I have to apply the method of variation of parameters to find a particular solution of the following system: $x' = A\vec x + \vec f(t)$ for
$$ A = \begin{bmatrix}
2 & -4 \\
1 & -2 \\
\end{bmatrix}
$$
and
$$ \vec f(t) = \begin{bmatrix}
\ln t \\
t\\
\end{bmatrix}
$$
To begin, we have to find the associated fundamental matrix, $\Phi (t)$. We calculate $\Phi (t)$ as follows:
$$det(\lambda I – A) = \begin{bmatrix}
\lambda – 2 & -4 \\
1 & \lambda + 2 \\
\end{bmatrix}$$
which yields the characteristic equation $(\lambda ^ 2 – 4) + 4 = \lambda ^2$; an eigenvalue of 0 with a defect of 1. The only eigenvector we have is that which corresponds to $\lambda = 0$, more specifically, $ \vec v_0 = [1 \quad \frac{1}{2}]^T$. To accommodate this defect, we seek out a length two chain, $(\vec v_1, \vec v_2)$, where each vector is linearly independent and the first lies in the eigenspace of $\vec v_0$. We find $\vec v_2$ by solving for $(A – \lambda I)^2 \vec v_2 = (A)^2 \vec v_2 = \vec 0$. $A^2$ is simply
$$\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\end{bmatrix}$$
so we can choose any arbitrary $\vec v_2$, so I chose $v_2 = [1 \quad 0]^T$. We now find $\vec v_1$ by computing $(A – \lambda I)\vec v_2 = A \vec v_2 = \vec v_1$, which is
$$\vec v_1 = \begin{bmatrix}
2\\
1\\
\end{bmatrix}$$
which does indeed lie in the eigenspace of $\vec v_0$ ($\vec v_1 = 2 * \vec v_0$). We know that our solutions to the linear system will be of the form:
$$x_1 = \vec v_1 e^{\lambda t} = \vec v_1$$
$$x_2 = (\vec v_1 t + \vec v_2)e^{\lambda t} = \vec v_1 t + \vec v_2$$
Thus, the column vectors of our fundamental matrix are $x_1$ and $x_2$, so we conclude
$$\Phi (t) = \begin{bmatrix}
2 & 2t+1 \\
1 & t \\
\end{bmatrix}$$
Did I mess up anywhere? Because I don't seem to be coming out with the correct answer. To find the particular solution, I then proceed to calculate $$x_p (t) = \Phi (t) \int_1^t {\Phi (t) }^{-1} f(t) dt$$ (provided that $x_p (a) = \vec 0$).
Best Answer
This sounds like a homework question, but you may or may not be aware of a cheat which has been used in systems theory for many years. In this case the equation $$\dot{x}(t) = Ax(t) + B(t)u(t),\ t\in[t,t_0]$$ is always solved by $$x(t) = e^{A(t-t_0)}x(t_0)+\int_{t_0}^te^{A(t-\tau)}B(\tau)u(\tau)\ d\tau,$$ for any time-invariant matrix $A$. If $A$ is diagonalizable, as is true in your case, then $$e^{At} = V \Lambda(t) V^{-1}$$ where $V$ is a matrix of the eigenvectors and $\Lambda$ is a diagonal matrix whose $i$th diagonal entry is $e^{\lambda_it}$, the eigenvalue for the $i$th eigenvector. In a more general case you can get $e^{At}$ through any one of a number of methods, though my favorite (and a very useful one) is the inverse Laplace transform of the resolvant: $$e^{At} = \mathcal{L}^{-1}(sI-A)^{-1}.$$ Furthermore you'll notice that the fundamental matrix here is exactly $\Phi(t) = e^{At}$, which is true for any constant $A$, and the state transition matrix is $\Psi(t,\tau) = e^{A(t-\tau)} = \Phi(t)\Phi(\tau)^{-1},$ as should be expected. You can thus verify your answer for $\Phi$ by cross checking it with another method of calculating or approximating $e^{At}$.