Let y=y(x) solve x_exdydx+y=x^2_ex, x1.

Question

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

Accepted Answer

$$\frac{dy}{dx}+\frac{y}{x\ln x} = x.$$

I.F. $=\ln x$.
$$y\ln x = \int x\ln x\,dx = \frac{x^2}{2}\ln x-\frac{x^2}{4}+c.$$

Using $y(2)=2$: $2\ln2 = 2\ln2-1+c \Rightarrow c=1$.

$$y(e) = \frac{e^2}{2}-\frac{e^2}{4}+1 = 1+\frac{e^2}{4} = \frac{4+e^2}{4}.$$

Solutions · 1