Class 12 NCERT Solutions- Mathematics Part II – Chapter 9 Differential Equations-Exercise -9.2

In each of the Questions 1 to 6 verify that the given functions (explicit) is a solution of the corresponding differential equation:

Question 1. y = e^x + 1          : y” – y’ = 0

Solution:

Given: y = e^x + 1

On differentiating we get

y’ = e^x          -(1)

Again differentiating we get

y” = e^x           -(2)

Now substitute the values from equation(1) and (2), in the differential equation 

y” – y’ = e^x – e^x = 0       

Hence verified

Question 2. y = x² + 2x + C          : y’ – 2x – 2 = 0

Solution:

Given: y = x² + 2x + C

On differentiating we get

y’ = 2x + 2      

y’ – 2x – 2 = 0   

Hence verified

Question 3. y = cosx +  c          : y’ + sin x = 0

Solution:

Given: y = cos x + c

On differentiating we get

y’ = -sin x        -(1)

Now substitute the values from equation(1), in the differential equation 

y’ + sin x = 0

-sin x + sin x = 0

0 = 0

Hence verified

Question 4.          : 

Solution:

Given: 

Question 5. y = Ax          : xy’ = y (x ≠ 0)

Solution:

Given: y = Ax

y/A = x          -(1)

On differentiating we get

y’ = A          -(2)

Now substitute the values from equation(1) and (2), in the differential equation 

xy’ = y

 = y

y = y

Hence verified

Question 6. y = x sin x          : xy’ = y + x(x ≠ 0 and x > y or x < -y)

Solution:

Given: y = x.sin x   

On differentiating we get

y’ = x cos x + sin x           -(1)

Now substitute the values from equation(1), in the differential equation 

Taking LHS

xy’ = x(x cos x + sin x)

= x² cos x + x sin x

= x²√1 – sin²x + y

= y + x²

= y + x²

= y + x

LHS = RHS

Hence verified

Question 7. xy = log y + C          : y’ = 

Solution:

Given: xy = logy + C          -(1)

On differentiating we get

xy’ + y = > y’

xyy’ + y² = y’

xyy’ – y’ = -y²

y'(xy – 1) = -y²

y’ = -y²/ (xy – 1)          -(2)

Now substitute the values from equation(1) and (2), in the differential equation 

y’ = 

Hence verified

Question 8. y – cos y = x          : (y sin y + cos y + x)y’ = y

Solution:

Given: y – cos y = x         -(1)

On differentiating we get

y’ – sin y.y’ = 1

y'(1 + sin y) = 1

          -(2)

Now substitute the values from equation(1) and (2), in the differential equation 

(y sin y + cos y + x)y’ = y

y = y

Hence verified

Question 9. x + y = tan^-1y          : y²y’ + y² + 1 = 0

Solution:

Given: x + y = tan^-1y

On differentiating we get

1 + y’ = 

          -(1)

Now substitute the values from equation(1), in the differential equation 

y²y’ + y² + 1 = 0

0 = 0

Hence verified

Question 10.         : 

Solution:

Given: 

We can also write as 

y² = a² – x²

Now on differentiating we get

2yy’ = -2x

y’ = -2x/2y

y’ = -x/y         -(1)

Now substitute the values from equation(1), in the differential equation 

x + y.

x + y (-x/y) = 0

x – x = 0

0 = 0

Hence verified

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

Solution:

(D) is correct answer because 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

Solution:

(D) is the correct answer because there are no arbitrary constants in a particular solution.