Question

use the method of variation of parameters to solve the initial value problem $x = ax + f(t), x(a) = x_a$ using the following values.
$a=\begin{bmatrix}5&-26\\1&-5end{bmatrix},f(t)=\begin{bmatrix}6t\\8end{bmatrix},x(0)=\begin{bmatrix}0\\0end{bmatrix},e^{at}=\begin{bmatrix}cos t + 5sin t&-26sin t\sin t&cos t - 5sin tend{bmatrix}$
$x(t)=\square$

Explanation:

Step1: Recall the formula for variation of parameters

The formula for the solution of the system \(\mathbf{x}' = A\mathbf{x}+\mathbf{f}(t)\) is \(\mathbf{x}(t)=e^{At}\mathbf{x}(0)+\int_{0}^{t}e^{A(t - s)}\mathbf{f}(s)ds\).

Step2: First, find \(e^{A(t - s)}\)

Since \(e^{At}=

$$\begin{bmatrix}\cos t+5\sin t&- 26\sin t\\\sin t&\cos t - 5\sin t\end{bmatrix}$$

\), then \(e^{A(t - s)}=

$$\begin{bmatrix}\cos(t - s)+5\sin(t - s)&-26\sin(t - s)\\\sin(t - s)&\cos(t - s)-5\sin(t - s)\end{bmatrix}$$

\)

Step3: Calculate the integral \(\int_{0}^{t}e^{A(t - s)}\mathbf{f}(s)ds\)

We have \(\mathbf{f}(s)=

$$\begin{bmatrix}6s\\8\end{bmatrix}$$

\)

$$
LATEXBLOCK3
$$

For the first - component:

$$
LATEXBLOCK4
$$

Using integration by parts: \(\int_{0}^{t}s\cos(t - s)ds=t\sin t+\cos t - 1\), \(\int_{0}^{t}s\sin(t - s)ds=-t\cos t+\sin t\), \(\int_{0}^{t}\sin(t - s)ds=1-\cos t\)

$$
LATEXBLOCK5
$$

For the second - component:

$$
LATEXBLOCK6
$$

Using integration by parts: \(\int_{0}^{t}s\sin(t - s)ds=-t\cos t+\sin t\), \(\int_{0}^{t}\cos(t - s)ds=\sin t\), \(\int_{0}^{t}\sin(t - s)ds=1-\cos t\)

$$
LATEXBLOCK7
$$

Step4: Calculate \(e^{At}\mathbf{x}(0)\)

Since \(\mathbf{x}(0)=

$$\begin{bmatrix}0\\0\end{bmatrix}$$

\), then \(e^{At}\mathbf{x}(0)=

$$\begin{bmatrix}0\\0\end{bmatrix}$$

\)

Step5: Combine the results

$$ \mathbf{x}(t)=
LATEXBLOCK10
$$

Answer:

\(\mathbf{x}(t)=

$$\begin{bmatrix}6t\sin t-30t\cos t + 30\sin t+214\cos t-214\\-6t\cos t+14\sin t+40\cos t-40\end{bmatrix}$$

\)