Question

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

Answer 1

Let $z=x+y$

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

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

Now,

$z^2(\frac{dz}{dx}-1)=a^2$

$z^2\frac{dz}{dx}-z^2=a^2$

$z^2\frac{dz}{dx}=z^2+a^2$

$\frac{z^{2}dz}{(z^2+a^2)}=dx$

On integration

$\int \frac{z^{2}dz}{(z^2+a^2)}=\int dx...........\left(1\right)$

$\int \frac{z^{2}dz}{(z^2+a^2)}=\int \frac{(z^2+a^2-a^2)dz}{(z^2+a^2)}$

$=\int dz-a^2\frac{1}{2}{\tan }^{-1}\frac{z}{a}$

$=z-a{\tan }^{-1}\frac{z}{a}$

Now, (1) become

$z-a{\tan }^{-1}\frac{z}{a}=x+c$

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

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