Question

Find the solution of the differential equation $\left(x\log x\right)\frac{dy}{dx}+y=2x\log x,(x\ge 1)$.

Answer 1

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

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

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

Let $\log =4=e^{\int \frac{1}{u}du}=e^{\log 4}=u=\log x$

$\frac{1}{x}dx=dx$

$y\times \log x=2\int \log xdx+C$

Using $ILATE$ rule

$y\times \log x=2(x\log x-x)+C$

$y\times \log x=2x\log x-2x+C$

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