Solving $\frac{dy}{dx} = ae^{-bx} - cy(x)$ How would you solve an equation in the form
$$\frac{dy}{dx} = ae^{-bx} - cy(x) $$
where $a, b, c$ are just constants. My ultimate goal is to find $y(x)$ without the derivative in there. 
My confusion comes from the fact that the right hand side has both $y(x)$ and $x$ itself in it. I tried using the integrating factor method and it gets me a similar form of solution that I want, but not completely. So is this the method I should be using or is there another that works for this type of equation?
 A: $$y'+cy = ae^{-bx} \tag 1$$
Solving with the variation of parameter method :
First, solve the associated homogeneous ODE 
$$\quad y'+cy = 0 \tag 2$$
The solution is :
$$y=\lambda e^{-cx}$$
where $\lambda$ is a constant with respect to $x$.
Second, apply the method of variation of parameter. This means that the constant $\lambda$ is now considered as a function of $x$.
$y=\lambda(x) e^{-cx}$ is no longer solution of Eq.$(2)$, but will be solution of Eq.$(1)$ :
$y'=\lambda'e^{-cx}-c\lambda e^{-cx}\quad$ Putting it into Eq.$(1)$ :
$$y'+cy = ae^{-bx}=(\lambda'e^{-cx}-c\lambda e^{-cx})+c(\lambda e^{-cx})$$
$$ae^{-bx}=\lambda'e^{-cx}$$
$$\lambda'=ae^{(c-b)x}$$
$$\lambda=\frac{a}{c-b}e^{(c-b)x}+C$$
$y=\left(\frac{a}{c-b}e^{(c-b)x}+C \right) e^{-cx}$
$$y(x)=\frac{a}{c-b}e^{-bx}+Ce^{-cx}$$
A: Hint :
It is :
$$y'(x) + cy(x) = ae^{-bx}$$
Find an integrating factor $\mu$ as 
$$\mu(x) = \exp\left( \int c\mathrm{d}x\right) = e^{cx}$$
and multiply both sides of the ODE with it. It yields :
$$e^{cx}y'(x) + ce^{cx}y(x) = ae^{cx-bx} $$
Now, observe that :
$$e^{cx}y'(x) + cy(x) = \Big( e^{cx}y(x) \Big)'$$
Can you finish ?
