Question

Find the general solution of the $x\log x\frac{dy}{dx}+y=\frac{2}{x}logx$

Answer 1

Given,

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

$\Rightarrow \frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x^2}$

$\frac{dy}{dx}+Py=Q$

$P=\frac{1}{x\log x}\to I.F=e^{\int Pdx}=e^{\int \frac{1}{x\log x}dx}=e^{\log \left(\log x\right)}=\log x$

$Q=\frac{2}{x^2}$

$y\times I.F=\int Q\times I.Fdx$

$y\times logx=\int \frac{2}{x^2}\times \log xdx$

$ylogx=2(-\frac{\log \left(x\right)}{x}-\frac{1}{x})+C$