If x=x(y) solves ydxdy=2x+y^3(y+1)e^y, x(1)=0, then x(e) is:

Question

If $x=x(y)$ solves $y\dfrac{dx}{dy}=2x+y^3(y+1)e^y$, $x(1)=0$, then $x(e)$ is:

Accepted Answer

$$\frac{dx}{dy}-\frac{2}{y}x = y^2(y+1)e^y.$$

I.F. $=1/y^2$.
$$\frac{x}{y^2} = \int(y+1)e^y\,dy = ye^y+c.$$

Using $x(1)=0$: $0=e+c \Rightarrow c=-e$.

At $y=e$: $\dfrac{x}{e^2} = e\cdot e^e - e \implies x = e^3(e^e-1)$.

Solutions · 1