Question

Find the unique solution to the following linear system X = (0 -1 2 3)X + (e' -e') X(0)=(5 4)

Answer 1

It looks like the system is supposed to be

$\mathbf x'=\begin{bmatrix}0&-1\\2&3\end{bmatrix}\mathbf x+\begin{bmatrix}e^t\\e^{-t}\end{bmatrix}$

with the initial condition, $\mathbf x(0)=\begin{bmatrix}5&4\end{bmatrix}^\top$.

Compute the eigensystem for the coefficient matrix:

$\begin{vmatrix}-\lambda&-1\\2&3-\lambda\end{vmatrix}=\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2)=0$

$\lambda_1=1\implies\begin{bmatrix}-1&-1\\2&2\end{vmatrix}\mathbf v_1=\mathbf 0\implies\mathbf v_1=\begin{bmatrix}1\\-1\end{bmatrix}$

$\lambda=2\implies\begin{bmatrix}-2&-2\\2&1\end{bmatrix}\mathbf v_2=\mathbf 0\implies\mathbf v_2=\begin{bmatrix}1\\-2\end{bmatrix}$

So we have the characteristic solution,

$\mathbf x_c=C_1e^{\lambda_1t}\mathbf v_1+C_2e^{\lambda_2t}\mathbf v_2$

$\mathbf x_c=C_1e^t\begin{bmatrix}1\\-1\end{bmatrix}+C_2e^{2t}\begin{bmatrix}1\\-2\end{bmatrix}$

For the non-homogeneous part, we can guess a particular solution of the form

$\mathbf x_p=te^t\begin{bmatrix}a\\b\end{bmatrix}$

(We might have started with $e^t$ instead, but that is already accounted for in the first characteristic solution.)

Its derivative is

${\mathbf x_p}'=e^t\begin{bmatrix}a\\b\end{bmatrix}+te^t\begin{bmatrix}a\\b\end{bmatrix}=e^t\begin{bmatrix}at+a\\bt+b\end{bmatrix}$

Substitute $\mathbf x_p$ into the system and solve for $a,b$:

$e^t\begin{bmatrix}at+a\\bt+b\end{bmatrix}=te^t\begin{bmatrix}0&-1\\2&3\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}+e^t\begin{bmatrix}1\\-1\end{bmatrix}$

$\begin{bmatrix}at+a\\bt+b\end{bmatrix}=\begin{bmatrix}-bt+1\\(2a+3b)t-1\end{bmatrix}$

$\implies a=1,b=-1$

and so the particular solution is

$\mathbf x_p=te^t\begin{bmatrix}1\\-1\end{bmatrix}$

The general solution to the system is then

$\mathbf x=\mathbf x_c+\mathbf x_p$

$\mathbf x=C_1e^t\begin{bmatrix}1\\-1\end{bmatrix}+C_2e^{2t}\begin{bmatrix}1\\-2\end{bmatrix}+te^t\begin{bmatrix}1\\-1\end{bmatrix}$

Use the given initial conditions to solve for the remaining coefficients.

$\mathbf x(0)=\begin{bmatrix}5\\4\end{bmatrix}$

$\implies\begin{bmatrix}5\\4\end{bmatrix}=C_1\begin{bmatrix}1\\-1\end{bmatrix}+C_2\begin{bmatrix}1\\-2\end{bmatrix}$

$\implies C_1=14,C_2=-9$

Then the solution to the initial value problem is

$\mathbf x=14e^t\begin{bmatrix}1\\-1\end{bmatrix}-9e^{2t}\begin{bmatrix}1\\-2\end{bmatrix}+te^t\begin{bmatrix}1\\-1\end{bmatrix}$

$\mathbf x=\begin{bmatrix}14e^t-9e^{2t}+te^t\\-14e^t-18e^{2t}-te^t\end{bmatrix}$