Question

The general solution of differential equation $\frac{dy}{dx}=\frac{x+y}{x-y}$is

Answer 1

$\frac{dy}{dx}=\frac{x+y}{x-y}$

Put $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}=\frac{1+v}{1-v}$

$x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v}$

$\frac{1-v}{1+v^2}dv=\frac{dx}{x}$

$\frac{v-1}{1+v^2}dv=\frac{-dx}{x}$

By integrating, we get

$\log (1+v^2)-{\tan }^{-1}v=-\log x-\log C$

$\log \left(Cx\right(1+v^2\left)\right)={\tan }^{-1}v$

$e^{{\tan }^{-1}\left(\frac{y}{x}\right)}=Cx(1+(\frac{y}{x}{)}^2)$

$x{e}^{{\tan }^{-1}\left(\frac{y}{x}\right)}=C(x^2+y^2)$