Question

Find the particular solution of the differential equation $(x-y)\frac{dy}{dx}=x+2y$given that when $x=1,y=0$

Answer 1

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

$(x-y)dy=(x+2y)dx$

$xdy-ydy=xdx+2ydx$

Integrate on both sides

$xy-\frac{y^2}{2}=\frac{x^2}{2}+2xy+c$

It is given that$x=1,$ when $y=0$

Substituting these.

$0-0=\frac{1}{2}+c\Rightarrow c=\frac{-1}{2}$

therefore equation is

$xy-\frac{y^2}{2}=\frac{x^2}{2}+2xy-\frac{1}{2}$

$\frac{x^2+y^2}{2}+xy-\frac{1}{2}=0$

$x^2+y^2+2xy-1=0$