Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& \frac{dy}{dx}+\frac{y}{x\ell \ln x}=\frac{3}{2x^2}\\ & \therefore \text{ I.F. }=e^{\int \frac{1}{x\ln x}dx}=e^{\ln (\ln (x))}=\ln x\\ & \therefore y\ell \ln x=\int \frac{3\ln x}{2x^2}dx\\ & =\frac{3\ln x}{2}\int x^{-2}dx-\int \left(\frac{3}{2x}\cdot \int x^{-2}dx\right)dx\\ & =\frac{3\ln x}{2}\left(-\frac{1}{x}\right)-\int \frac{3}{2x}\left(-\frac{1}{x}\right)dx\\ & \text{ y. }\ln x=\frac{-3\ell nx}{2x}-\frac{3}{2x}+C\end{array}$
$\begin{array}{rl}& \because \mathrm{y}\left({\mathrm{e}}^{-1}\right)=0\\ & \therefore 0(-1)=\frac{3\mathrm{e}}{2}-\frac{3\mathrm{e}}{2}+\mathrm{C}\Rightarrow \mathrm{C}=0\\ & \therefore \mathrm{y}=\frac{-3\ell \mathrm{nx}}{2\mathrm{x}}-\frac{3}{2\mathrm{x}}\\ & \therefore \mathrm{y}(\mathrm{e})=\frac{-3}{2\mathrm{e}}-\frac{3}{2\mathrm{e}}=\frac{-3}{\mathrm{e}}\end{array}$