000.

⇒ 1000 = e^C … (2)

At t = 10, equation (1) becomes:

Hence, after 10 years the amount will worth Rs 1648.

Question 22:

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

Answer

Let y be the number of bacteria at any instant t.

It is given that the rate of growth of the bacteria is proportional to the number present.

Integrating both sides, we get:

Let y₀ be the number of bacteria at t = 0.

⇒ log y₀ = C

Substituting the value of C in equation (1), we get:

Also, it is given that the number of bacteria increases by 10% in 2 hours.

Substituting this value in equation (2), we get:

Therefore, equation (2) becomes:

Now, let the time when the number of bacteria increases from 100000 to 200000 be t₁.

⇒ y = 2y₀ at t = t₁

From equation (4), we get:

Hence, in                hours the number of bacteria increases from 100000 to 200000.

Question 23:

The general solution of the differential equation

A.

B.

C.

D.

Answer

Integrating both sides, we get:

Hence, the correct answer is A.

Exercise 9.5

Question 1:

Answer

The given differential equation i.e., (x² + xy) dy = (x² + y²) dx can be written as:

This shows that equation (1) is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

Substituting the values of v and       in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 2:

Answer

The given differential equation is:

Thus, the given equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 3:

Answer

The given differential equation is:

Thus, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 4:

Answer

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 5:

Answer

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Integrating both sides, we get:

This is the required solution for the given differential equation.

Question 6:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of v and  in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 7:

Answer

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 8:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 9:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and   in equation (1), we get:

Integrating both sides, we get:

Therefore, equation (1) becomes:

This is the required solution of the given differential equation.

Question 10:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

x = vy

Substituting the values of x and  in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 11:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

Now, y = 1 at x = 1.

Substituting the value of 2k in equation (2), we get:

This is the required solution of the given differential equation.

Question 12:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

Now, y = 1 at x = 1.

in equation (2), we get:

This is the required solution of the given differential equation.

Question 13:

Answer

Therefore, the given differential equation is a homogeneous equation. To solve this differential equation, we make the substitution as:

y = vx

Integrating both sides, we get:

Now,

.

Substituting C = e in equation (2), we get:

This is the required solution of the given differential equation.