NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations Miscellaneous Exercise

Page No 419:

Question 1:

For each of the differential equations given below, indicate its order and degree (if defined).

(i) 

(ii) 

(iii) 

Answer:

(i) The differential equation is given as:

The highest order derivative present in the differential equation is. Thus, its order is two. The highest power raised to is one. Hence, its degree is one.

(ii) The differential equation is given as:

The highest order derivative present in the differential equation is. Thus, its order is one. The highest power raised to is three. Hence, its degree is three.

(iii) The differential equation is given as:

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

However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

Page No 420:

Question 2:

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) 

(ii) 

(iii) 

(iv) 

Answer:

(i) 

Differentiating both sides with respect to x, we get:

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

Now, on substituting the values of and in the differential equation, we get:

⇒ L.H.S. ≠ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii) 

Differentiating both sides with respect to x, we get:

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

Now, on substituting the values of and in the L.H.S. of the given differential equation, we get:

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

(iii) 

Differentiating both sides with respect to x, we get:

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

Substituting the value of  in the L.H.S. of the given differential equation, we get:

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

(iv) 

Differentiating both sides with respect to x, we get:

Substituting the value of in the L.H.S. of the given differential equation, we get:

Hence, the given function is a 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.

Answer:

Differentiating with respect to x, we get:

From equation (1), we get:

On substituting this value in equation (3), we get:

Hence, the differential equation of the family of curves is given as 

Question 4:

Prove that is the general solution of differential equation, where cis a parameter.

Answer:

This is a homogeneous equation. To simplify it, we need to make the substitution as:

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

Integrating both sides, we get:

Substituting the values of I₁ and I₂ in equation (3), we get:

Therefore, equation (2) becomes:

Hence, the given result is proved.

Question 5:

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

Answer:

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

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

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

Hence, the required differential equation of the family of circles is 

Question 6:

Find the general solution of the differential equation 

Answer:

Integrating both sides, we get:

Question 7:

Show that the general solution of the differential equation is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter

Answer:

Integrating both sides, we get:

Hence, the given result is proved.

Question 8:

Find the equation of the curve passing through the point whose differential equation is, 

Answer:

The differential equation of the given curve is:

Integrating both sides, we get:

The curve passes through point 

On substituting in equation (1), we get:

Hence, the required equation of the curve is 

Question 9:

Find the particular solution of the differential equation

, given that y = 1 when x = 0

Answer:

Integrating both sides, we get:

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

Now, y = 1 at x = 0.

Therefore, equation (2) becomes:

Substituting in equation (2), we get:

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

Question 10:

Solve the differential equation 

Answer:

Differentiating it with respect to y, we get:

From equation (1) and equation (2), we get:

Integrating both sides, we get:

Question 11:

Find a particular solution of the differential equation, given that y = – 1, when x = 0 (Hint: put x – y = t)

Answer:

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

Integrating both sides, we get:

Now, y = –1 at x = 0.

Therefore, equation (3) becomes:

log 1 = 0 – 1 + C

⇒ C = 1

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

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

Page No 421:

Question 12:

Solve the differential equation 

Answer:

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Question 13:

Find a particular solution of the differential equation , given that y = 0 when 

Answer:

The given differential equation is:

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Now,

Therefore, equation (1) becomes:

Substituting in equation (1), we get:

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

Question 14:

Find a particular solution of the differential equation, given that y = 0 when x = 0

Answer:

Integrating both sides, we get:

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

Now, at x = 0 and y = 0, equation (2) becomes:

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

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

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 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

Integrating both sides, we get:

log y = kt + C … (1)

In the year 1999, t = 0 and y = 20000.

Therefore, we get:

log 20000 = C … (2)

In the year 2004, t = 5 and y = 25000.

Therefore, we get:

In the year 2009, t = 10 years.

Now, on substituting the values of t, k, and C in equation (1), we get:

Hence, the population of the village in 2009 will be 31250.

Question 16:

The general solution of the differential equation is

A. xy = C

B. x = Cy²

C. y = Cx

D. y = Cx²

Answer:

The given differential equation is:

Integrating both sides, we get:

Hence, the correct answer is C.

Question 17:

The general solution of a differential equation of the type is

A. 

B. 

C. 

D. 

Answer:

The integrating factor of the given differential equation

The general solution of the differential equation is given by,

Hence, the correct answer is C.

Question 18:

The general solution of the differential equation  is

A. xe^y + x² = C

B. xe^y + y² = C

C. ye^x + x² = C

D. ye^y + x² = C

Answer:

The given differential equation is:

This is a linear differential equation of the form

The general solution of the given differential equation is given by,

Hence, the correct answer is C.