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

### Question 1. y = ex + 1          : y” – y’ = 0

Solution:

Given: y = ex + 1

On differentiating we get

y’ = ex          -(1)

Again differentiating we get

y” = ex           -(2)

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

y” – y’ = ex – ex = 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

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

### (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.

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

Solution:

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

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