For x > 1, a function y satisfies (x ln x)dy/dx + y = 2 ln x, and y = 4 when x = e…

Step 1. Standard form. $\dfrac{dy}{dx} + \dfrac{y}{x\ln x} = \dfrac{2}{x}$.

Step 2. Integrating factor. $\displaystyle\int \dfrac{dx}{x\ln x} = \ln(\ln x)$, so $I = \ln x$.

Step 3. Integrate. $y\ln x = \displaystyle\int \dfrac{2\ln x}{x}\,dx = (\ln x)^2 + C$. So $y = \ln x + \dfrac{C}{\ln x}$.

Step 4. The constant. At $x = e$: $4 = 1 + \dfrac{C}{1}$, so $C = 3$.

Step 5. Evaluate. At $x = e^2$: $\ln x = 2$, so $y = 2 + \dfrac{3}{2} = 3.5$.