Let y = y ( x ) be the solution of the differential equation x log e x d y d x + y = x 2 log e x , ( x >…

Given,

$x\ln x\frac{dy}{dx}+y=x^2\ln x$

$\Rightarrow \frac{dy}{dx}+\frac{y}{x\ln x}=x$

This is a linear differential equation.

Now, finding integrating factor we get

$\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \frac{1}{\mathrm{xlnx}}\mathrm{dx}}={\mathrm{e}}^{\ln \left(\mathrm{lnx}\right)}=\mathrm{lnx}$

Now, solution of the differential equation is given by

$y\times \mathrm{I}.\mathrm{F}=\int \left(\mathrm{x}\times \mathrm{I}.\mathrm{F}\right)dx$

$\Rightarrow y\times \ln x=\int \mathrm{xlnx} dx$

$\Rightarrow y\times \ln x=\ln x·\frac{x^2}{2}-\frac{x^2}{4}+c$

Now given $y\left(2\right)=2$,

$2\times \ln 2=\ln 2·\frac{2^2}{2}-\frac{2^2}{4}+c$

$\Rightarrow c=1$

So, equation will be $y\times \ln x=\ln x·\frac{x^2}{2}-\frac{x^2}{4}+1$

Now finding $y\left(e\right)$ we get,

$y\times \ln e=\ln e·\frac{e^2}{2}-\frac{e^2}{4}+1$

$\Rightarrow y\times 1=1·\frac{e^2}{2}-\frac{e^2}{4}+1$

$\Rightarrow y=\frac{4+e^2}{4}$