use the method of variation of parameters to solve the initial value problem $x = ax + f(t), x(a) = x_a$…

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}) [ \begin{align*} \int_{0}^{t}e^{A(t - s)}\mathbf{f}(s)ds&=\int_{0}^{t}\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}\begin{bmatrix}6s\8\end{bmatrix}ds\ &=\int_{0}^{t}\begin{bmatrix}6s(\cos(t - s)+5\sin(t - s))-208\sin(t - s)\6s\sin(t - s)+8(\cos(t - s)-5\sin(t - s))\end{bmatrix}ds \end{align*} ] For the first - component: [ \begin{align*} &\int_{0}^{t}(6s\cos(t - s)+30s\sin(t - s)-208\sin(t - s))ds\ =&6\int_{0}^{t}s\cos(t - s)ds+30\int_{0}^{t}s\sin(t - s)ds-208\int_{0}^{t}\sin(t - s)ds \end{align*} ] 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) [ \begin{align*} 6(t\sin t+\cos t - 1)+30(-t\cos t+\sin t)-208(1 - \cos t)&=6t\sin t+6\cos t-6-30t\cos t + 30\sin t-208 + 208\cos t\ &=6t\sin t-30t\cos t+30\sin t+214\cos t-214 \end{align*} ] For the second - component: [ \begin{align*} &\int_{0}^{t}(6s\sin(t - s)+8\cos(t - s)-40\sin(t - s))ds\ =&6\int_{0}^{t}s\sin(t - s)ds+8\int_{0}^{t}\cos(t - s)ds-40\int_{0}^{t}\sin(t - s)ds \end{align*} ] 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) [ \begin{align*} 6(-t\cos t+\sin t)+8\sin t-40(1 - \cos t)&=-6t\cos t+6\sin t+8\sin t-40 + 40\cos t\ &=-6t\cos t+14\sin t+40\cos t-40 \end{align*} ]

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)=\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} ]

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})