Show that

Solution:

The given differential equation is:

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

$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$

This equation is in the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q$ (where $p=\frac{2}{x}$ and $Q=x \log x$ )

Now, I.F $=e^{\int p d x}=e^{\int_{x}^{2} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}$

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+\mathrm{C}$

$\Rightarrow y \cdot x^{2}=\int\left(x \log x \cdot x^{2}\right) d x+\mathrm{C}$

$\Rightarrow x^{2} y=\int\left(x^{3} \log x\right) d x+\mathrm{C}$

$\Rightarrow x^{2} y=\log x \cdot \int x^{3} d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^{3} d x\right] d x+\mathrm{C}$

$y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+\mathrm{C}$

$\Rightarrow y \cdot x^{2}=\int\left(x \log x \cdot x^{2}\right) d x+\mathrm{C}$

$\Rightarrow x^{2} y=\int\left(x^{3} \log x\right) d x+\mathrm{C}$

$\Rightarrow x^{2} y=\log x \cdot \int x^{3} d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^{3} d x\right] d x+\mathrm{C}$

$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int\left(\frac{1}{x} \cdot \frac{x^{4}}{4}\right) d x+\mathrm{C}$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+\mathrm{C}$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+\mathrm{C}$

$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+\mathrm{C}$

$\Rightarrow y=\frac{1}{16} x^{2}(4 \log x-1)+\mathrm{C} x^{-2}$