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?

$$(x-3) \frac{dy}{dx}+y=0 \Rightarrow \frac{dy}{dx}=\frac{-y}{x-3} \Rightarrow -\frac{dy}{y}=\frac{dx}{x-3} \Rightarrow \int \left ( \frac{-1}{y} \right)dy=\int \frac{1}{x-3} dx \\ \Rightarrow -\ln |y| =\ln |x-3|+c \Rightarrow e^{-\ln |y|}=e^{\ln |x-3|+c} \Rightarrow \frac{1}{|y|}=C |x-3| \Rightarrow |y|=\frac{c'}{|x-3|}\\ \Rightarrow y= \pm \frac{c'}{x-3} \Rightarrow y=\frac{A}{x-3} $$
So, for our non-homogeneous problem we have:
$$y(x)=\frac{A(x)}{x-3}$$
$$(x-3) \frac{dy}{dx}+y=6e^x \Rightarrow (x-3)\frac{A'(x)(x-3)-A(x)}{(x-3)^2}+\frac{A(x)}{x-3}=6e^x \\ \Rightarrow A'(x)-\frac{A(x)}{x-3}+\frac{A(x)}{x-3}=6e^x \\ \Rightarrow A'(x)=6e^x \Rightarrow A(x)=6e^x+D$$
Therefore,the solution is:
$$y(x)=\frac{6e^x+D}{x-3}+\frac{A}{x-3}=\frac{6e^x+E}{x-3}, \text{where E=A+D is a constant}$$
