Let $y=y(x)$ be the solution of the differential equation $\left(2 x \log _e x\right) \frac{d y}{d x}+2…
Let $y=y(x)$ be the solution of the differential equation $\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$ and $y\left(e^{-1}\right)=0$. Then, $y(e)$ is equal to
$-\frac{3}{\mathrm{e}}$
$-\frac{3}{2 e}$
$-\frac{2}{3 e}$
$-\frac{2}{\mathrm{e}}$
Solution
$\begin{aligned} & \frac{d y}{d x}+\frac{y}{x \ell \ln x}=\frac{3}{2 x^2} \\ & \therefore \text { I.F. }=e^{\int \frac{1}{x \ln x} d x}=e^{\ln (\ln (x))}=\ln x \\ & \therefore y \ell \ln x=\int \frac{3 \ln x}{2 x^2} d x \\ & =\frac{3 \ln x}{2} \int x^{-2} d x-\int\left(\frac{3}{2 x} \cdot \int x^{-2} d x\right) d x \\ & =\frac{3 \ln x}{2}\left(-\frac{1}{x}\right)-\int \frac{3}{2 x}\left(-\frac{1}{x}\right) d x \\ & \text { y. } \ln x=\frac{-3 \ell n x}{2 x}-\frac{3}{2 x}+C\end{aligned}$ $\begin{aligned} & \because \mathrm{y}\left(\mathrm{e}^{-1}\right)=0 \\ & \therefore 0(-1)=\frac{3 \mathrm{e}}{2}-\frac{3 \mathrm{e}}{2}+\mathrm{C} \Rightarrow \mathrm{C}=0 \\ & \therefore \mathrm{y}=\frac{-3 \ell \mathrm{nx}}{2 \mathrm{x}}-\frac{3}{2 \mathrm{x}} \\ & \therefore \mathrm{y}(\mathrm{e})=\frac{-3}{2 \mathrm{e}}-\frac{3}{2 \mathrm{e}}=\frac{-3}{\mathrm{e}}\end{aligned}$