Question

$(x^2+y^2)\frac{dy}{dx}=xy.$ Solving this we get $ky=e^{\frac{x^r}{m{y}^{k+1}}}$.Find $r+m+k$ ?

Answer 1

$\frac{dy}{dx}=\frac{xy}{x^2+y^2}$
Put $y=vx...\left(1\right)\therefore \frac{dy}{dx}=v+x\frac{dv}{dx}...\left(2\right)$
$\therefore v+x\frac{dv}{dx}=\frac{x.vx}{x^2+v^2{x}^2}=\frac{v}{1+v^2}...\left(3\right)$
$x\frac{dv}{dx}=\frac{v}{1+v^2}-v,...\left(4\right)$
$x\frac{dv}{dx}=\frac{v-v-v^3}{1+v^2}=\frac{-v^3}{1+v^2}\therefore \left(\frac{1+v^2}{-v^3}\right)dv=\frac{dx}{x}$
or $(\frac{-1}{v^3}-\frac{1}{v})dv=\frac{dx}{x}$
Integrate both sides, $\frac{1}{2v^2}-\log v=\log x+c,$
or $\frac{1}{2v^2}=\log vxk,$ where $c=\log k$
or
$\frac{1}{2}\frac{x^2}{y^2}=\log \frac{y}{x}x.k=\log ky.\therefore e^{x^2/2y^2}=ky$