If dy dx= xy x^2+y^2;y(1)=1; then a value of x satisfying y(x)=e is:

Put $y=vx$y=vx

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

$$v+x\frac{dv}{dx}=\frac{v{x}^{2}}{{x}^{2}+{v}^{2}{x}^{2}}$$v+xdxdv​=x2+v2x2vx2​

$$\Rightarrow \frac{1+{v}^{2}}{{v}^{3}}dv=-\frac{1}{x}dx$$⇒v31+v2​dv=−x1​dx

$$\Rightarrow \int (\frac{1}{{v}^{3}}+\frac{1}{v})dv=\int -\frac{1}{x}dx$$⇒∫(v31​+v1​)dv=∫−x1​dx

$$\Rightarrow \frac{-1}{2}\frac{1}{{v}^{2}}+lnv=-lnx+c$$⇒2−1​v21​+lnv=−lnx+c

$$\Rightarrow -\frac{{x}^{2}}{2{y}^{2}}=-lny+c$$⇒−2y2x2​=−lny+c

When $x=1,y=1$x=1,y=1 then

$$-\frac{1}{2}=c$$−21​=c

$$\Rightarrow {x}^{2}={y}^{2}(1+2lny)$$⇒x2=y2(1+2lny)

$$\Rightarrow {x}^{2}={e}^{2}(3)$$⇒x2=e2(3)