A forced system $\mathbf{x}' = A\mathbf{x} + \mathbf{f}(t)$ has the same structure as a forced single equation: every solution is one particular solution plus the general solution of $\mathbf{x}' = A\mathbf{x}$ from eigenvalue solutions. And the particular solution can often be guessed, as in undetermined coefficients.
Key idea
$\mathbf{x} = \mathbf{x}_p + c_1e^{\lambda_1t}\mathbf{v}_1 + c_2e^{\lambda_2t}\mathbf{v}_2$. For a constant forcing $\mathbf{b}$, the equilibrium $\mathbf{x}_p = -A^{-1}\mathbf{b}$ works. For $e^{kt}\mathbf{g}$ with $k$ not an eigenvalue, try $\mathbf{x}_p = e^{kt}\mathbf{a}$ with $(kI - A)\mathbf{a} = \mathbf{g}$.
Why it works
If $\mathbf{x}$ and $\mathbf{x}_p$ both solve the forced system, their difference solves $\mathbf{x}' = A\mathbf{x}$, by linearity. A constant $\mathbf{x}_p$ has $\mathbf{x}_p' = \mathbf{0}$, so it needs $A\mathbf{x}_p + \mathbf{b} = \mathbf{0}$, solvable when $A$ is invertible (no zero eigenvalue). For $e^{kt}\mathbf{a}$: substituting gives $ke^{kt}\mathbf{a} = e^{kt}A\mathbf{a} + e^{kt}\mathbf{g}$, so $(kI - A)\mathbf{a} = \mathbf{g}$, which has a unique solution unless $k$ is an eigenvalue (the resonance case, where extra terms with $t$ are needed).
Worked example: Constant forcing
Find the equilibrium of $\mathbf{x}' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\mathbf{x} + \begin{bmatrix} -5 \\ -4 \end{bmatrix}$.
- Solve $A\mathbf{x}_p = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$: $x + 2y = 5$, $2x + y = 4$.
- Subtracting twice the first from the second: $-3y = -6$, so $y = 2$, $x = 1$.
Answer: $\mathbf{x}_p = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$
Worked example: The full solution
Solve that system with $\mathbf{x}(0) = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$.
- From the eigen-solutions lesson: $\mathbf{x}_h = c_1e^{3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2e^{-t}\begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
- At $t = 0$: $c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \mathbf{x}(0) - \mathbf{x}_p$ $= \begin{bmatrix} 3 \\ 1 \end{bmatrix}$, so $c_1 = 2$, $c_2 = 1$.
Answer: $x_1 = 1 + 2e^{3t} + e^{-t}$, $x_2 = 2 + 2e^{3t} - e^{-t}$
Worked example: Exponential forcing
Find $\mathbf{x}_p = e^{2t}\mathbf{a}$ for $\mathbf{x}' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\mathbf{x} + e^{2t}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
- $2I - A = \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}$ (invertible: 2 is not an eigenvalue).
- Solve $a_1 - 2a_2 = 1$, $-2a_1 + a_2 = 0$: $a_2 = 2a_1$, so $-3a_1 = 1$.
Answer: $\mathbf{a} = \begin{bmatrix} -\frac{1}{3} \\ -\frac{2}{3} \end{bmatrix}$
Common mistake
Fitting the initial condition to $\mathbf{x}_h$ alone: the constants must make $\mathbf{x}_p(0) + \mathbf{x}_h(0) = \mathbf{x}(0)$, so subtract $\mathbf{x}_p(0)$ first. Also check the sign in the equilibrium: it solves $A\mathbf{x}_p = -\mathbf{b}$, not $A\mathbf{x}_p = \mathbf{b}$.