ODE problem involving differential operator, substitution I am not sure how to even begin approaching this problem. My professor mostly only  introduced the idea of the differential operator, but didn't solve anything similar. Could someone just point me in the right direction or explain the idea to solve? 
By using the differential operator $$D = \frac{d}{dt}$$ we can express $$y''-4y'+4y = te^{2t}$$ as $$(D-2)^2y = te^{2t}$$ 
Solve this differential equation by setting $$ u = (D-2)y $$ and solving $$(D-2)u = te^{2t} $$ first. 
My attempt:
By solving $$(D-2)u = te^{2t} $$ I found that $$u=(D-2)y=\frac{1}{2}t^2e^{2t}+ce^{2t}$$
I'll call this equation 1. 
I first tried to find (D-2) by dividing both sides of Eq. 1 by y. Then substituting what I found into below:
$$(D-2)(D-2)y=te^{2t}$$ 
But I found
$$y=e^{2t}(\frac{1}{4}t^4+ct+c^2t^{-1})$$ 
Which Wolfram Alpha seems incorrect. What am I doing wrong here?

Note the following equivalence:
$$(D-2)u\equiv \frac{du}{dt}-2u$$
Start by writing:
$$(D-2)u=te^{2t} \implies \frac{du}{dt}-2u=te^{2t} \tag{1}$$
This is a linear ODE, which you can solve with an integrating factor:
$$\mu(t)=e^{\int -2~dt}=e^{-2t}$$
Hence:
$$e^{-2t}\cdot \frac{du}{dt}-2e^{-2t}\cdot u=t$$
$$\int \frac{d}{dt}(e^{-2t}\cdot u)~dt=\int t~dt$$
$$e^{-2t}\cdot u=\frac{t^2}{2}+c_1$$
This implies that the general solution for $u$ is:
$$u=c_1e^{2t}+\frac{1}{2}e^{2t}t^2 \tag{2}$$
Now, we must substitute for $u=(D-2)y=\frac{dy}{dt}-2y$ to obtain the general solution for $y(t)$:
$$\frac{dy}{dt}-2y=c_1e^{2t}+\frac{1}{2}e^{2t}t^2 \tag{3}$$
This can be solved again using an integrating factor (In a very similar way). Can you continue?

After your edit to include your attempt:
Instead of substituting into the second order ODE, substitute $u=(D-2)y$ into $(2)$ instead to obtain a linear differential equation as on $(3)$. You can solve $(3)$ using an integrating factor.
You can find the integrating factor to be the same as last time:
$$\mu_2(t)=e^{\int -2~dt}=e^{-2t}$$
$$e^{-2t}\cdot \frac{dy}{dt}-2e^{-2t}\cdot y=c_1+\frac{1}{2}t^2$$
$$\int \frac{d}{dt}(e^{-2t}\cdot y)~dt=\int \left(c_1+\frac{1}{2}t^2\right)~dt$$
Can you continue? I checked Wolfram|Alpha just to be sure, and the solution I obtain is the same.
