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

$x^{2}dy=(2xy+y^2)dx$

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

which is homogeneous $....\left(i\right)$

Put $y=vx$

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

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

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

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

$\log \left(v\right)-\log (v+1)=\log x+logc$

$\log \left|\frac{v}{v+1}\right|=\log cx$

$\frac{y}{y+x}=cx...\left(ii\right)$

When $y=1,x=1$ then

$\frac{1}{1+1}=c$

$\Rightarrow c=\frac{1}{2}$

Put the value of c in eq. (ii)

$\frac{y}{y+x}=\frac{1}{2}x$

$\Rightarrow 2y=xy+x^2$

$\Rightarrow x^2+xy-2y=0$