Correct option is (3) \(-\frac{3}{\mathrm{e}}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x} \ln \mathrm{x}}=\frac{3}{2 \mathrm{x}^{2}}\)
\(\therefore I.F. =\mathrm{e}^{\int \frac{1}{x \ln x} \mathrm{dx}}=\mathrm{e}^{\ln (\ln (\mathrm{x}))}=\ln \mathrm{x}\)
\(\therefore \mathrm{y} \ell \mathrm{nx}=\int \frac{3 \ln \mathrm{x}}{2 \mathrm{x}^{2}} \mathrm{dx}\)
\(=\frac{3 \ln x}{2} \int x^{-2} d x-\int\left(\frac{3}{2 x} \cdot \int x^{-2} d x\right) d x\)
\(=\frac{3 \ln \mathrm{x}}{2}\left(-\frac{1}{\mathrm{x}}\right)-\int \frac{3}{2 \mathrm{x}}\left(-\frac{1}{\mathrm{x}}\right) \mathrm{dx}\)
\(
y. \ln x=\frac{-3 \ln x}{2 x}-\frac{3}{2 x}+C\)
\(\because y\left(\mathrm{e}^{-1}\right)=0\)
\(\therefore 0(-1)=\frac{3 \mathrm{e}}{2}-\frac{3 \mathrm{e}}{2}+\mathrm{C} \Rightarrow \mathrm{C}=0\)
\(\therefore \mathrm{y}=\frac{-3 \ln \mathrm{x}}{2 \mathrm{x}}-\frac{3}{2 \mathrm{x}}\)
\(\therefore y(e)=\frac{-3}{2 e}-\frac{3}{2 e}=\frac{-3}{e}\)
