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?
 A: $$(x-3) \frac{dy}{dx}+y=0 \Rightarrow \frac{dy}{dx}=\frac{-y}{x-3} \Rightarrow -\frac{dy}{y}=\frac{dx}{x-3} \Rightarrow \int \left ( \frac{-1}{y} \right)dy=\int \frac{1}{x-3} dx \\ \Rightarrow -\ln |y| =\ln |x-3|+c \Rightarrow e^{-\ln |y|}=e^{\ln |x-3|+c} \Rightarrow \frac{1}{|y|}=C |x-3| \Rightarrow |y|=\frac{c'}{|x-3|}\\ \Rightarrow y= \pm \frac{c'}{x-3} \Rightarrow y=\frac{A}{x-3} $$
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:
$$y(x)=\frac{6e^x+D}{x-3}+\frac{A}{x-3}=\frac{6e^x+E}{x-3}, \text{where E=A+D is a constant}$$
A: This is a standard linear ordinary differential equation
To solve it we note:
$$ \frac{dy}{dx} + \frac{1}{x - 3}y = \frac{1}{x-3}6e^x$$
We now seek an integration factor $w(x)$ which can be multiplied on both sides to form (just observing left hand side)
$$w(x) \frac{dy}{dx} + \frac{w(x)}{x-3}y = p(x)y' + p'(x)y = (p(x)y)'$$
It becomes clear that $p(x) = w(x)$ and that if $w(x) = e^{\int \frac{1}{x-3}} = x - 3$ we will be good to go
The resulting expression is after integration
$$((x-3)y)' = 6e^x \rightarrow (x-3)y = 6e^x + C$$
And finally:
$$y = \frac{6e^x + C}{x-3}$$
look up this general method, its very standard
A: $$(x-3)(\frac{dy}{dx})+y=6e^x \Rightarrow \frac{dy}{dx}+\frac{1}{x-3}y=\frac{6 e^x}{x-3} \Rightarrow y'(x)+\frac{1}{x-3}y(x)=\frac{6 e^x}{x-3} $$
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 $$y=y_h+y_p$$
A: There are plenty of good standard-type answers here about solving the general linear differential equation of the form you are presented with; I just want to make a small note that, from a calculus perspective, finding a specific solution you needn't bother with the high-powered approach of integrating factors, but rather can recognize $(x-3) \frac{dy}{dx} + y = (x-3) \frac{dy}{dx} + \frac{d (x-3)}{dx} y$ as the product expansion form of $\frac{d}{dx} \left( (x-3) \cdot y \right)$. Then, the equation becomes
$$
\frac{d}{dx} \left( (x-3) \cdot y \right)
=
6 e^x
= 6 \frac{d}{dx} e^x
= \frac{d}{dx} 6 \cdot e^x
.
$$
Then via unique anti-derivatives (up to a constant of integration) or simply the fundamental theorem of calculus, we have that 
$$
(x-3)y 
= 6 \cdot e^x + C
\implies
y
=
\frac{6 e^x + C}{x-3}
.
$$
This method is more directed towards a calculus student who has been given this problem without seeing the theory related to solving ordinary differential equations, and wants to know how they might be able to bring their calculus orientation, perspective, and techniques in order to solve the problem.
To connect this with the other solutions offering the complete solution consisting of the so-called "homogenous solution" and the "particular solution", note that this solution is to that of the particular equation. The idea of the homogenous solution is that it is both consistent with the original equation (because, the homogenous solution is when the left hand of the equation vanishes, and so if you add that 'solution' to the solution satisfying the original equation, it still works), but tells you more information about what the solutions of the differential equation are/look like, than just the particular solution alone - it "completes the picture", if you will, by telling you on what functions the differential expression on the left vanishes.
