<s>[INST] Differential Equation (Non linear to linear differential equation) Show that the substitution $u=\frac{1}{y}$ transform the non-linear differential equation $$\frac{dy}{dx}+\frac{y}{x}=y^2\ln (x)$$ into the linear differential equation $$\frac{du}{dx}-\frac{u}{x}=-\ln (x)$$. Solve this linear differential equation, and hence obtain $y$ in terms of $x$, given that $y=\frac{1}{2}$ when $x=1$
My attempt, 
$\frac{dy}{dx}+\frac{y}{x}=y^2\ln (x)$
$\frac{-\frac{dy}{dx}}{y^2}-\frac{1}{xy}=-\ln (x)$
Let $u=\frac{1}{y}$ $\frac{du}{dx}=\frac{-\frac{dy}{dx}}{y^2}$
$\frac{du}{dx}-\frac{y}{x}=-\ln (x)$
$e^{\int -\frac{1}{x}dx}=\frac{1}{x}$
$\frac{\frac{du}{dx}}{x}-\frac{u}{x^2}=-\frac{\ln (x)}{x}$
$\frac{\frac{du}{dx}}{x}+\frac{d}{dx}(\frac{1}{x})u=-\frac{\ln (x)}{x}$
$\frac{d}{dx}(\frac{u}{x})=\frac{-\ln (x)}{x}$
$\int \frac{d}{dx}(\frac{u}{x})dx=\int \frac{-\ln (x)}{x}dx$
$\frac{u}{x}=-\frac{1}{2}\ln ^2(x)+c_1$
$u=x(-\frac{1}{2}\ln ^2(x)+c_1)$
$y=\frac{1}{u}=\frac{2}{-x\ln ^2(x)+2c_1x}$
$y=\frac{2}{-x\ln ^2(x)+c_1x}$
When $x=1$, $y=\frac{1}{2}$
I got $c_1=4$
Therefore, $y=\frac{2}{-x\ln ^2(x)+4x}$
Is this working right? Is there another method to solve this question?
 [/INST] $$
y' + \frac{y}{x} = y^2\ln x
$$
lets multiply by $x$ we find
$$
xy' + y = \frac{d}{dx}(xy) = xy^2\ln x
$$
let $u = xy$ we find
$$
u' = \frac{u^2}{x}\ln x
$$
thus we find
$$
\int\frac{du}{u^2} = \int \frac{\ln x}{x}dx = \int \ln x\dfrac{d}{dx}\ln x dx = \int \dfrac{d}{dx}\frac{(\ln x)^2}{2}dx = \frac{(\ln x)^2}{2} +C
$$
so we find
$$
-\frac{1}{u} = \frac{(\ln x)^2}{2} + C
$$
so we get
$$
u = \frac{-1}{ \frac{(\ln x)^2}{2} + C} 
$$
thus we lead to
$$
y = \frac{1}{x}\frac{-1}{ \frac{(\ln x)^2}{2} + C} = \frac{2}{-x\ln^2x -2Cx}
$$
Is this easier, I can not say.
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