Question

$x\frac{dy}{dx}+y=x\log x$

Answer 1

We have,
$x\frac{dy}{dx}+y=x\log x$
Dividing both sides by x, we get
$$\begin{gathered} \frac{dy}{dx}+\frac{y}{x}=\log x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{x} \\ Q=\log x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int Pdx}=e^{\int \frac{1}{x}dx} \\ =e^{\log \left|x\right|} \\ =x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow xy=\int \underset{\mathrm{II}}{x}\underset{\mathrm{I}}{\log x}dx+C \\ \Rightarrow xy=\log x\int xdx-\int \left[\frac{d}{dx}\left(\log x\right)\int xdx\right]dx+C \\ \Rightarrow xy=\frac{x^2\log x}{2}-\int \frac{x}{2}dx+C \\ \Rightarrow xy=\frac{x^2\log x}{2}-\frac{x^2}{4}+C \\ \Rightarrow 4xy=2x^2\log x-x^2+K\left(\mathrm{where},K=2C\right) \end{gathered}$$