Finding the particular solution to a inhomogenous system of differential equations

Question

Given \begin{equation} x' = \begin{bmatrix} 2&-1\\3&-2 \end{bmatrix}x+\begin{bmatrix} 1\\-1 \end{bmatrix}e^t \end{equation}

how do I find the particular solution $x_{p}$ via method of undetermined coefficients? I know where to start, but I'm stuck.

I start with $x_{p}$ = $\vec c$$e^t$ + $\vec d$t$e^t$, which I substitute back into the original equation. After I add the vectors together, I get

\begin{equation} \begin{bmatrix} -c_1+c_2+d_1\\-3c_1+3c_2+d_2 \end{bmatrix}e^t+\begin{bmatrix} -d_1+d_2\\-2d_1+3d_2 \end{bmatrix}te^t=\begin{bmatrix} 1\\-1 \end{bmatrix}e^t \end{equation} From this, I get $d_1$ = 2 = $d_2$, so \begin{equation} \vec d = \begin{bmatrix} 2\\2 \end{bmatrix}\end{equation} However, I am then left with -$c_1$+$c_2$ = -1 and -3$c_1$+3$c_2$ = -3.

Since they are multiples of each other, how can I accurately determine what the vector c is? I assume I can't arbitrarily choose two numbers, ex. $c_1$ = 7 and $c_2$ = 6...

Answer 1

If we calculate the eigenvalues of $M = \begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}$, we get $\lambda = \pm 1$, which means that $e^t$ is actually a component of the solution of the homogeneous DE $x' = Mx$, and indeed $\begin{bmatrix}1 \\ -1\end{bmatrix}$ is its corresponding eigenvector. So what that means is that any choice of $c_1$ (and corresponding value of $c_2$) will give you a valid particular solution to the DE. The difference between the solutions produced by two different choices of $c_1$ will show up in a different coefficient of $\begin{bmatrix}1 \\ -1\end{bmatrix}e^t$ in the homogeneous component of the solution.

Answer 2

Giving that

$M = \left( \begin{array}{cc} 2 & -1 \\ 3 & -2 \\ \end{array} \right) = T\cdot\Lambda\cdot T^{-1},\ \ \ T = \left( \begin{array}{cc} 1 & 1 \\ 3 & 1 \\ \end{array} \right),\ \ \Lambda = \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} \right) $

we have

$$ T^{-1}\dot x= \Lambda\cdot T^{-1}x+T^{-1}\cdot B e^t $$

so calling $y = T^{-1}x$ we follow with

$$ \dot y = \Lambda y + T^{-1}\cdot B e^t $$

The homogeneous have as solution

$$ y_h = e^{\Lambda t}c_0 $$

now proposing as a particular solution $y_p = e^{\Lambda t}c_0(t)$ after substitution into the complete ode we have

$$ \Lambda \cdot e^{\Lambda t}c_0(t) + e^{\Lambda t}\dot c_0(t)=\Lambda \cdot e^{\Lambda t}c_0(t)+T^{-1}\cdot B e^t $$

or

$$ \dot c_0(t) = e^{-\Lambda t}\cdot T^{-1}\cdot B e^t $$

now $e^{-\Lambda t} = \left( \begin{array}{cc} e^t & 0 \\ 0 & e^{-t} \\ \end{array} \right)$ and with $B=\left( \begin{array}{c} 1\\ -1 \\ \end{array} \right)$ we get $\dot c_0(t) = \left( \begin{array}{c} -e^{2 t} \\ 2 \\ \end{array} \right)$ then $c_0(t) = \left( \begin{array}{c} -\frac{e^{2 t}}{2} \\ 2 t \\ \end{array} \right)$ so

$$ y_p = \left( \begin{array}{cc} e^{-t} & 0 \\ 0 & e^{t} \\ \end{array} \right)\left( \begin{array}{c} -\frac{e^{2 t}}{2} \\ 2 t \\ \end{array} \right)= \left( \begin{array}{c} -\frac{e^{t}}{2} \\ 2 t e^t\\ \end{array} \right) = T^{-1}x_p $$

and finally

$$ x_p = T\cdot \left( \begin{array}{c} -\frac{e^{t}}{2} \\ 2 t e^t\\ \end{array} \right) =\left( \begin{array}{c} \frac{1}{2} e^t (4 t-1) \\ \frac{1}{2} e^t (4 t-3) \\ \end{array} \right) $$