Question

The solution of (y + x + 5) dy = (y – x + 1) dx is –

• A. log ((y + 3)² + (x + 2)²) + tan^–1$\frac{y + 3}{x + 2}$ = C
• B. log ((y + 3)² + (x – 2)²) + tan^–1$\frac{y–3}{x–2}$ = C
• C. log ((y + 3)² + (x + 2)²) + 2 tan^–1$\frac{y + 3}{x + 2}$ = C
• D. log ((y + 3)² + (x + 2)²) – 2 tan^–1$\frac{y + 3}{x + 2}$ = C

Answer 1

Answer: (C)

log ((y + 3)² + (x + 2)²) + 2 tan^–1$\frac{y + 3}{x + 2}$ = C

Answer 2

The intersection of y – x + 1 = 0 and

y + x + 5 = 0 is (–2, – 3). Put x = X – 2,

y = Y – 3. The given equation reduces to

$\frac{dY}{dX}$ = $\frac{Y - X}{Y + X}$. This is a homogeneous equation, so putting, so putting Y = vX, we get

X = – $\frac{v^{2} + 1}{v + 1}$ Ž $\left( \frac{- v}{v^{2} + 1} - \frac{1}{v^{2} + 1} \right)$ dv = $\frac{dX}{X}$

Ž – $\frac{1}{2}$log (v² + 1) – tan^–1 v = log |X| + Const

Ž log (Y² + X²) + 2 tan^–1 $\frac{Y}{X}$ = Const

Ž log ((y + 3)² + (x + 2)²) + 2 tan^–1 $\frac{y + 3}{x + 2}$ = C