If d y d x = x y x 2 + y 2 ; y 1 = 1 ; then a value of x satisfying y x = e is:

Put $y=vx$

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

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

$\Rightarrow \frac{1+v^2}{v^3}dv=-\frac{1}{x}dx$

$\Rightarrow \int \left(\frac{1}{v^3}+\frac{1}{v}\right)dv=\int -\frac{1}{x}dx$

$\Rightarrow \frac{-1}{2}\frac{1}{v^2}+\mathcal{l}nv=-\mathcal{l}nx+c$

$\Rightarrow -\frac{x^2}{2y^2}=-\mathcal{l}ny+c$

When $x=1,y=1$ then

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

$\Rightarrow x^2=y^2\left(1+2\mathcal{l}ny\right)$

$\Rightarrow x^2=e^2\left(3\right)$