Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.5

In each of the questions 1 to 10, show that the given differential equation is homogeneous and solve each of them:

Question 1.

(x^{2} + x dy) = (x^{2} + y^{2}) dx

Solution:

(x^{2} + xy) dy = (x^{2} + y^{2})dx

or \(\frac{dy}{dx}\) = \(\frac{x^{2}+y^{2}}{x^{2}+x y}\) = f(x, y) (say) …………….. (1)

where f(x, y) = \(\frac{x^{2}+y^{2}}{x^{2}+x y}\)

Replacing by Î»x and y by Î»y, we get

Hence, f(x, y) is a homogeneous function of degree zer0.

To solve it, put y = vx or

\(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\)

Equation (1) may be written as

Integrating, we get

Put v = \(\frac{y}{x}\), we get

âˆ´ Solution is (x – y)^{2} = Cxe^{-y/x}.

Question 2.

y’ = \(\frac{x+y}{x}\)

Solution:

y’ = \(\frac{x+y}{x}\)

âˆ´ \(\frac{dy}{dx}\) = f(x, y), where f(x, y) = \(\frac{x+y}{x}\).

Replacing x by Î»x and y by Î»y,

âˆ´ f(x, y) is a homogeneous function of degree zero.

Putting v = \(\frac{y}{x}\), we get \(\frac{y}{x}\) = log |x| + C or y = x log |x| + Cx.

Question 3.

(x – y)dy – (x + y)dx = 0

Solution:

(x – y) dy – (x + y)dx = 0

or \(\frac{dy}{dx}\) = \(\frac{x+y}{x-y}\) = f(x, y)

âˆ´ f(x, y) = \(\frac{x+y}{x-y}\)

Replacing x by Î»x and y by Î»y,

â‡’ f(x, y) is a homogeneous function of degree zero.

Put y = vx so that

Integrating, we get

Question 4.

(x^{2} – y^{2})dx + 2xy dy = 0

Solution:

Replacing x by Î»x and y by Î»y, we get

Hence, f(x, y) is a homogeneous functoin of degree zero.

Integrating, we get

Question 5.

x^{2} \(\frac{dy}{dx}\) = x^{2} – 2y^{2} + xy

Solution:

Replacing x by Î»x and y by Î»y in f(x, y), we get

âˆ´ f(x, y) is a homogenous function of degree zero.

Putting these values in (1), we get

Integrating, we have:

Put \(\sqrt{2}\)v = t so that \(\sqrt{2}\)dv = dt.

Put v = \(\frac{y}{x}\), we get the solution as

Question 6.

x dy – y dx = \(\sqrt{x^{2}+y^{2}}\) dx

Solution:

Replacing x by Î»x and y by Î»y in f(x, y), we get

âˆ´ f(x, y) is a homogenous function of degree zero.

Putting these values in (1), we get

Integrating, we have:

Question 7.

{x cos (\(\frac{y}{x}\)) + y sin (\(\frac{y}{x}\))} y dx = {y sin (\(\frac{y}{x}\)) – x cos (\(\frac{y}{x}\))} x dy

Solution:

Replacing x by Î»x and y by Î»y in f(x, y), we get

âˆ´ f(x, y) is a homogenous function of degree zero.

Putting in these values in (1), we get

Transposing v to R.H.S., we get

Integrating, we get

Question 8.

x \(\frac{dy}{dx}\) – y + x sin \(\frac{y}{x}\) = 0

Solution:

Replacing x by Î»x and y by Î»y in f(x, y), we get

âˆ´ f(x, y) is a homogenous function of degree zero.

Question 9.

y dx + xlog (\(\frac{y}{x}\))dy – 2x dy = 0

Solution:

Replacing x by Î»x and y by Î»y in f(x, y), we get

Put y = vx, so that

\(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\).

Putting these values in (2), we get

Integrating, we get

Put log v – 1 = t, so that \(\frac{1}{v}\)dv = dt

Therefore, from (3), we get

– log v + âˆ« \(\frac{dt}{t}\) = log x + log C

â‡’ – log v + log t = log x + log C

or – log v + log (log v – 1) = log x + log C

or log (log v – 1) = log v + log x + log C

= log v Cx

Put v = \(\frac{y}{x}\). Then the solution is

Question 10.

(1 + e^{x/y})dx + e^{x/y}(1 – \(\frac{x}{y}\))dy = 0

Solution:

Replacing x by Î»x and y by Î»y and obtain:

Hence, f(x, y) is a homogeneous function of degree zero.

Integrating, we get

âˆ´ Required solution is x + y e^{x/y} = Cy.

For each of the differential equations in questions 11 to 15, find the particular solution satisfying the given condition:

Question 11.

(x + y)dy + (x – y)dx = 0, y = 1, when x = 1.

Solution:

(x + y)dy + (x – y)dx

or (x + y) dy = – (x – y) dx

âˆ´ \(\frac{dy}{dx}\) = – \(\frac{x-y}{x+y}\) = f(x, y).

f(x, y) is a homogeneous function.

Put v^{2} + 1 = t so that 2v dv = dt

or \(\frac{1}{2}\) âˆ«\(\frac{dt}{t}\) + tan^{-1} v = – log x + C

or \(\frac{1}{2}\)log t + tan^{-1}v = – log x + C

or \(\frac{1}{2}\)log(v^{2} + 1) + tan^{-1} v = – log x + C

Put v = \(\frac{y}{x}\) and obtain:

Put x = 1, y = 1 and obtain:

\(\frac{1}{2}\)log(1 + 1) + tan^{-1}\(\frac{1}{1}\) = C

or \(\frac{1}{2}\)log 2 + \(\frac{Ï€}{4}\) = C

Putting value of C in (1). The particular solution is

\(\frac{1}{2}\)log(x^{2} + y^{2}) + tan^{-1}(\(\frac{y}{x}\))

= \(\frac{1}{2}\) log 2 + \(\frac{Ï€}{4}\).

Question 12.

x^{2}dy + (xy + y^{2})dx = 0, y = 1, when x = 1.

Solution:

x^{2}dy + (xy + y^{2}) dx = 0

âˆ´ \(\frac{dy}{dx}\) = \(\frac{x y+y^{2}}{x^{2}}\) = f(x, y)

f(x, y) is homogeneous

âˆ´ Put y = vx, so that \(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\).

Integrating, we get

Putting v = \(\frac{y}{x}\), we get

Putting x = 1, y = 1, we get

1 = C^{2}(1 + 2) âˆ´ C^{2} = \(\frac{1}{3}\).

Putting C^{2} = \(\frac{1}{3}\) in (1), we get

x^{2}y = \(\frac{1}{3}\)(y + 2x)

âˆ´ Particular solution is 3x^{2}y = y + 2x.

Question 13.

(x sin^{2} \(\frac{y}{x}\) – y) dx + x dy = 0, y = \(\frac{Ï€}{4}\), when x = 1.

Solution:

[x sin^{2} \(\frac{y}{x}\) – y] dx + x dy = 0

which is homogeneous

â‡’ – cot v = – log x + C

â‡’ + log x – cot v = C

Putting v = \(\frac{y}{x}\), we get the general solution as

log x – cot \(\frac{y}{x}\) = C.

Putting x = 1 and y = \(\frac{Ï€}{4}\), we get

log 1 – cot \(\frac{Ï€}{4}\) = C or 0 – 1 = C â‡’ C = – 1.

âˆ´ Particular solution is

log x – cot \(\frac{y}{x}\) = – 1 or cot \(\frac{y}{x}\) – log x = 1.

Question 14.

\(\frac{dy}{dx}\) – \(\frac{y}{x}\) + cosec (\(\frac{y}{x}\)) = 0, y = 0, when x = 1.

Solution:

We have:

\(\frac{dy}{dx}\) – \(\frac{y}{x}\) + cosec \(\frac{y}{x}\) = 0 ……………. (1)

Put y = vx, so that \(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\). ………………….. (2)

From (1) and (2), we get

Integrating both sides, we get

âˆ«(- sin v)dv = âˆ«\(\frac{dx}{x}\)

â‡’ cos v = log |x| + C

â‡’ cos \(\frac{y}{x}\) = log|x| + C

It is given that y(1) = 0, i.e; when x = 1, y = 0.

âˆ´ cos 0 = log|1| + C

â‡’ 1 = 0 + C â‡’ C = 1.

âˆ´ cos \(\frac{y}{x}\) = log |x| + 1

â‡’ log |x| = cos \(\frac{y}{x}\) – 1, (x â‰ 0)

which is the required particular solution.

Question 15.

2 xy + y^{2} – 2x^{2} \(\frac{dy}{dx}\) = 0, y = 2, when x = 1.

Solution:

We have: 2xy + y^{2} – 2x^{2} \(\frac{dy}{dx}\) = 0

which is a homogeneous differential equation.

Put y = vx. Then \(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\).

âˆ´ (1) becomes v + x \(\frac{dv}{dx}\) = v + \(\frac{1}{2}\)v^{2}

â‡’ x \(\frac{dv}{dx}\) = \(\frac{1}{2}\) v^{2} â‡’ \(\frac{2}{v^{2}}\)dv = \(\frac{dx}{x}\).

Integrating both sides, we get

It is given that y(1) = 2, i.e., when x = 1, y = 2.

which is the required particular solution.

Choose the correct answers in the following questions 16 and 17:

Question 16.

A homogeneous differential equation of the form \(\frac{dx}{dy}\) = h(\(\frac{x}{y}\)) can be solved by making the substitution.

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

Solution:

For solving the homogeneous equation of the form \(\frac{dx}{dy}\) = h(\(\frac{x}{y}\)), we make the substitution x = vy.

âˆ´ Part (C) is the correct answer.

Question 17.

Which of the following is a homogeneous differential equation?

(A) (4x + 6y + 5)dy – (3y + 2x + 4) dx = 0

(B) xy dx – (x^{3} + y^{3})dy = 0

(C) (x^{3} + 2y^{2}) dx + 2xy dy = 0

(D) y^{2}dx + (x^{2} – xy – y^{2}) dy = 0

Solution:

Consider the differential equation

Replacing x by Î»x by Î»y, we get

âˆ´ f(x, y) is the homogeneous function of degree zero.

Hence, part (D) is the correct answer.