Let x=x(y) solve 2ye^x/y^2,dx+(y^2-4xe^x/y^2),dy=0, x(1)=0.

Question

Let $x=x(y)$ solve $2ye^{x/y^2}\,dx+(y^2-4xe^{x/y^2})\,dy=0$, $x(1)=0$. Then $x(e)$ is:

Accepted Answer

Rewriting: $2e^{x/y^2}\left(\dfrac{2y^2\,dx-4xy\,dy}{y^4}\right)=-\dfrac{1}{y}\,dy$ leads, after setting $u=x/y^2$, to
$$2e^u\,du = -\frac1y\,dy \implies 2e^{x/y^2} = -\ln y+c.$$

Using $x(1)=0$: $c=2$. At $y=e$: $2e^{x/e^2}=-1+2=1 \implies x(e)=-e^2\ln2$.

Solutions · 1