Question

Solve (x + y)²$\frac{dy}{dx}$ = a²

• A. y = a tan^–1$\left( \frac{x + y}{a} \right)$ + C
• B. y = a tan^–1$\left( \frac{x + y}{a} \right)$ – C
• C. y = a tan^–1$\left( \frac{x–y}{a} \right)$ + C
• D. None

Answer 1

Answer: (B)

y = a tan^–1$\left( \frac{x + y}{a} \right)$ – C

Answer 2

(x + y)² $\frac{dy}{dx}$ = a² ... (i)

Put x + y = t Ž 1 + $\frac{dy}{dx}$ = $\frac{dt}{dx}$

Ž $\frac{dy}{dx}$ = $\left( \frac{dt}{dx} - 1 \right)$

\Equation (i) reduces to,

t² $\left\{ \frac{dt}{dx} - 1 \right\}$ = a² Ž t² $\frac{dt}{dx}$ = a² + t²,

Separating the variable and integrating.

$\int_{}^{}{dx}$ = $\int_{}^{}\frac{t^{2}}{a^{2} + t^{2}}$dt = $\int_{}^{}\left( 1 - \frac{a^{2}}{a^{2} + t^{2}} \right)$dt

\ x = t – a tan^–1 $\left( \frac{t}{a} \right)$ + c

Ž x = x + y – a tan^–1 $\left( \frac{x + y}{a} \right)$ + c

Ž y = a tan^–1 $\left( \frac{x + y}{a} \right)$ – c,

Which is the required general solution.