Differential Equation Special Case I have a DEQ that has a special case.
$$\dfrac{d^2y}{dx^2} +4 \dfrac{dy}{dx} + 4y = e^{-2x}$$
I understand my general solution will be: $ Ae^{-2x} + Bxe^{-2x} $, however my problem is dealing with the particular for the $e^{-2x} $ on the RHS. I thought of this: $ Ce^{-2x} + Dxe^{-2x} + Ex^2e^{-2x}$. How would I deal with the $e$ on the RHS and solve the DE EQ?
 A: You have already correctly shown the homogeneous solution to be $$y_h = Ae^{-2x}+Bxe^{-2x}$$ But when you're looking to build a particular solution you might first think that $$y_p = Ce^{-2x}$$ But that isn't right since there is an $e^{-2x}$ term in the homoegeneous, so now multiply by $x$, and you know see that this too exists in the homoegeneous solution, now finally multiply by another $x$ and you have the particular solution you're looking for $$y_p = Cx^2e^{-2x}$$
A: $$\dfrac{d^2y}{dx^2} +4 \dfrac{dy}{dx} + 4y = e^{-2x}$$
You found the general solution  $ Y(x)=Ae^{-2x} + Bxe^{-2x}$ of the homogeneous equation : $\dfrac{d^2Y}{dx^2} +4 \dfrac{dY}{dx} + 4Y = 0$
If you look for only one particular solution $y_p$ of the inhomogeneous equation, no need for $ Ae^{-2x}$ and $Bxe^{-2x}$ Only one is sufficient : $ Ae^{-2x}$ or $Bxe^{-2x}$.
For example, consider $ Ae^{-2x}$ and replace the constant $A$ by an unknown function $f(x)$, i.e.: $$y_p=f(x)e^{-2x}$$
$y_p'=(f'-2f)e^{-2x}$ and $y_p''=(f''-4f'+4f)e^{-2x}$
puting them into the inhomogeneous equation gives :
$(f''-4f'+4f+4(f'-2f)+4f)e^{-2x}=e^{-2x}$ which is simplified to :
$$f''=1$$
$$f=\frac{x^2}{2}+c_1 x+c_2$$
$$y=(\frac{x^2}{2}+c_1 x+c_2)e^{-2x}$$
A: You can write $(\frac{d}{dx} + 2)f = e^{-2x}\frac{d}{dx}(e^{2x}f)$. Your equation is
$$
        \left(\frac{d}{dx}+2\right)^2f = e^{-2x} \\
           \left(e^{-2x}\frac{d}{dx}e^{2x}\right)^2f = e^{-2x}
$$
Now apply another factor of $(\frac{d}{dx}+2)$ to both sides:
$$
         \left(e^{-2x}\frac{d}{dx}e^{2x}\right)^3f = \left(e^{-2x}\frac{d}{dx}e^{2x}\right)e^{-2x}=0 \\
                  e^{-2x}\frac{d^3}{dx^3}(e^{2x}f)=0, \\
            e^{2x}f = A+Bx+Cx^2 \\
               f = (A+Bx+Cx^2)e^{-2x}
$$
When you plug into the original equation
$$
         \left(e^{-2x}\frac{d}{dx}e^{2x}\right)^2\left[(A+Bx+Cx^2)e^{-2x}\right] \\
     = e^{-2x}\frac{d^2}{dx^2}(A+Bx+Cx^2)
     = 2Ce^{-2x}
$$
So you want $2C=1$ or $C=1/2$. The general solution of your problem is
$$
                f = \left(A+Bx+\frac{1}{2}x^2\right)e^{-2x}
$$
A: You can write $(\frac{d}{dx} + 2)f = e^{-2x}\frac{d}{dx}(e^{2x}f)$. Your equation is
$$
        \left(\frac{d}{dx}+2\right)^2f = e^{-2x} \\
           \left(e^{-2x}\frac{d}{dx}e^{2x}\right)^2f = e^{-2x}
$$
Now apply $(\frac{d}{dx}+2)$ to both sides:
$$
         \left(e^{-2x}\frac{d}{dx}e^{2x}\right)^3f = \left(e^{-2x}\frac{d}{dx}e^{2x}\right)e^{-2x}=0 \\
                  e^{-2x}\frac{d^3}{dx^3}(e^{2x}f)=0, \\
            e^{2x}f = A+Bx+Cx^2 \\
               f = (A+Bx+Cx^2)e^{-2x}
$$
When you plug into the original equation
$$
         \left(e^{-2x}\frac{d}{dx}e^{2x}\right)^2\left[(A+Bx+Cx^2)e^{-2x}\right] \\
     = e^{-2x}\frac{d^2}{dx^2}(A+Bx+Cx^2)
     = 2Ce^{-2x}
$$
So you want $2C=1$ or $C=1/2$. The general solution of your problem is
$$
                f = \left(A+Bx+\frac{1}{2}x^2\right)e^{-2x}
$$
A: I would opt to use differential operators in this case.
This consist of finding $y_p$ (particular solution) and $y_c$ (complementary function).
Attempting $y_p$
The formula is as follows
$\frac{1}{(D-a)^r}$$e^{ax}$=$\frac{x^r}{r!}$$e^{ax}$
So for $y_p$($D^2$+$4D$+$4$)=$e^{-2x}$
$y_p$=$\frac{e^{-2x}}{D^2+4D+4}$=$\frac{x^2}{2!}$$e^{-2x}$
=$\frac{x^2}{2}$$e^{-2x}$
Finding $y_c$ is (A+B$x$)$e^{-2x}$
Hence $y$=$y_p$+$y_c$= (A+B$x$+$\frac{x^2}{2}$)$e^{-2x}$
