Differential Equation $y'' - 4y' + 4y = 0$ [1] $y'' - 4y' + 4y = 0$
Usually problem like these will have the answer in the form $C_1e^a + C_2e^b ... $
where $a $ and $b$ are the roots of the characteristic equation $e^{rt}$
$$ y = e^{rt} $$
$$ y' = re^{rt}.. y'' = r^2 e^{rt} $$ 
$$ r^2e^{rt} - 4e^{rt} + 4e^{rt} = 0$$
$$ e^{rt}(r-2)(r-2) = 0$$
$$ y= C_1e^{2t} + C_2e^{2t}$$
However, this is not correct as the answer is $ y = C_1e^{2t} + C_2te^{2t}$ ! I just don't know why.
[2] For a similar problem , $y'' + 3y' - 4y = 0 $ I did the exact same thing and the answer is $$y = c_1e^t + c_2e^{-4t} $$
How are [1] and [2] different? They look the same, why does [2]'s solution have an extra factor of t.
 A: Consider $\rm D$ as the derivation operator on the space of functions you want to study. This just says that $\mathrm D f = f'$. 
Then your differential equation can be written as "Find the kernel of the linear operator $\rm D^2 - 4 \rm D + 4$". The theory of differential equation tells you that this kernel has dimension 2.
Let's find a basis of it, $\rm D^2 - 4 D + 4$ can be rewritten as $(\rm D - 2)^2$, so if I know a basis of the kernel of $\rm D - 2$ it will be part of a basis we want to find. Here it is the function $x \mapsto e^{2x}$. But this is not enough, because the kernel of $\rm D -2$ has only dimension 1. Then we need to add $x \to x e^{2x}$ which is not leaving in the kernel of $\rm D - 2$.
To compare with your second example, here you want to find the kernel of $\rm D^2 + 3 D - 4$ which is $(\rm D - 1)(D + 4)$.
Now, as before, the kernel has dimension 2, but as the characteristic equation has two distinct roots, you can take a vector from the basis of $\rm D -1$ and complete with another vector of the basis of the kernel of $\rm D + 4$.
In conclusion : it all depends on whether the characteristic equation has multiple roots or not.
A: You have the following differential equation:
$$ \frac{d^{2}y}{dt^{2}}-4\frac{dy}{dt}+4y=0$$
Applying the laplace transform:
$$ Y(s)(s^{2}-4s+4)=sy(0)+\frac{\text{d}y}{\text{d}x}(0)-y(0) $$
$$ \frac{\text{d}y}{\text{d}x}(0)+y(0)$$ is a constant
$$ y(0)$$ a constant as well.
Rearranging:
$$ Y(s)=\frac{sK_{1}+K_{2}}{(s-2)^{2}} $$
I can divide Y(s) into simple fractions:
$$ Y(s)=\frac{C_{1}}{(s-2)^{2}}+\frac{C_{2}}{(s-2)}$$
Doing the inverse laplace transform:
$$ y(t)=C_{1}te^{2t}+C_{2}e^{2t}$$
A: The difference is that [1] has a repeated root. When you have a root $r$ that is repeated $n$ times, its contribution to the solution is of the form:
$$
(c_0 + c_1 t + \cdots + c_n t^n) e^{r t}
$$
When $n = 1$, this becomes:
$$
(c_0 + c_1 t) e^{r t}
$$
A: As the root is repeated,
$$C_1e^{rt}+C_2e^{rt}$$ can be regrouped as 
$$(C_1+C_2)e^{rt}=Ce^{rt}$$
which is lacking one constant, and an extra term must be found (because the solution requires two linearly independent functions).
Let us assume for a moment that the roots are close but differby $h$, and the solution is
$$C_1e^{rt}+C_2e^{(r+h)t}=e^{rt}(C_1+C_2e^{ht})\approx e^{rt}\left(C_1+C_2+C_2ht+\frac{C_2}2h^2t^2+\frac{C_2}{3!}h^3t^3+\cdots\right)$$
If we let $h$ tend to $0$, and at the same time keep $C_2h=K_2$ and $C_1+C_2=K_1$, the function tends to
$$e^{rt}(K_1+K_2t).$$
As we can check,
$$(e^{rt}(K_1+K_2t))''-2r(e^{rt}(K_1+K_2t))'+r^2(e^{rt}(K_1+K_2t))=\\
(r^2-2r^2+r^2)e^{rt}K_1+(2r-2r+((r^2-2r^2+r^2)t)e^{rt}K_2=0.$$
