Finding a particular solution for $y'''+3y''-4y=e^{-2x}$ Given the ODE $y'''+3y''-4y=e^{-2x}$ i am trying to find a particular solution.
My Approach was $y_p(x)=Ae^{-2x}$ with
$$y_p'(x)=-2Ae^{-2x}$$
$$y_p''(x)=4Ae^{-2x}$$
$$y_p'''(x)=-8Ae^{-2x}$$
But inserting this into the equation i get 0=1
According to WolframAlpha the particular solution is
$$y_p(x)= -\frac{1}6 e^{-2x}x^2$$
My Question is: Where does the $x^2$ come from?
Thanks in advance
 A: $$y'''+3y''-4y=\color {red}{e^{-2x}}$$
Characteristic polynomial is;
$$r^3+3r^2-4=0$$
Since $r=-2$ and $r=1$ are solutions you can easiy factorize:
$$(r-1)(r+2)^2=0$$
Then the solution to the homogeneous equation is :
$$y_h=c_1e^{x}+e^{-2x}(c_2+c_3x)$$
Your guess for the particular solution should be :
$$y_p=Ax^2e^{-2x}$$
A: There are many ways to formulate this but I find it nicer to talk about a non-homogeneous, constant coefficient, linear ODE like this in terms of its differential operator. That is, write your equation as:
$$L[y] = e^{2x},\,\, L = D^3+3D^2-4 = (D-1)(D+2)^2$$
where $Dy = y'$ is the differential operator. Therefore, you can write your original ODE as:
$$(D-1)(D+2)^2[y] = e^{-2x}.$$
But then you notice that $g(x) = e^{-2x}$ is itself annihilated by the operator $D+2$ and so any solution to your original ODE would be a solution to:
$$(D-1)(D+2)^3[y] = (D+2)[e^{-2x}] = 0.$$
But this one is homogeneous and you can write down the general solution easily:
$$y_h = c_1e^x+c_2e^{-2x}+c_3xe^{-2x}+c_3x^2e^{-2x}.$$
However, we know that the first three terms of the above constitute the general solution to the homogeneous version of your original ODE and therefore you can now be certain that $$y_p = C_3x^2e^{-2x}$$
must work.
Note that if you reason this way, then you do not have to memorize the different "guesses" for different non-homogeneous terms given on the right hand side.
