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

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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Class 12 NCERT Solutions- Mathematics Part II - Chapter 9 Differential Equations-Exercise -9.1

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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 = ex + 1 : y” – y’ = 0

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

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

Question 4. :

Question 5. y = Ax : xy’ = y (x ≠ 0)

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

Question 7. xy = log y + C : y’ =

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

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

Question 10. :

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

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

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Question 1. y = e^x+ 1 : y” – y’ = 0

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

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

Question 6. y = x sin x : xy’ = y + x $\sqrt{x^2-y^2}$ (x ≠ 0 and x > y or x < -y)

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

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

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