General solution of second-order differential equation I'm working on finding the general solution of a differential equation. But I'm not getting too far. This is a question from my exam papers, can someone work it out, with simple steps so I can understand it, please. 

Thank you.
 A: Step 1) find a general solution of the homogeneous problem ($y'' -4y' +4y = 0$)
Step 2) apply the "variation of constant" formula to find a particular solution of the inhomogeneous problem.
Step 3) [what you have found in step 1)] + [what you have found in step 2)]
Have you tried this approach?
EDIT:
As you checked $y(t) = c_1e^{2t} + c_2te^{2t}$ is the general solution you have to find in step 1).
Theorem: Let $y(t) = \int_{t_0}^t Y(t - s)f(s)\ ds$, where $Y(t)$ is the solution of the problem 
$$Y'' +aY' + bY = 0$$
$$Y(0) = 0,\ Y'(0) = 1.$$
Then $y(t)$ is a solution of the equation $y'' + ay' + by = f(t)$. Moreover, it is the only solution such that $y(t_0) = y'(t_0) = 0$.
The proof is very easy because you are just asked to calculate the derivatives of $y(t)$ as it is defined and substitute in the equation keeping in mind that $Y$ is the solution of a known problem.
We are going to apply this to your problem!
Let's find $Y$!
It is the solution of
$$Y'' - 4Y' + 4Y = 0$$
$$Y(0) = 0,\ Y'(0) = 1.$$
Therefore $Y(t) = c_1e^{2t} + c_2te^{2t}$, where
$$0 = Y(0) = c_1$$
$$1 = Y'(0) = 2c_2.$$
This gives $Y(t) = \frac{t}{2}e^{2t}$.
According with the theorem above a solutin of the inhomogeneous problem is
$$y(t) = \int_{t_0}^t Y(t - s)f(s)\ ds = \int_{t_0}^t \frac{(t - s)}{2}e^{2(t - s)}se^{5s}\ ds.$$
This ends spep 2)
Then, the general solution of your equation is 
$$y(t) = c_1e^{2t} + c_2te^{2t} + \int_{t_0}^t \frac{(t - s)}{2}e^{2(t - s)}se^{5s}\ ds.$$
You can of course calculate the last integral. You can also choose $t_0 = 0$ if you wish so. A particular choise for $t_0$ is required only when you have to impose cauchy's conditions.
I hope this helps!!!
