Question

Solve the differential equation : $y+x\frac{dy}{dx}=x-y\frac{dy}{dx}$

Answer 1

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

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

$\frac{dy}{dx}=\frac{x-y}{x+y}$ ………..$\left(1\right)$

Let $y=vx$

on differentiating we get $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now from equation $\left(1\right)$ we get

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

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

$v+v^2-1+v$

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

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

On integrating both side we get

$\int \frac{1}{(V+1{)}^2-(\sqrt{2}{)}^2}dv=\int \frac{-1}{x}dx$

$\frac{1}{2\sqrt{2}}ln\left|\frac{v+1-\sqrt{2}}{v+1+\sqrt{2}}\right|=-lnx+c$

$\frac{1}{2\sqrt{2}}ln\left|\frac{y+x(1-\sqrt{2})}{y+x(1+\sqrt{2})}\right|=-lnx+c$.