Question

$xy\frac{dy}{dx}=x^2+2y^2;y\left(1\right)=0$

Answer 1

Solution -

$xy\frac{dy}{dx}=x^2+2y^2$

$\frac{dy}{dx}=\frac{x}{y}+\frac{2y}{x}$

Let $y=vx$

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

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

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

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

put $v^2=t$ $2vdv=dt$

$\frac{1}{2}\frac{dt}{t+1}=lnx+c$

$\frac{1}{2}ln|v^2+1|=lnx+c$

Put $v=y/x$

$\frac{1}{2}ln\left|\frac{y^2+x^2}{x^2}\right|=ln\left|x\right|+c$

$\frac{1}{2}ln{|y^2+x|}^2-ln\left|x\right|=ln\left|x\right|+c$

$y\left(1\right)=0$

$c=0$

$ln\left|\frac{y^2+x^2}{x^4}\right|=0$

$y^2+x^2=x^4$

$y=\sqrt{x^4-x^2}$