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The generalized exponential response formula tells us that a particular solution for
\begin{equation}
p(D)y=e^{\alpha t}
\end{equation}
is either
\begin{equation}
y_p=\frac{e^{\alpha t}}{p(\alpha)}
\end{equation}
or if $\alpha$ is a root of the characteristic polynomial, then there's a positive integer $m$ such that $p(\alpha)=p'(\alpha)=\ldots=p^{(m-1)}(\alpha)=0$, and $p^{(m)}(\alpha)\neq 0$. The solution is then
\begin{equation}
y_p=\frac{t^me^{\alpha t}}{p^{(m)}(\alpha)}
\end{equation}
tl;dr: multiply by $t$ and differentiate the characteristic polynomial until it works.

Similarly, we examine the differential equation (called an Euler equation)
\begin{equation}
a_0y+a_2xy'+\ldots+a_nx^ny^{(n)}=x^\beta
\end{equation}
define $q(r)$ to be the Euler characteristic polynomial, or indicial polynomial (which can be calculated by substituting $x^r$ into the differential equation)
\begin{equation}
q(r)=a_0+a_1r+a_2r(r-1)+\ldots+a_nr(r-1)\ldots(r-n+1)
\end{equation}
\begin{equation}
q(r)=a_0(r)_0+a_1(r)_1+a_2(r)_2+\ldots+a_n(r)_n
\end{equation}
The solution to the Euler equation will be either
\begin{equation}
y_p=\frac{x^\beta}{q(\beta)}
\end{equation}
or if $\beta$ is a root of the indicial equation, then there's a positive integer $m$ such that $q(\beta)=q'(\beta)=\ldots=q^{(m-1)}(\beta)=0$, and $q^{(m)}(\beta)\neq 0$. The solution is then
\begin{equation}
y_p=\frac{\ln(x)^mx^\beta}{q^{(m)}(\beta)}
\end{equation}
tl;dr: multiply by $\ln(x)$ and differentiate the indicial equation until it works.

Sat, 12 Dec 2020 03:38 GMT