I've been asked to find the general solution of this system:
$\begin{cases} x_1'= -x_2 + 2 \\ x_2'= 2x_1 +3x_2 +t\end{cases}$
So, we have been working on this with a friend and were wondering if there's a better method to solve it.
We wanted to used variations of parameters. To start with, we found the eigenvalues:
$\lambda_1= \frac {3+\sqrt17}{2}$
$\lambda_2=\frac {3-\sqrt17}{2}$
Then, we computed the eigenvectors and found the $X_h$, the homogenous solution. With that, we set out to find $X_p(t$), by solving $Q(t).C'(t)=b(t)$ where $X_p$ stands for a particular solution.
We learned that we need to find $C(t)$ to get $X_p(t)= Q(t).C(t)$ in order to, finally, get the general solution, $X_g$, where $X_g= X_p + X_h$.
In our case, we had:
$ e^{\frac{3}{2}t}\left(\begin{matrix} 2e^{\frac {\sqrt 17}{2}t} & 2e^{-\frac {\sqrt 17}{2}t} \\ (3+\sqrt 17) e^{\frac {\sqrt 17}{2} t}&(3-\sqrt 17) e^{-\frac {\sqrt 17}{2}t}\\ \end{matrix}\right)$. $\left(\begin{matrix} c_1' \\ c_2'\\ \end{matrix}\right) =\left(\begin{matrix} 2 \\ t\\ \end{matrix}\right)$
I prefer to avoid the rest of the computations, I believe they are unnecessarily complicated and you don't need to see them to understand the point.
In short, we found ourselves dealing with a pretty complicated integral to find $C(t)$ and very long computations and started wondering if we were mistaking somewhere or if maybe there's a better method than this one.
Thanks for the help!
(P.S. : How can I write the braces for the system of equations? I tried many times and failed)
EDIT Solved the problem with the braces
There is a sign error in the matrix you used. You used
$$A = \begin{pmatrix} 0 & 1 \\ 2 & 3 \\ \end{pmatrix}$$
You should have used
$$A = \begin{pmatrix} 0 & -1 \\ 2 & 3 \\ \end{pmatrix} \implies \lambda_{1, 2} = 1, 2, v_1 = (-1, 2), v_2 = (2, 3)$$
Please retry your method.
Alternate approaches include Undetermined Coefficients, Laplace Transforms, Matrix Exponential, Fundamental Matrix...
Another approach is to convert the system to a higher order equation by noting from equation $1$:
$$x_2 = 2 - x_1' \implies x_2 ' = - x_1''$$
Substituting those two into equation $2$:
$$x_1'' - 3 x_1' + 2 x_1 = -(t + 6)$$
Hopefully you can take it from here.