JEE 2023 — Mathematics PYQ

Solution: Standard form: $\frac{dy}{dx}+\frac{1}{x\log x}y=x$.

Integrating Factor (I.F.) $=e^{\int \frac{1}{x\log x}dx}=e^{\log (\log x)}=\log x$.

Solution: $y\log x=\int x\log xdx=\frac{x^2}{2}\log x-\frac{x^2}{4}+C$.

Using $y(2)=2$: $2\log 2=2\log 2-1+C\Rightarrow C=1$.

At $x=e$: $y\log e=\frac{e^2}{2}\log e-\frac{e^2}{4}+1\Rightarrow y=\frac{e^2}{4}+1=\frac{4+e^2}{4}$.

Correct Option: (2)