Ways for the differential equation $y' + {{y\ln(y)}\over x}= xy$ I've been stuck on one of my homework numbers. 
The number precise that the following equation is a non-linear equation of order 1 with x>0.
$$y' + {{y\ln(y)}\over x}= xy$$
So far, I tried 2 different methods to solve them. As suggested by internet (link below): Bernoulli, and as an inexact one. However, I get stuck every time. 
Bernoulli sample
**As a small side note, can someone explain to me the following from the link: how, by passing ln(y) to the right side, we get y^1+1 ? 
For the Bernoulli equation, when I try to solve it manually, I get in a kind of infinite integral, no matter how many time I integrate.
Am I missing something?
 A: $$y' + {{y\ln(y)}\over x}= xy$$
$$\dfrac {y'}{y}+\dfrac {\ln y}{x}=x$$
$$(\ln y)'+\dfrac {\ln y}{x}=x$$
It's a linear first order DE. Substitute $u=\ln y$
$$u'+\dfrac {u}{x}=x$$
$$xu'+u=x^2$$
$$(xu)'=x^2$$
Integrate.
$$\ln y =\dfrac {x^2}3+\dfrac {c_1}{x}$$
This answer dosent agree with Symbolab link but it agrees with Wolfram Alpha's answer
$$ y(x) =\exp {\left (\dfrac {x^2}3+\dfrac {c_1}{x}\right)}$$
A: Well, we have the following first-order nonlinear ordinary differential equation:
$$\text{y}'\left(x\right)+\frac{\text{y}\left(x\right)\ln\left(\text{y}\left(x\right)\right)}{x}=x\text{y}\left(x\right)\tag1$$
This can be rewritten in the following form:
$$x^2\text{y}\left(x\right)-\text{y}\left(x\right)\ln\left(\text{y}\left(x\right)\right)-x\text{y}'\left(x\right)=0\tag2$$
Now, let $\text{R}\left(x,\text{y}\right)=x^2\text{y}-\text{y}\ln\left(\text{y}\right)$ and $\text{S}\left(x,\text{y}\right)=-x$. This is not an exact equation, because:
$$\left(\frac{\partial\text{R}\left(x,\text{y}\right)}{\partial\text{y}}=x^2-\ln\left(\text{y}\right)-1\right)\ne\left(-1=\frac{\partial\text{S}\left(x,\text{y}\right)}{\partial\text{y}}\right)\tag3$$
Find an integrating factor $\mu(\text{y})$ such that $\mu(\text{y})\text{R}\left(x,\text{y}\right)+\text{y}'\left(x\right)\text{S}\left(x,\text{y}\right)=0$ is exact. This means:
$$\frac{\partial}{\partial\text{y}}\left(\mu(\text{y})\text{R}\left(x,\text{y}\right)\right)=\frac{\partial}{\partial x}\left(\mu(\text{y})\text{S}\left(x,\text{y}\right)\right)\tag4$$
Which gives:
$$\frac{\text{d}\mu(\text{y})}{\text{d}\text{y}}\left(\text{y}x^2-\text{y}\ln(\text{y})\right)+\mu(\text{y})\left(x^2-\ln(\text{y})-1\right)=-\mu(\text{y})\tag5$$
Isolte $\mu(\text{y})$ to the left-hand side:
$$\frac{\partial\mu(\text{y})}{\partial\text{y}}\cdot\frac{1}{\mu(\text{y})}=-\frac{1}{\text{y}}\tag6$$
Integrate both sides with respect to $\text{y}$:
$$\ln\left|\mu(\text{y})\right|=-\ln\left|\text{y}\right|\space\Longrightarrow\space\mu(\text{y})=\frac{1}{\text{y}}\tag7$$
So:
$$x^2-\ln\left(\text{y}\left(x\right)\right)-\frac{\text{d}\text{y}\left(x\right)}{\text{d}x}\cdot\frac{x}{\text{y}\left(x\right)}=0\tag8$$

I let you finish.

