Answer
\[y(x)=\frac{x}{m+1}\ln x-\frac{x}{(m+1)^2}+\frac{C}{x^m}\]
Work Step by Step
$y'+mx^{-1}y=\ln x$ ____(1)
(1) is Linear Differential Equation
Integrating Factor:-
\[I(x)=e^{\int mx^{-1} dx}=e^{m\int\frac{1}{x} dx}=e^{m\ln x}=x^m\]
Multiply (1) by $I(x)$
\[x^m\frac{dy}{dx}+mx^{m-1}y=x^m \ln x\]
\[\frac{d}{dx}(yx^m)=x^m\ln x\]
Integrating
\[yx^m=\int x^m\ln x dx+C\]
$C$ is constant of integration
\[yx^m=\ln x\left(\frac{x^{m+1}}{m+1}\right)-\int\frac{1}{x}\left(\frac{x^{m+1}}{m+1}\right)dx+C\]
\[yx^m=\ln x\left(\frac{x^{m+1}}{m+1}\right)-\frac{x^{m+1}}{(m+1)^2}+C\]
\[y(x)=\ln x\left(\frac{x}{m+1}\right)-\frac{x}{(m+1)^2}+\frac{C}{x^m}\]
Hence General Solution of (1) is $y(x)=\frac{x}{m+1}\ln x-\frac{x}{(m+1)^2}+\frac{C}{x^m}$.
