Question

For each the differential equations given, find the general solution :
$xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

Answer 1

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

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

$I.F=c^{\int pdx}=c^{\int \frac{1}{x\log x}dx}$

Let $\log x=t\frac{1}{x}dx=dt$

$\Rightarrow I.F=c^{\int \frac{1}{t}dt}=t=\log x$

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

$\log x=t\frac{1}{x}dx=dt$

$yt=2\int e^{-t}tdt$

$yt=2(-c^{-t}(1+t))$

$y\log x=\frac{2}{x}dx=(1+\log x)$.