Solving a first order differential equation - possibly using integrating factor? Question Find the general solution for 
$$xy' - y = -x .e^{\frac{y}{x}}$$
My attempt:
The only two methods I have learnt in regards to solving First order ODE's are using an integrating factor and separation of variables. I tried to go with the first method and got:
$$y' - \frac{1}{x}y= -e^\frac{y}{x}$$
But I cannot advance from here as there is a y term in the exponential.

Let $y(x)=xr(x)$ which gives $y'(x)=r(x)+xr'(x)$:
$$xy'(x)-y(x)=-xe^{\frac{y(x)}{x}}\Longleftrightarrow x\left(r(x)+xr'(x)\right)-xr(x)=-xe^{\frac{xr(x)}{x}}\Longleftrightarrow\frac{r'(x)}{e^{r(x)}}=-\frac{1}{x}$$
Now, integrate with respect to $x$:
$$\int\frac{r'(x)}{e^{r(x)}}\space\text{d}x=\int-\frac{1}{x}\space\text{d}x$$
And use:


*

*Substitute $u=r(x)$ and $\text{d}u=r'(x)\space\text{d}x$:
$$\int\frac{r'(x)}{e^{r(x)}}\space\text{d}x=\int\frac{1}{e^u}\space\text{d}u=\int e^{-u}\space\text{d}u=-\frac{1}{e^u}+\text{C}=\text{C}-\frac{1}{e^{r(x)}}$$

*$$\int-\frac{1}{x}\space\text{d}x=-\int\frac{1}{x}\space\text{d}x=\text{C}-\ln\left|x\right|$$


So, we get (set $r(x)=\frac{y(x)}{x}$ back):
$$-\frac{1}{e^{\frac{y(x)}{x}}}=\text{C}-\ln\left|x\right|\Longleftrightarrow e^{\frac{y(x)}{x}}=\frac{1}{\ln\left|x\right|+\text{C}}$$
