Find the integrating factor of the differential equation $(y\log y)dx = (\log y-x)dy$ I am familiar with linear differential equations of the form
$$\frac{dy}{dx} + P(x)y = Q(x)$$
But this equation doesn’t seem to be of that form, or any other ones I know. It could be converted into the linear form, but I don’t know how. How should I simplify?
 A: Hint: $$\frac{dx}{dy}+\frac x{y\log y}=\frac1y$$is first order linear ODE in $x$.
A: $$(y\log y)dx = (\log y-x)dy$$
Rewrite the DE as:
$$\log y\,dx = (\log y-x)\dfrac 1 ydy$$
$$\log y \,dx = (\log y-x)d ( \log y)$$
$$\log y \,dx +x d \log y= \log yd ( \log y)$$
$$d(x\log y )= \log yd ( \log y)$$
$$d(x\log y )= \dfrac {d ( \log^2 y)}2$$
Inetegrate.
A: With appropriate substitution(s) one can extract a separable ODE.
$$y\log y\,\mathrm dx=(\log y-x)\,\mathrm dy$$
Substitute $v(x)=\log y(x)\iff y(x)=e^{v(x)}$ and $\mathrm dv=\frac{\mathrm dy}y$:
$$\begin{align}
ve^v\,\mathrm dx&=(v-x)e^v\,\mathrm dv\\
v\,\mathrm dx&=(v-x)\,\mathrm dv\\[1ex]
\frac{\mathrm dv}{\mathrm dx}&=\frac v{v-x}
\end{align}$$
Substitute $w(x)=v(x)-x\iff v(x)=w(x)+x$ and $\mathrm dw=\mathrm dv-1$:
$$\begin{align}
\frac{\mathrm dw}{\mathrm dx}+1&=\frac{w-x}w\\[1ex]
\frac{\mathrm dw}{\mathrm dx}&=\frac xw\\[1ex]
w\,\mathrm dw&=x\,\mathrm dx
\end{align}$$
Now integrate both sides and solve:
$$\begin{align}
\frac{w^2}2&=\frac{x^2}2+C\\[1ex]
w^2&=x^2+C\\[1ex]
(v-x)^2&=x^2+C\\[1ex]
(\log y-x)^2&=x^2+C
\end{align}$$
