I want to solve the following D.E.
$$xy'+(lnx)y =lnx$$
First I need to bring it to the form $y'+a(x)y=b(x)$, so it becomes: $$y'+\frac{lnx}{x}y=\frac{lnx}{x} $$
After this, I need to calculate the $\int a(x)dx$ which will give me: $$\int{\frac{lnx}{x}dx} = \frac{(lnx)^2}{2} +c$$
So the Euler multiplier is $e^\frac{(lnx)^2}{2}$. Now I multiply my equation with this, at each side and get on the left side the product rule:
$$(ye^\frac{(lnx)^2}{2})'=e^\frac{(lnx)^2}{2}\frac{lnx}{x}$$
So the question that I have is this: Am I allowed to do the following?
$$ye^\frac{(lnx)^2}{2}=\int{e^\frac{(lnx)^2}{2}\frac{lnx}{x}}dx$$
If no, why?
Edit: If this is correct, then if I simplify I get:
$$ yx=\int{lnx}dx $$
and by computing the integral I would end up with this:
$$yx= xlnx -x +c \Rightarrow y=lnx -1+c$$
But if it is done otherwise, like this:
$$(ye^\frac{(lnx)^2}{2})'=(e^\frac{(lnx)^2}{2})' \Rightarrow ye^\frac{(lnx)^2}{2}=e^\frac{(lnx)^2}{2}$$
I get this result:
$$ y= 1 +c *e^{-\frac{(lnx)^2}{2}} $$
which is different from the other result.
write your equation in the form $$\frac{dy}{1-y}=\frac{\log(x)}{x}dx$$