Question

General solution of $\left(x+y\right)^{2} \frac{dy}{dx} = a^{2}, a \ne 0$ is (c is an arbitrary constant)

• A. $\frac{x}{a} = tan \frac{y}{a} + c$
• B. $tanxy = c$
• C. $tan \left(x + y\right) = c$
• D. $tan \frac{y+c}{a} = \frac{x+y}{a}$

Answer 1

Answer: (D)

$tan \frac{y+c}{a} = \frac{x+y}{a}$

Answer 2

$x + y = z, \Rightarrow \frac{dy}{dx} = \frac{dz}{dx} - 1$
$\Rightarrow\, \frac{z^{2}dz}{a^{2}+z^{2}} = dx$
Integrating
$\Rightarrow\,x+y-a\,tan^{-1} \frac{x+y}{a} = x+c, \quad\Rightarrow\,tan \left(\frac{y-c_{1}}{a}\right) = \frac{x+y}{a}$
$\Rightarrow\,tan\left(\frac{y+c}{a}\right) = \frac{x+y}{a}\quad\quad\left(c_{1} = -c\right)$