How do I solve this first order linear ordinary differential equation with factorization?

Question

I'm getting an equation $$(\ln y - x)\frac{dy}{dx} - y\ln y = 0$$

Which I try to factorize and bring over to get: $$\frac{dy}{dx} = \frac{y\ln y}{(\ln y - x)}$$

But this cannot be factorized further. I am intending to use either the the substitution $u=y/x$ or finding the integrating factor, but this form makes it hard for me to try either method. How should I proceed?

Answer 1

Hint

Substitute $y=e^z$ $$(\ln y - x)\frac{dy}{dx} - y\ln y = 0$$ $$(z - x)\frac{dy}{dz} \frac{dz}{dx}- e^zz = 0$$ $$(z - x)e^z z'- e^zz = 0$$ $$(z - x) z'-z = 0$$ $$z'=\frac z {z-x}$$ Then consider $\frac {dx}{dz}$ instead: $$x'+\frac x z=1$$ Multiply by z $(z \ne 0)$ $$ x'z+x=z$$ Note the derivative of $xz$ $$(xz)'=z$$ Now integrate $$ x=\frac 1 z \int zdz$$ Evaluate the integral then substitute $z=\ln(y)$

Answer 2

Hint...substitute $u=\ln y$ and you will arrive at a linear differential equation requiring an integrating factor.