Question

Solve :

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

Answer 1

Let $x+y=y=v$. Then,

$1+\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{dy}{dx}=\frac{dv}{dx}-1$

Therefore,

$v^2(\frac{dv}{dx}-1)=a^2$

$v^2\frac{dv}{dx}=a^2+v^2$

$v^{2}dv=(a^2+v^2)dx$

$\frac{v^2}{v^2+a^2}dv=dx$

$(1-\frac{a^2}{v^2+a^2})dv=dx$

$\int dv-a^2\int \frac{1}{v^2+a^2}dv=\int dx+C$

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

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