Question

Solve the differential equation, $(x^2+xy)dy=(x^2+y^2)dx$.

Answer 1

Different equation $(x^2+xy)dy=(x^2+y^2)dx$
$\Rightarrow \frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}$
by putting $y=vx$
$\therefore \frac{dy}{dx}=v+x.\frac{dv}{dx}$
$\therefore v+x.\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{x^2+v{x}^2}$
$\Rightarrow v+x.\frac{dv}{dx}=\frac{x^2(1+v^2)}{x^2(1+v)}$
$\Rightarrow x.\frac{dv}{dx}=\frac{(1+v^2)}{(1+v)}-v$
$\Rightarrow x.\frac{dv}{dx}=\frac{1+v^2-v-v^2}{1+v}$
$\Rightarrow x.\frac{dv}{dx}=\frac{1-v}{1+v}$
$\Rightarrow \left(\frac{1+v}{1-v}\right)dv=\frac{1}{x}.dx$
$\Rightarrow (-1+\frac{2}{1-v})dv=\frac{1}{x}.dx$
On integration both side
$\therefore -\int 1dv+2\int \frac{1}{1-v}dv=\int \frac{1}{x}dx$
$\Rightarrow \therefore -v+\frac{2lo{g}_e(1-v)}{-1}={\log }_{e}x+c$
$\Rightarrow {\log }_{e}x+2{\log }_e(1-v)+v+c=0$
$\Rightarrow {\log }_{e}x+2{\log }_e(1-\frac{y}{x})+\frac{y}{x}+c=0$.