Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.2

In each of the questions 1 to 10, verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

Question 1.

y = e^{x} + 1 : y” – y’ = 0

Solution:

y = e^{x} + 1

Differentiating w.r.t. x,

y’ = e^{x} + 0 ……………. (1)

Again differentiating,

y” = e^{x} ……………….. (2)

Subtracting (1) from (2),

y” – y’ = e^{x} – e^{x} = 0.

∴ Required differential equation is y” – y’ = 0.

Question 2.

y = x^{2} + 2x + C : y’ – 2x – 2 = 0

Solution:

y = x^{2} + 2x + C

Differentiating w.r.t.x,

y’ = 2x + 2

⇒ y’ – 2x – 2 = 0

∴ Required differential equation is y’ – 2x – 2 = 0.

Question 3.

y = cos x + C : y’ + sin x = 0

Solution:

y = cos x + C

Differentiating w.r.t. x,

y’ = – sin x or y + sin x = 0

∴ Required differential equation is y’ + sin x = 0

Question 4.

y = \(\sqrt{1+x^{2}}\) : y’ = \(\frac{x y}{\sqrt{1+x^{2}}}\)

Solution:

Hence, the required differential equation is y’ = \(\frac{x y}{1+x^{2}}\).

Question 5.

y = A x : xy’ = y, (x ≠ 0)

Solution:

y = Ax ………….. (1)

Differentiating w.r.t. x,

y’ = A …………….. (2)

Eliminating A from equations (1) and (2),

y = y’x

∴ The required differential equation is xy’ = y. (x ≠ 0)

Question 6.

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

Solution:

y = x sin x …………….. (1)

Differentiating w.r.t. x,

y’ = 1.sin x + x cos x

= sin x + x cos x [∵ sin^{2}x + cos^{2}x = 1] ……………… (2)

From (1), sin x = \(\frac{y}{x}\)

Putting this value sin x in (2), we get

Multiplying by x, we get

xy’ = y + x\(\sqrt{x^{2}-y^{2}}\)

Which is the required differential equation.

Question 7.

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

Solution:

xy = log x + C

Differentiating w.r.t. x,

1.y + xy’ = \(\frac{1}{y}\)y’

or y^{2} + xy y’ = y’

or y^{2} = y’ – xy y’ = y'(1 – xy)

y’ = \(\frac{y^{2}}{1-x y}\).

which is the required differential equation.

Question 8.

y – cos y = x : (y sin y + cos y + x)y’ = y

Solution:

y – cos y = x ……………… (1)

Differentiating w.r.t. x,

(1 + sin y)y’ = x

Multiplying both sides by y,

(y + sin y)y’ = y ………………… (2)

From (1), y = x + cos y.

Putting it in (2), we get

(x + cos y + y sin y)y’ = y

or (y sin y + cos y + x)y’ = y,

which is the required differential equation.

Question 9.

x + y = tan^{-1}y : y^{2}y’ + y^{2} + 1 = 0

Solution:

x + y = tan^{-1}y

Differentiating w.r.t. x,

1 + y’ = \(\frac{y^{\prime}}{1+y^{2}}\)

(1 + y^{2}) + y'(1 + y^{2}) = y’

or 1 + y^{2} + y’ (1 + y^{2} – 1) = 0

or y^{2}y’ + y^{2} + 1 = 0,

which is the required differential equation.

Question 10.

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

Solution:

y = \(\sqrt{a^{2}-x^{2}}\)

Squaring both sides, y^{2} = a^{2} – x^{2}

or x^{2} + y^{2} = a^{2}

Differentiating w.r.t. x,

2x + 2yy’ = 0

Dividing by 2, x + y \(\frac{dy}{dx}\) = 0,

which is the required differential equation.

Choose the correct answers in the following questions 11 and 12:

Question 11.

The number of arbitarary constants in the general solution of a differential equation of fourth order is

(A) 0

(B) 2

(C) 3

(D) 4

Solution:

Number of constants in general solution = order of the differential equation.

The differential equation is of fourth order.

∴ Number of arbitarary constants = 4

∴ Part (D) is the correct answer.

Question 12.

The number of arbitarary constants in the particular solution of a differential equation of third order is

(A) 3

(B) 2

(C) 1

(D) 0

Solution:

In a particular solution of a differential equation, there is no arbitarary constant.

∴ Part (D) is the correct answer.