Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c)

I.

Question 1.

Express x dy – ydx = \(\sqrt{x^2+y^2}\) dx in the form F (\(\frac{y}{x}\)) = (\(\frac{dy}{dx}\)).

Solution:

x. dy – y dx = \(\sqrt{x^2+y^2}\) dx

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

Question 2.

Express (x – y Tan^{-1} \(\frac{y}{x}\))dx + x Tan^{-1}\(\frac{y}{x}\) dy = 0 in the form F(\(\frac{y}{x}\)) = \(\frac{dy}{dx}\).

Solution:

Question 3.

Express x\(\frac{dy}{dx}\) = y(log y – log x + 1) in the from F (\(\frac{y}{x}\)) = \(\frac{dy}{dx}\)

Solution:

x . \(\frac{dy}{dx}\) = y(log y – log x + 1)

\(\frac{dy}{dx}\) = \(\frac{y}{x}\)(log\(\frac{y}{x}\) + 1)

II. Solve the following differential equations.

Question 1.

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

Solution:

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

Put y = vx

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

Question 2.

(x² + y²) dy = 2xy dx

Solution:

(x² + y²) dy = 2xy dx

1 + v² = A(1 – v²) + Bv(1 – v) + Cv(1 + v)

v = 0 ⇒ = A

v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1

v = -1 ⇒ 1 + 1 = B(-1) (2) ⇒ 2 = -2B

B = -1

Question 3.

\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)

Solution:

\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)

Put y = vx

Multiplying with (v + 1)³

3 + v² = A(v + 1)² + B(v + 1) + C

v = – 1 ⇒ 3 + 1 = C ⇒ C = 4

Equating the co-efficients of v²

A = 1

Equating the co-efficients of v

0 = 2A + B

B = -2A = -2

Question 4.

y²dx + (x² – xy)dy = 0

Solution:

v – log v = log x + log k

v = log v + log x + log k

= log k (vx)

\(\frac{y}{x}\) = log ky

Solution is ky = e^{y/x}

Question 5.

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

Solution:

Question 6.

(x² – y²)dx – xy dy = 0

Solution:

(x² – y²)dx – xy dy = 0

(x² – y²)dx = xy . dy

Question 7.

(x²y – 2xy²)dx = (x³ – 3x²y)dy

Solution:

(x²y – 2xy²)dx = (x³ – 3x²y)dy

Question 8.

y²dx + (x² – xy + y²) dy = 0

Solution:

y²dx = – (x² – xy + y²)dy

1 – v + v² = A(1 + v²) + (Bv + C)v

v = 0 ⇒ 1 = A

Equating the co-efficients of v²

1 = A + .B ⇒ B = 0

Equating the co-efficients of v

-1 = C

Question 9.

(y² – 2xy)dx + (2xy – x²)dy = 0

Solution:

(y² – 2xy)dx + (2xy – x²)dy = 0

(2xy – x²)dy = – (y² – 2xy)dx

2v – 1 = A(1 – v) + Bv

v = 0 ⇒ -1 = A ⇒ A = -1

v = 1 ⇒ 1 = B ⇒ B = 1

Question 10.

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

Solution:

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

Question 11.

x dy – y dx = \(\sqrt{x^2+y^2}\)

Solution:

x dy – y dx = \(\sqrt{x^2+y^2}\)

Question 12.

(2x – y)dy = (2y – x)dx

Solution:

⇒ (y – x)² = c²(y + x)²(y² – x²)

⇒ y – x = c²(y + x)³

⇒ (x + y)³ = c(x – y) where c = \(\frac{-1}{c^2}\)

∴ (x + y)³ = c(x – y)

Question 13.

(x² – y²)\(\frac{dy}{dx}\) = xy

Solution:

Question 14.

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

Solution:

Put y = vx

(y – x)² = y²c²x

Solution is y²x = c(x – y)²

III.

Question 1.

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

Solution:

Put x = vy

Question 2.

Solve : x sin \(\frac{y}{x}.\frac{dy}{dx}\) = y sin \(\frac{y}{x}\) – x

Solution:

Dividing with x, we have

Question 3.

Solve : x dy = (y + x cos² \(\frac{y}{x}\))dx

Solution:

x.\(\frac{dy}{dx}\) = y + xcos² \(\frac{y}{x}\)

Question 4.

Solve : (x – y log y + y log x)dx + x(log y – log x)dy = 0

Solution:

Dividing with x. dx we get

Question 5.

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

Solution:

The given equation can be written as

∴ This is a homogeneous equation

Question 6.

Find the equation of a curve whose gradient is \(\frac{dy}{dx}=\frac{y}{x}-\cos^2\frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).

Solution:

tan v = – log |x| + c

This curve passes through (1, \(\frac{\pi}{4}\))

tan \(\frac{\pi}{4}\) = c – log 1

c = 1

∴ Equation of the curve is

tan v = 1 – log |x|

tan \(\frac{y}{x}\) = 1 – log |x|