Answer
$y(x)=xe^{x^2}$
Work Step by Step
We are given:
$$y^{-1}y'-2x^{-1}y\ln y=x^{-1}(1-2\ln x)$$
Let $t= \ln y \rightarrow t'=y^{-1}y$
Substituting:
$$t'-2x^{-1}t=x^{-1}(1-2 \ln x)$$
We have $p(x)=-2x^{-1} q(x)=x^{-1}(1-2\ln x)$
The solution can be found by obtaning:
$$t=e^{-\int p(x)dx}[\int e^{p(x)dx}q(x)dx+C$$
$$t=e^{2x^{-1}dx}[\int e^{-2x^{-1}dx}x^{-1}1-2\ln x)dx+C$$
$$t=e^{2\ln x}[\int e^{-2\ln x}x^{-1}(1-2\ln x)dx + C$$
$$t=x^2[\int (x^{-3}-2x^{-3}\ln x)dx+C$$
$$t=x^2[\frac{x^{-2}}{-2}+\frac{\ln x}{x^2}+\frac{x^{-2}}{2}+C$$
$$t=\ln x +Cx^2$$
Since $t= \ln y\rightarrow y=e^t$
Hence $y=e^{\ln x+Cx^2}=xe^{Cx^2}$
We are given: $y(1)=e$
Substituting:
$1e^C=e \rightarrow C=1$
The solution is:
$$y(x)=xe^{x^2}$$
