Let y=y(x) solve xdydx+y=x_ex, x1.

Question

Let $y=y(x)$ solve $x\dfrac{dy}{dx}+y=x\log_ex$, $x>1$. If $2y(2)=\log_e4-1$, then $y(e)$ equals:

Accepted Answer

I.F.$=x$. $xy = \dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}+c$.

Using $2y(2)=\ln4-1$: $c=0$. So $y=\dfrac{x}{2}\ln x-\dfrac{x}{4}$. At $x=e$: $y=\dfrac e2-\dfrac e4=\dfrac e4$.

Solutions · 1