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 puttig Y = uX, we get

X$\frac{d\upsilon}{dX}$ = $- \frac{\upsilon^{2} + 1}{\upsilon + 1}$

Ž $\left( - \frac{\upsilon}{\upsilon^{2} + 1} - \frac{1}{\upsilon^{2} + 1} \right)$du = $\frac{dX}{X}$

Ž –$\frac{1}{2}$log (u² + 1) – tan^–1 u = log | X | + C

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

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