What ansatz should I try when $t^{-n}$ is in the non-homogeneous equation of a system of ODEs? So I know how to find particular solutions for many systems of ODE, but I cannot solve a particular type using the method of undetermined coefficients because I do not know what ansatz to use.
Basically, in examples like the ones below:
$$\frac{d\vec{x}}{dt}=\left( {\begin{array}{cc}
   4 & -2 \\
   8 & -4 \\
  \end{array} } \right)\vec{x}+\left( {\begin{array}{cc}
   1 \\
   0 \\
  \end{array} } \right)t^{-3}+\left( {\begin{array}{cc}
   0 \\
   -1 \\
  \end{array} } \right)t^{-2},\;\;\; t>0$$
the particular solution is $\vec{y}_p(t)=-\frac{1}{2}\left( {\begin{array}{cc}
   1 \\
   0 \\
  \end{array} } \right) t^{-2}+\left( {\begin{array}{cc}
   2 \\
   5 \\
  \end{array} } \right)t^{-1}-2\left( {\begin{array}{cc}
   1 \\
   2 \\
  \end{array} } \right)\ln(t)-2\left( {\begin{array}{cc}
   1 \\
   2 \\
  \end{array} } \right)$, but I do not know where $\ln(t)$ comes from. How do I find the ansatz to find the particular solution when negative powers are being used?
Thanks a lot!
 A: Don't try any ansatz, just solve. Namely, find two solutions $\vec{x}_1$, $\vec{x}_2$ of the homogeneous problem and form out of them the fundamental matrix $X_h$. Next looking for the solution of the inhomogeneous problem $\vec{x}'=A\vec{x}+\vec{f}$ in the form $X_h \vec{c}(t)$, one obtains an equation for $\vec{c}(t)$:
$$X_h\vec{c}\,'=\vec{f}\quad \Longrightarrow\quad \vec{c}=\int^t X_h^{-1}(s)\vec{f}(s)\,ds.$$
In your case $\vec{x}_1=\left(\begin{array}{c}1+4t \\ 8t & \end{array}\right)$, $\vec{x}_2=\left(\begin{array}{cc} -2t \\  1-4t\end{array}\right)$, hence
$$X_h=\left(\begin{array}{cc}1+4t & -2t \\ 8t & 1-4t\end{array}\right)\quad \Longrightarrow \quad X_h^{-1}=\left(\begin{array}{cc}1-4t & 2t \\ -8t & 1+4t\end{array}\right),$$
and therefore the complete solution - say, for the non-homogeneity of the form $\left(\begin{array}{c}0  \\ -1 \end{array}\right)t^{-2}$  - will have the following form:
\begin{align}
\vec{x}(t)&=X_h(t)\int^t \left(\begin{array}{cc}1-4s & 2s \\ -8s & 1+4s\end{array}\right)\left(\begin{array}{c}0  \\ -1 \end{array}\right)s^{-2}ds=\\
&=X_h(t)\int^t\left(\begin{array}{c} -2s^{-1}  \\ -s^{-2}-4s^{-1} \end{array}\right)ds=\\
&=\left(\begin{array}{cc}1+4t & -2t \\ 8t & 1-4t\end{array}\right)\left(\begin{array}{c} -2\ln t+c_1  \\ t^{-1}-4\ln t+c_2 \end{array}\right).
\end{align}

Remark:  If you absolutely want to do this exercise with undetermined coefficients, the approach is as follows. For constant $A$ one obtains the same powers of $t$ in $X^{−1}_h$ as in $X_h$ (here $1$ and $t$). By multiplying them by the nonhomogeneity of the form $t^{−2}$ and integrating the only terms that can arise are $t^{−1}$ and $\ln t$. Hence to find a particular solution you should try their linear combination. For the non-homogeneity of the form $t^{−3}$ you would similarly have to try a linear combination of $t^{−2}$ and $t^{−1}$.
