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Answer: $y=\frac{x^{2}}{4}+\frac{x^{2}}{16}+C$

Give:  $x \frac{d y}{d x}+2 y=x^{2} \log x$

Hint: Using integration by parts and  $\int \frac{1}{x} d x$

Explanation:  $x \frac{d y}{d x}+2 y=x^{2} \log x$

Divide by x

$=\frac{d y}{d x}+\left(\frac{2}{x}\right) y=x \log x$

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=\frac{2}{x} \text { and } Q=x \log x \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{2}{x}^{2} d x} \\ &=e^{2 \int \frac{1}{x} d x} \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \\ &=e^{2 \log |x|} \\ &=e^{\log \left|x^{2}\right|} \\ &=x^{2} \quad\left[e^{\log e^{x}}=x\right] \end{aligned}

Hence, the solution of different equation is

$y I f=\int Q I f d x+C$

\begin{aligned} &=y x^{2}=\int x \log x x^{2} d x+C \\ &=y x^{2}=\int x^{3} \log x d x+C \ldots(i) \end{aligned}

We have  \begin{aligned} & \int x^{3} \log x d x=\log x \frac{x^{4}}{4}-\int \frac{1}{x} \frac{x^{4}}{4} d x+C\\ & \end{aligned}

\begin{aligned} & =\log x \frac{x^{4}}{4}-\frac{1}{4} \int x^{3} d x+C\\ &=\log x \frac{x^{4}}{4}-\frac{1}{4} \frac{x^{4}}{4}+C\\ &=\log x \frac{x^{4}}{4}-\frac{x^{4}}{16}+C \end{aligned}

From i

$=y x^{2}=\log x \frac{x^{4}}{4}-\frac{x^{4}}{16}+C$

Divide by $x^{2}$

\begin{aligned} &=y=\frac{1}{x^{2}} \log x \frac{x^{4}}{4}-\frac{1}{x^{2}} \frac{x^{4}}{16}+C \\\\ &=y=\frac{x^{2}}{4} \log x-\frac{x^{2}}{16}+C \end{aligned}

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