Let y = y ( x ) be the solution of the differential equation x _ e...

Given,

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

$⇒ \dfrac{dy}{dx} + \dfrac{y}{xlnx} = x$

This is a linear differential equation.

Now, finding integrating factor we get

$I . F = {e}^{∫ \dfrac{1}{xlnx}dx} = {e}^{\ln \left(\ln x\right)} = \ln x$

Now, solution of the differential equation is given by

$y \times I . F = ∫ \left(x \times I . F\right)dx$

$⇒ y \times \ln x = ∫ xlnxdx$

$⇒ y \times \ln x = \ln x \cdot \dfrac{{x}^{2}}{2} - \dfrac{{x}^{2}}{4} + c$

Now given $y\left(2\right) = 2$,

$2 \times \ln 2 = \ln 2 \cdot \dfrac{{2}^{2}}{2} - \dfrac{{2}^{2}}{4} + c$

$⇒ c = 1$

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

Now finding $y\left(e\right)$ we get,

$y \times lne = lne \cdot \dfrac{{e}^{2}}{2} - \dfrac{{e}^{2}}{4} + 1$

$⇒ y \times 1 = 1 \cdot \dfrac{{e}^{2}}{2} - \dfrac{{e}^{2}}{4} + 1$

$⇒ y = \dfrac{4 + {e}^{2}}{4}$