How to solve $(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$ 
Solve $$(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$$

I have a very similar problem like this on my homework, and I have no clue how to set it up or even start. How could I set this up?

$$(x-3)(\frac{dy}{dx})+y=6e^x \Rightarrow \frac{dy}{dx}+\frac{1}{x-3}y=\frac{6 e^x}{x-3} \Rightarrow y'(x)+\frac{1}{x-3}y(x)=\frac{6 e^x}{x-3} $$
The solution of the homogeneous problem is the following:
$$y'_h(x)+\frac{1}{x-3}y_h(x)=0 \Rightarrow \frac{dy_h}{dx}=-\frac{1}{x-3}y_h \Rightarrow \frac{dy_h}{y_h}=-\frac{dx}{x-3} \Rightarrow \int \frac{dy_h}{y_h}=-\int \frac{dx}{x-3} \\ \Rightarrow \ln y_h=-\ln{|x-3|} +c \Rightarrow \ln {|y_h|}=\ln{(|x-3|)^{-1}} +c \Rightarrow e^{\ln {|y_h|}}=e^{\ln{(|x-3|)^{-1}} +c} \\ \Rightarrow |y_h|=e^c \frac{1}{|x-3|}  \Rightarrow y_h=\pm e^c \frac{1}{x-3} \overset{\pm e^c=C}{\Rightarrow} y_h(x)=\frac{C}{x-3}$$
The solution of the non-homogengeneous problem is the following:
We suppose that the solution is of the form: $$y_p(x)=\frac{C(x)}{x-3}$$
Replacing this at the problem we get:
$$y'_p(x)+\frac{1}{x-3}y_p(x)=\frac{6 e^x}{x-3} \Rightarrow \frac{C'(x)(x-3)-C(x)}{(x-3)^2}+\frac{1}{x-3}\frac{C(x)}{x-3}=\frac{6 e^x}{x-3} \Rightarrow \frac{C'(x)(x-3)-C(x)+C(x)}{(x-3)^2}=\frac{6 e^x}{x-3} \Rightarrow C'(x)=6 e^x \Rightarrow C(x)=6 e^x+c_2$$
So, $$y_p(x)=\frac{6e^x+c_2}{x-3}$$
The solution of the initial problem is equal to the sum of the homogeneous and the non-homogeneous problem $$y=y_h+y_p$$
