ODEs with constant coefficients For what value(s) of $A$, if any, will $y = At\mathrm{e}^{-2t}$ be a solution of the differential equation
$$
2y'+ 4y = 3\mathrm{e}^{-2t}?
$$
For what value(s) of $B$, if any, will $y = B\mathrm{e}^{-2t}$ be a solution?
When attempting to find $A$ I got the equation
    $$-4At+At+3=0$$ 
When attempting to find $B$ I got the equation
     $$-4B+4B-3=0$$  
according to my results the answer would be there is no value of A but I feel that its wrong.
Can someone help?
 A: Careful when differentiating.  With $y=Ate^{-2t}$ I got,
$$y'=-2Ate^{-2t}+Ae^{-2t} $$
and so when we plug that into the differential equation we are looking for values of $A$ such that the following equation is satisfied,
$$ 2(-2Ate^{-2t}+Ae^{-2t})+4Ate^{-2t}=3e^{-2t} $$
So we are looking for $2A=3$.  So I do think there is a solution for $A$.  You do have the correct solution method though!  Then do the same thing for $B$.
A: Letting $D$ denote the differentiation operator, your equation is
$$
               (D+2)y=\frac{1}{2}e^{-2t}.
$$
$(D+2)$ annhilates $e^{-2t}$. So $Be^{-2t}$ is never a solution by itself, but it can be added to any particular solution of the above to get another. The general solution of the above must have the form $Ate^{-2t}+Be^{-2t}$ because any solution of the above is also a solution of
$$
           (D+2)^{2}y=\frac{1}{2}(D+2)e^{-2t}=0.
$$
The constant $B$ is arbitrary, but the constant $A$ can be only one thing. There exists a constant $A$, and it must be unique. As you say, $A=3/2$. So the general solution is
$$
                  y(t)=\frac{3}{2}te^{-2t}+Be^{-2t},
$$
where $B$ can be anything.
A: You are right there are no values of $A$ and $B$ that for which your $y$'s are solutions. Actually the solution is a linear function $y = 3t +C$.
