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Class 12 NCERT Solutions- Mathematics Part II – Chapter 9 Differential Equations-Exercise -9.2
  • Last Updated : 12 Mar, 2021

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 = 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

Question 4. y=\sqrt{1+x^{2}}          : y'=\frac{xy}{1+x^{2}}

Solution:

Given: y=\sqrt{1+x^{2}}  

y'=\frac{1}{2\sqrt{1+x^2}}.2x=\frac{x}{\sqrt{1+x^2}}=\frac{x.\sqrt{1+x^2}}{1+x^2}=\frac{xy}{1+x^2}

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

\frac{y}{A}A = y

y = y

Hence verified

Question 6. y = x sin x          : xy’ = y + x\sqrt{x^2-y^2} (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\sqrt{1 - (\frac{y}{x})^2}

= y + x2\frac{\sqrt{x^2 - y^2}}{x}

= y + x\frac{\sqrt{x^2 - y^2}}{x}

LHS = RHS

Hence verified

Question 7. xy = log y + C          : y’ = \frac{y^{2}}{1-xy}(xy≠1)

Solution:

Given: xy = logy + C          -(1)

On differentiating we get

xy’ + y = \frac{1}{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’ = \frac{y^{2}}{1-xy}

\frac{-y^2}{-(1 - xy)} = \frac{y^{2}}{1-xy}

\frac{y^2}{1 - xy} = \frac{y^{2}}{1-xy}

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

y' = \frac{1}{(1 + sin y)}          -(2)

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

(y sin y + cos y + x)y’ = y

(y sin y + cos y + y - cosy)\frac{1}{1+siny} = y

(y sin y + y)\frac{1}{1+siny} = y

y(sin y + 1)\frac{1}{1+siny} = 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’ = \frac{1}{1+y^2}y'

y'[\frac{1}{1+y^2}-1]=1

y'[\frac{1-1+y^2}{1+y^2}]=1

y'[\frac{-y^2}{1+y^2}]=1

y'=\frac{1}{\frac{-y^2}{1+y^2}}

y'=\frac{-(1+y^2)}{y^2}          -(1)

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

y2y’ + y2 + 1 = 0

y^2[\frac{-(1+y^2)}{y^2}] + y^2 + 1 = 0

-1-y^2 + y^2 + 1 = 0

0 = 0

Hence verified

Question 10. y=\sqrt{a^2-x^2}x∈(-a,a)         : x+y\frac{dy}{dx}=0(y≠0)

Solution:

Given: y=\sqrt{a^2-x^2}

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.\frac{dy}{dx}=0

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.

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