Question

Find the particular solution of the differential equation $x^{2}dy=(2xy+y^2)dx$, given that $y=1$ when $x=1$.

Answer 1

$x^{2}dy=(2xy+y^2)dx$
$\frac{dy}{dx}=\frac{2xy+y^2}{x^2}$
Put $y=\nu x$
$\frac{d}{dx}\left(\nu x\right)=\frac{2x\left(\nu x\right)+{\left(\nu x\right)}^2}{x^2}$
$x\frac{d\nu }{dx}+\nu =2\nu +{\nu }^2$
$x\frac{d\nu }{dx}+\nu =\nu +{\nu }^2$
$\frac{d\nu }{{\nu }^2+\nu }=\frac{dx}{x}$
Intregrate both sides
$\int \frac{d\nu }{({\nu }^2+\nu )}=\int \frac{dx}{x}$
$\int \frac{d\nu }{{(\nu +\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}=lnx+C$
$\frac{1}{2\times \frac{1}{2}}ln\left|\frac{\nu +\frac{1}{2}-\frac{1}{2}}{\nu ++\frac{1}{2}+\frac{1}{2}}\right|=lnx+C$
$ln\left|\frac{\nu }{\nu +1}\right|=lnx+C$
$ln\left|\frac{y/x}{y/x+1}\right|=lnx+C$
$ln\left|\frac{y}{x+y}\right|=lnx+C$
$ln\left(\frac{y}{x+y}\right)-ln\left(x\right)=C\Rightarrow ln\left(\frac{y}{x(x+y)}\right)=C$
$y=1,x=1\Rightarrow ln\left(\frac{1}{2}\right)=C$
$\therefore \frac{y}{x(x+y)}=\frac{1}{2}\Rightarrow 2y=x(x+y)\Rightarrow y=\frac{x^2}{2-x}$