Stuck at second order differential equation Solve the equation 
\begin{equation}
y''-2y'+y=t^3\\
y(0)=1\\
y'(0)=0
\end{equation}
Solved the homogeneous equation $y''-2y
+y=0$, which gives the general solution
$y_0(t)=Ae^t+Bte^t\\$
Then, the solution supposedly involves a $y_p$ such that 
\begin{equation}y(t)=y_p+Ae^t+Bte^t\end{equation}
I thought the solution could be on the form $at^3+bt^2+ct+d$ since this is of the same order, but plugging it into the equation just gives me gibberish back. Am I at least on the right track? Thank you for your feedback :)
edit:
All right so I went all the way through with the polynomial of degree 3: Assuming that
\begin{equation}
y_p=at^3+bt^2+ct+d
\end{equation}
I plug it into the equation and get 
\begin{equation}
(6at+2b)-2(3at^2+2bt+c)+at^3+bt^2+ct+d=t^3
\end{equation}
Rearranging, I get
\begin{equation}
at^3+(b-6a)t^2+(6a-4b+c)t+(2b-2c+d)=t^3
\end{equation} which implies that
$a=1$, $b=6$, $c=18$, $d=24$ (assuming I did the calculations correctly)
Plugging back in, I get
\begin{equation}y(t)=t^3+6t^2+18t+24+Ae^t+Bte^t\end{equation} Remembering our initial conditions, we get $A=-23$ and $B=5$ if I'm not mistaken. Then, the solution sounds:
\begin{equation}
y(t)=\underline{t^3+6t^2+18t+24-23e^t+5te^t}
\end{equation}
Did I do this correctly? Again, thank you for your wonderful responses :)
edit 2: fixed bad algebra
 A: make for $y_p$ the ansatz $$y_p=At^3+Bt^2+Ct+D$$ where $$A,B,C,D$$ are real variables
A: that is the correct approach.  double check your differentiation, and remember to collect like powers
y = at^3 + bt^2 + ct + d
y' = 3at^2 + 2bt + c
y'' = 6at + 2b
y'' -2y' + y = (6at + 2b) - 2(3at^2 + 2bt + c) + at^3 + bt^2 + ct + d
= at^3 + (-6a + b)t^2 + (6a -4b + c)t - 2c+d = t^3
so you get
d-2c  = 6a-4b + c = -6a + b =0
a = 1
b = 6
c =  4*b - 6*a = 18
d =36
double check my arithmetic, but that's the process
A: Your approach is correct.
Start by letting:
$$y_p=at^3+bt^2+ct+d$$
Therefore, you must substitute the following in your differential equation:
$$\frac{dy_p}{dt}=3at^2+2bt+c$$
$$\frac{d^2y_p}{dt^2}=6at+2b$$
To obtain:
$$(6at+2b)-2(3at^2+2bt+c)+(at^3+bt^2+ct+d)=t^3$$
Equate the coeficcients of $1$, $t$, $t^2$ and $t^3$ on both sides of the equation, and solve for each of the coefficients $a$, $b$, $c$ and $d$.
Edit: Your trial after substituting into the differential equation
Your coefficients for $a$ and $b$ are correct. However, I think you've made a mistake somewhere when gathering coefficients of $t$ since it should be:
$$at^3+(b-6a)t^2+\color{#bb0000}{(6a-4b+c)t}+(2b-2c+d)=t^3$$
Perhaps you could try evaluating the constants again and let us know what you get.
A: You could also use the method of "variation of parameters".  Since $e^t$ and $te^t$ are solutions to the associated homogeneous equation, we should try $y= u(t)e^t+ v(t)te^t$.  Then $y'= u'e^t+ ue^t+ v'te^t+ ve^t+ vte^t$.  Since there are, in fact, an infinite number of such solutions, we can simplify and reduce the number of possible solutions by requiring that $u'e^t+ v'te^t= 0$.  That leaves $y'= ue^t+ ve^t+ vte^t$.  Then $y''= u'e^t+ ue^t+ v'e^t+ 2ve^t+ vte^t$.  
So $y''- 2y'+ y= u'e^t+ ue^t+ v'e^t+ ve^t+ vte^t- 2ue^t- 2ve^t- 2vte^t+ ue^t+ vte^t= u'e^t+ v'te^t= u'e^t+ v'e^t= t^3$.
We now have the two equations $u'e^t+ v'te^t= 0$ and $u'e^t+ v'e^t= t^3$ to solve for u and v.  Subtracting the first equation from the second, the u' terms cancel leaving $v'(t+ 1)e^t= t^3$ and $v'= \frac{t^3}{e^t(t+ 1)}$.  Integrate that to find v, then put that v' back into either equation and solve for u.
