GSEB Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.5

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.
(x² + x dy) = (x² + y²) dx
Solution:
(x² + xy) dy = (x² + y²)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 λx and y by λy, 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

∴ Solution is (x – y)² = Cxe^-y/x.

Question 2.
y’ = $\frac{x+y}{x}$
Solution:
y’ = $\frac{x+y}{x}$
∴ $\frac{dy}{dx}$ = f(x, y), where f(x, y) = $\frac{x+y}{x}$.
Replacing x by λx and y by λy,

∴ 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)
∴ f(x, y) = $\frac{x+y}{x-y}$
Replacing x by λx and y by λy,

⇒ f(x, y) is a homogeneous function of degree zero.
Put y = vx so that

Integrating, we get

Question 4.
(x² – y²)dx + 2xy dy = 0
Solution:

Replacing x by λx and y by λy, we get

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

Integrating, we get

Question 5.
x² $\frac{dy}{dx}$ = x² – 2y² + xy
Solution:

Replacing x by λx and y by λy in f(x, y), we get

∴ 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 λx and y by λy in f(x, y), we get

∴ 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 λx and y by λy in f(x, y), we get

∴ 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 λx and y by λy in f(x, y), we get

∴ 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 λx and y by λy 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 + ∫ $\frac{dt}{t}$ = log x + log C
⇒ – 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 λx and y by λy and obtain:

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

Integrating, we get

∴ 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
∴ $\frac{dy}{dx}$ = – $\frac{x-y}{x+y}$ = f(x, y).
f(x, y) is a homogeneous function.

Put v² + 1 = t so that 2v dv = dt
or $\frac{1}{2}$ ∫$\frac{dt}{t}$ + tan^-1 v = – log x + C
or $\frac{1}{2}$log t + tan^-1v = – log x + C
or $\frac{1}{2}$log(v² + 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 + $\frac{π}{4}$ = C
Putting value of C in (1). The particular solution is
$\frac{1}{2}$log(x² + y²) + tan^-1($\frac{y}{x}$)
= $\frac{1}{2}$ log 2 + $\frac{π}{4}$.

Question 12.
x²dy + (xy + y²)dx = 0, y = 1, when x = 1.
Solution:
x²dy + (xy + y²) dx = 0
∴ $\frac{dy}{dx}$ = $\frac{x y+y^{2}}{x^{2}}$ = f(x, y)
f(x, y) is homogeneous
∴ 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 = C²(1 + 2) ∴ C² = $\frac{1}{3}$.
Putting C² = $\frac{1}{3}$ in (1), we get
x²y = $\frac{1}{3}$(y + 2x)
∴ Particular solution is 3x²y = y + 2x.

Question 13.
(x sin² $\frac{y}{x}$ – y) dx + x dy = 0, y = $\frac{π}{4}$, when x = 1.
Solution:
[x sin² $\frac{y}{x}$ – y] dx + x dy = 0

which is homogeneous

⇒ – cot v = – log x + C
⇒ + 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 = $\frac{π}{4}$, we get
log 1 – cot $\frac{π}{4}$ = C or 0 – 1 = C ⇒ C = – 1.
∴ 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
∫(- sin v)dv = ∫$\frac{dx}{x}$
⇒ cos v = log |x| + C
⇒ cos $\frac{y}{x}$ = log|x| + C
It is given that y(1) = 0, i.e; when x = 1, y = 0.
∴ cos 0 = log|1| + C
⇒ 1 = 0 + C ⇒ C = 1.
∴ cos $\frac{y}{x}$ = log |x| + 1
⇒ log |x| = cos $\frac{y}{x}$ – 1, (x ≠ 0)
which is the required particular solution.

Question 15.
2 xy + y² – 2x² $\frac{dy}{dx}$ = 0, y = 2, when x = 1.
Solution:
We have: 2xy + y² – 2x² $\frac{dy}{dx}$ = 0

which is a homogeneous differential equation.
Put y = vx. Then $\frac{dy}{dx}$ = v + x $\frac{dv}{dx}$.
∴ (1) becomes v + x $\frac{dv}{dx}$ = v + $\frac{1}{2}$v²
⇒ x $\frac{dv}{dx}$ = $\frac{1}{2}$ v² ⇒ $\frac{2}{v^{2}}$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.
∴ 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 – (x³ + y³)dy = 0
(C) (x³ + 2y²) dx + 2xy dy = 0
(D) y²dx + (x² – xy – y²) dy = 0
Solution:
Consider the differential equation

Replacing x by λx by λy, we get

∴ f(x, y) is the homogeneous function of degree zero.
Hence, part (D) is the correct answer.