$\mathbf{x'}=\begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}\mathbf{x}+\begin{bmatrix} e^{3t}\\e^{3t}\end{bmatrix}$
My attempt:
We first find a solution to the associated homogeneous system $\mathbf{x'}=\begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}\mathbf{x}$.
My answer is $\mathbf{x}_c(t)=a_0\begin{bmatrix}1 \\ 0\end{bmatrix}e^{3t}+b_0\left(\begin{bmatrix}1 \\ 0\end{bmatrix}t+\begin{bmatrix}0 \\ 1\end{bmatrix}\right)e^{3t}$
In order to eliminate duplication, let $\mathbf{x}_p(t)=\begin{bmatrix}a_1 \\ a_2\end{bmatrix}t^2e^{3t}+\begin{bmatrix}b_1 \\ b_2\end{bmatrix}te^{3t}+\begin{bmatrix}c_1 \\ c_2\end{bmatrix}e^{3t}$
When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do
I proceed?
Best Answer
$$X'=AX+g(t)$$ Your particular solution is correct but since the matrix A is already in Jordan form. $$\pmatrix {x \\y}'=\pmatrix{3 & 1 \\ 0 & 3}\pmatrix {x \\ y}+\begin{bmatrix} e^{3t}\\e^{3t}\end{bmatrix}$$ $$y'=3y+e^{3t} \implies (ye^{-3t})'=1$$ $$y(t)=e^{3t}(t+K)$$ $$x'=3x+y+e^{3t}$$ $$x'=3x+e^{3t}(t+K+1) \implies (xe^{-3t})'=t+K+1$$ $$x(t)=(\frac {t^2}2+(K+1)t+C)e^{3t}$$