solving ODE: $\ln(y') + (x)y' -y = 0$ How to solve this ODE
$$\ln(y') + (x)y' -y = 0$$
I can solve using power series, but it's not a very elegant approach. I'm wondering if there is a better way.
 A: Clearly, $y' \ne 0$.
Let's differentiate this equation once more:
$$\frac{y''}{y'}+xy''+y'-y'=\frac{y''}{y'}+xy''.$$
Suppose for the moment that $y''\ne0$, then we get $$1+xy'=0,$$
which gives $y(x) = Const -\ln |x|$. Let's put this expression to the initial equation:
$$\ln (-\frac 1x)-1-Const+\ln |x|=0.$$
Apparently, it has sense only for $x<0$ and implies $Const = -1$. So we have the first solution
$$y(x) = -1-\ln |x|,\quad x<0.$$
The second family of solutions comes from the option $y''=0$ or, in other words, $y(x)=ax+b$. We put it into the equation to obtain
$$\ln a+xa-xa-b=0$$or $a=e^b$; thus, the second solution has the form
$$y(x) = e^bx+b,\quad b\in \Bbb R.$$
A: If you try $y=a x + b,$ you will be able to solve for $a$ and $b$ and be happy.
A: Actually, it is an example of a Clairaut equation, which is in the form $y=xy'+f(y')$ with $f(y')=\ln y'$ and the general solution will be all the sraight lines $y=cx+f(c)$, where $c$ is an arbitrary constant. In your case, $f(c)=\ln c$, $c>0$, so $y=cx+\ln c$.
