Answer: (3)

Solution

$\frac{dy}{dx} + \frac{y}{x} = \ln x$. I.F. $= x$. Solution: $xy = \int x\ln x,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + c$. $2y(2) = \ln 4 - 1$: $2 \cdot 2 \cdot y(2) = 4y(2) = 2\ln 2 \cdot 2 - 1 + \frac{2c}{…}$. Actually: $xy = \frac{x^2}{2}\ln x - \frac{x^2}{4} + c$. $2y(2) = \ln 4 - 1$, so $y(2) = \frac{\ln 4 - 1}{2}$. Then $2y(2) = 2\ln 2 - 1 = 4\ln 2 - 2 + 2c/…$. Hmm. From $2y(2) = \frac{4}{2}\ln 2 - 1 + \frac{c}{1}$: $2 \cdot y(2) = 2\ln 2 - 1 + c$. Given $2y(2) = \ln 4 - 1 = 2\ln 2 - 1$: $c = 0$. At $x = e$: $ey(e) = \frac{e^2}{2} - \frac{e^2}{4} = \frac{e^2}{4}$. So $y(e) = \frac{e}{4}$.