For nonhomogeneous linear systems of the form \acute{x} = Ax +b where A is a matrix with constant entries, state the natural guess to use for x_{p} when the vector b is
(a) b =\begin{bmatrix} e^{-t} \\ 2e^{-t}\end{bmatrix}, (b) b =\begin{bmatrix} 1 \\ t\end{bmatrix}, (c) b =\begin{bmatrix} t^{2} \\ 0\end{bmatrix}, (d) b =\begin{bmatrix} e^{-3t} \\ -2\end{bmatrix}
(a) With b = [e^{−t} 2e^{−t} ]^{T}, it is natural to expect that any particular solution must involve e^{−t} in its components. Specifically, we make the guess that
x_{p} =\begin{bmatrix} Ae^{-t} \\ Be^{-t}\end{bmatrix}
and substitute directly into x´= Ax +b in order to attempt to find values of A and B for which x_{p} satisfies the given system.^{3}
(b) Given b = [1 t ]^{T}, we must account for the fact that x_{p} and its derivative can involve constant and linear functions of t . In particular, we suppose that
x_{p}=\begin{bmatrix} At +B \\ Ct +D\end{bmatrix}
and substitute appropriately in an effort to determine A, B, C, and D.
(c) For b = [t^{2} 0]^{T}, with one quadratic term present in b, it is necessary to include quadratic terms in each entry of x_{p}. But since the derivative of x_{p} will be taken, linear terms must be included as well. Finally, once linear terms are included, for the same reason we must permit the possibility that constant terms can be present in x_{p}. Therefore, we guess the form
x_{p}=\begin{bmatrix} At^{2} +Bt +C \\ Dt^{2} +Et +F\end{bmatrix}
d) With b = [e^{−3t} −2]^{T} having both an exponential and constant term present, we account for both of these scalar functions and their derivative by assuming that
x_{p}=\begin{bmatrix} Ae^{−3t} +B \\ Ce^{−3t} +D\end{bmatrix}
3:It is possible that the guess can fail to work, in which case a modified form for x_{p} is required. One setting where this may occur is when λ=−1 is an eigenvalue of A, whereby a vector involving e^{−t} already appears in the complementary solution x_{h}. See exercise 8 for further investigation of this issue.