Let y=y(x) be the solution of the differential equation (2 x log _e x ) d y d x+2 y= 3 x log _e x, x>0 and y (e^-1 )=0. Then, y(e) is equal to

$$\begin{aligned} & \frac{d y}{d x}+\frac{y}{x \ell \ln x}=\frac{3}{2 x^2} \\ & \therefore \text { I.F. }=e^{\int \frac{1}{x \ln x} d x}=e^{\ln (\ln (x))}=\ln x \\ & \therefore y \ell \ln x=\int \frac{3 \ln x}{2 x^2} d x \\ & =\frac{3 \ln x}{2} \int x^{-2} d x-\int\left(\frac{3}{2 x} \cdot \int x^{-2} d x\right) d x \\ & =\frac{3 \ln x}{2}\left(-\frac{1}{x}\right)-\int \frac{3}{2 x}\left(-\frac{1}{x}\right) d x \\ & \text { y. } \ln x=\frac{-3 \ell n x}{2 x}-\frac{3}{2 x}+C\end{aligned}$$​dxdy​+xℓlnxy​=2x23​∴ I.F. =e∫xlnx1​dx=eln(ln(x))=lnx∴yℓlnx=∫2x23lnx​dx=23lnx​∫x−2dx−∫(2x3​⋅∫x−2dx)dx=23lnx​(−x1​)−∫2x3​(−x1​)dx y. lnx=2x−3ℓnx​−2x3​+C​

$$\begin{aligned} & \because \mathrm{y}\left(\mathrm{e}^{-1}\right)=0 \\ & \therefore 0(-1)=\frac{3 \mathrm{e}}{2}-\frac{3 \mathrm{e}}{2}+\mathrm{C} \Rightarrow \mathrm{C}=0 \\ & \therefore \mathrm{y}=\frac{-3 \ell \mathrm{nx}}{2 \mathrm{x}}-\frac{3}{2 \mathrm{x}} \\ & \therefore \mathrm{y}(\mathrm{e})=\frac{-3}{2 \mathrm{e}}-\frac{3}{2 \mathrm{e}}=\frac{-3}{\mathrm{e}}\end{aligned}$$​∵y(e−1)=0∴0(−1)=23e​−23e​+C⇒C=0∴y=2x−3ℓnx​−2x3​∴y(e)=2e−3​−2e3​=e−3​​