Trial Solutions for Differential Equations

Question

I have a question from my Differential Equations & Linear Algebra class. When you're trying to find the general solution to an nth order linear non-homogeneous differential equation, you have to find a trial solution to solve it (at least until you get to variation of parameters later in the same chapter) and I assume that the lack of information is due to people usually preferring variation of parameters, but can someone just explain to me the general way to find the form of a trial solution for a problem? Say you have the equation:

 y''-y'-2y = 5e^(2x)

I know the complementary solution would be

 Yc = c1*e^(-x)+c2*e^(2x)

when you're treating the left-hand side as a homogeneous equation, but then once you get to the right hand side, I'm pretty lost here. I know it has the annihilator (D - 2) where D is the differential operator, but how do I turn that into the trial solution? Sorry for the length, I just wanted to provide as much information as possible for accuracy. Thanks in advance everyone.

Answer 1

The equation is $$ (D-2)(D+1)y = 5e^{2x} $$ $D-2$ annihilates the right side, which gives $$ (D-2)^{2}(D+1)y = 0. $$ Applying $D-2$ introduces additional solutions, but the general solution of the second equation has the form $$ y=Ae^{2x}+Bxe^{2x}+Ce^{-x} $$ Plug this back into the original equation, and notice that $(D-2)(D+1)$ annhilates the terms with $A$ and $C$. So $B$ is determined by the equation $$ (D-2)(D+1)Bxe^{2x}=5e^{2x} $$ A convenient trick is to write $D+1=D-2+3$: \begin{align} (D-2)(D+1)Bxe^{2x}&=\{(D-2)^{2}+3(D-2)\}Bxe^{2x}\\ &=3(D-2)Bxe^{2x}=3(e^{2x}De^{-2x})Bxe^{2x} \\ &=3Be^{2x}=5e^{2x} \implies B=5/3. \end{align} The general solution is $$ y=Ae^{2x}+\frac{5}{3}xe^{2x}+Be^{-x}. $$

Answer 2

Lets try a particular solution of the form $Axe^{2x}$.

$y_p(x) = Axe^{2x}$

$y_p'(x) = A e^{2x} + 2Axe^{2x}$

$y_p''(x) = 2A e^{2x} + 2Ae^{2x} + 4Axe^{2x} = 4Ae^{2x} + 4A xe^{2x}$

$y_p'' - y_p' -2y_p = (4A - 2A - 2A) xe^{2x} + (4A - A)e^{2x} = 3Ae^{2x} = 5e^{2x}$

After which I can solve for $A$, and declare the general solution:

$y = C_1 e^{-x} + C_2 e^{2x} + \frac{5}{3} xe^{2x}$.

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The intuition behind this is the following: normally I'd try $e^{2x}$ for the form of the particular equation, but this is already a solution to the homogeneous one. When the characteristic equation admits roots with multiplicity $>1$, the standard ansatz is $e^{ax}, xe^{ax}, \dots$, so let's try that, and see where we get. In this case this works pretty well. I'm sure someone else can point to exactly how general this can get, but it certainly will work for equations of the form above. There's an obvious extension, that the initial ansatz should be $Ax^n e^{ax}$, where both $n$ and $A$ are to be determined.