Question

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

• A. $kx=e^{-y/x}$
• B. $kx=e^{y/x}$
• C. $ky=e^{x/y}$
• D. $ky=e^{-x/y}$

Answer 1

The correct option is A $kx=e^{-y/x}$

$y-x\frac{dy}{dx}=x+y\frac{dy}{dx}$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore -x(x\frac{dv}{dx}+v)+xv=x+x(x\frac{dv}{dx}+v)v$
$\Rightarrow \frac{dv}{dx}=\frac{-v^2-1}{x(v+1)}\Rightarrow \frac{v+1}{-v^2-1}\frac{dv}{dx}=\frac{1}{x}$
Integrating both sides w.r.t $x$ we get
$\int \frac{v+1}{-v^2-1}\frac{dv}{dx}dx=\int \frac{1}{x}dx$
$\Rightarrow {\tan }^{-1}v-\frac{1}{2}\log (v^2+1)=\log x+c\Rightarrow kx=e^{-\frac{y}{x}}$