How do you get a complete general solution for a differential like this?
$y^{\prime\prime}+6y^{\prime}+9y=14e^{-3x}$
This is what I have so far for the first part of the problem: $yp=Ce^{-3x}, yp'=-3Ce^{-3x}, yp'' = 9Ce^{-3x}$ and I plug these into:
$ (9Ce^{-3x})+6*-3Ce^{-3x}+9Ce^{-3x}$ for solving for C which gives C=0 which doesnt seem correct at all. Have I made a mistake or is there something I am missing here?
On
Rather, you should find that $$y_p''+6y_p'+9y_p=9Ce^{-3x}-18Ce^{-3x}+9Ce^{-3x}=(9-18+9)Ce^{-3x}=0\cdot Ce^{-3x}=0,$$ and so any choice of constant $C$ will give you a solution to the homogeneous differential equation $$y''+6y'+9y=0.\tag{$\star$}$$
In other words, $Ce^{-3x}$ is a solution to $(\star),$ not a particular solution to the ODE $$y''+6y'+9y=14e^{-3x}.$$ Moreover, $Ce^{-3x}$ is not the general solution to $(\star),$ because the characteristic equation has a repeated root. Instead, your general (and particular) solution should be some polynomial multiplied by $e^{-3x}.$
The equation can be written in terms of $D=\frac{d}{dx}$ as $$ (D+3)^{2}y =14e^{-3x} $$ The annihilator of $e^{-3x}$ is (D+3). Therefore, $$ (D+3)^{3}y = 0. $$ The general solution is $$ y = Ae^{-3x}+Bxe^{-3x}+Cx^{2}e^{-3x} $$ Plugging back into the original equation, and using $(D+3)^{2}(e^{-3x}f)=e^{-3x}D^{2}f$ $$ (D+3)^{2}(Ae^{-3x}+Bxe^{-3x}+Cx^{2}e^{-3x})=14e^{-3x}\\ C(D+3)^{2}(x^{2}e^{-3x}) = 14e^{-3x} \\ Ce^{-3x}D^{2}x^{2} = 14e^{-3x} \\ 2Ce^{-3x} = 14e^{-3x} \\ C = 7. $$ So the general solution involves two arbitrary constants $A$ and $B$, and is given by $$ y(x) = Ae^{-3x}+Bxe^{-3x}+7x^{2}e^{-3x}. $$