use the method of variation of parameters to solve the initial value problem ( mathbf{x}^{prime}=mathbf{a…

use the method of variation of parameters to solve the initial value problem ( mathbf{x}^{prime}=mathbf{a x}+mathbf{f}(t), mathbf{x}(a)=mathbf{x}_{a} ) using the following values.\n\n( mathbf{a}=left\begin{array}{cc}2 & -5 \\ 1 & -2end{array}\right, mathbf{f}(t)=left\begin{array}{c}2 t \\ 6end{array}\right, mathbf{x}(0)=left\begin{array}{l}0 \\ 0end{array}\right, e^{mathbf{a} t}=left\begin{array}{cc}cos t+2 sin t & -5 sin t \\ sin t & cos t-2 sin tend{array}\right )\n\n( mathbf{x}(t)= )

use the method of variation of parameters to solve the initial value problem ( mathbf{x}^{prime}=mathbf{a x}+mathbf{f}(t), mathbf{x}(a)=mathbf{x}_{a} ) using the following values.\n\n( mathbf{a}=left\begin{array}{cc}2 & -5 \\ 1 & -2end{array}\right, mathbf{f}(t)=left\begin{array}{c}2 t \\ 6end{array}\right, mathbf{x}(0)=left\begin{array}{l}0 \\ 0end{array}\right, e^{mathbf{a} t}=left\begin{array}{cc}cos t+2 sin t & -5 sin t \\ sin t & cos t-2 sin tend{array}\right )\n\n( mathbf{x}(t)= )

Answer

Explanation:

Step1: Recall the formula for variation of parameters

The solution of (x' = Ax+f(t)) is (x(t)=e^{At}x_0+\int_{0}^{t}e^{A(t - s)}f(s)ds).

Step2: Substitute the given values

We know that (x_0=\begin{bmatrix}0\0\end{bmatrix}), so (e^{At}x_0=\begin{bmatrix}0\0\end{bmatrix}). Now we need to calculate (\int_{0}^{t}e^{A(t - s)}f(s)ds).

Since (e^{A(t - s)}=\begin{bmatrix}\cos(t - s)+2\sin(t - s)&- 5\sin(t - s)\\sin(t - s)&\cos(t - s)-2\sin(t - s)\end{bmatrix}) and (f(s)=\begin{bmatrix}2s\6\end{bmatrix})

[ \begin{align*} \int_{0}^{t}e^{A(t - s)}f(s)ds&=\int_{0}^{t}\begin{bmatrix}\cos(t - s)+2\sin(t - s)&- 5\sin(t - s)\\sin(t - s)&\cos(t - s)-2\sin(t - s)\end{bmatrix}\begin{bmatrix}2s\6\end{bmatrix}ds\ &=\int_{0}^{t}\begin{bmatrix}2s(\cos(t - s)+2\sin(t - s))-30\sin(t - s)\2s\sin(t - s)+6(\cos(t - s)-2\sin(t - s))\end{bmatrix}ds \end{align*} ]

Calculate the first - component integral

Let (u = t - s), then (s=t - u) and (ds=-du) [ \begin{align*} \int_{0}^{t}[2s(\cos(t - s)+2\sin(t - s))-30\sin(t - s)]ds&=\int_{0}^{t}(2s\cos(t - s)+4s\sin(t - s)-30\sin(t - s))ds\ &=\int_{0}^{t}2s\cos(t - s)ds+\int_{0}^{t}(4s - 30)\sin(t - s)ds \end{align*} ]

Using integration by parts: For (\int_{0}^{t}2s\cos(t - s)ds), let (u = 2s), (dv=\cos(t - s)ds), then (du = 2ds), (v=-\sin(t - s)) [ \begin{align*} \int_{0}^{t}2s\cos(t - s)ds&=\left[-2s\sin(t - s)\right]{0}^{t}+2\int{0}^{t}\sin(t - s)ds\ &=0 + 2\left[\cos(t - s)\right]_{0}^{t}\ &=2(1-\cos t) \end{align*} ]

For (\int_{0}^{t}(4s - 30)\sin(t - s)ds), let (u = 4s-30), (dv=\sin(t - s)ds), then (du = 4ds), (v=\cos(t - s)) [ \begin{align*} \int_{0}^{t}(4s - 30)\sin(t - s)ds&=\left[(4s - 30)\cos(t - s)\right]{0}^{t}-4\int{0}^{t}\cos(t - s)ds\ &=(4t - 30)-(-30\cos t)-4\left[\sin(t - s)\right]_{0}^{t}\ &=4t-30 + 30\cos t \end{align*} ]

The first - component is (2(1-\cos t)+4t-30 + 30\cos t=4t + 28\cos t-28)

Calculate the second - component integral

[ \begin{align*} \int_{0}^{t}[2s\sin(t - s)+6(\cos(t - s)-2\sin(t - s))]ds&=\int_{0}^{t}2s\sin(t - s)ds+\int_{0}^{t}(6\cos(t - s)-12\sin(t - s))ds \end{align*} ]

For (\int_{0}^{t}2s\sin(t - s)ds), let (u = 2s), (dv=\sin(t - s)ds), then (du = 2ds), (v=\cos(t - s)) [ \begin{align*} \int_{0}^{t}2s\sin(t - s)ds&=\left[2s\cos(t - s)\right]{0}^{t}-2\int{0}^{t}\cos(t - s)ds\ &=2t-2\left[\sin(t - s)\right]_{0}^{t}\ &=2t \end{align*} ]

For (\int_{0}^{t}(6\cos(t - s)-12\sin(t - s))ds=\left[6\sin(t - s)+12\cos(t - s)\right]_{0}^{t}=12 - 12\cos t-6\sin t)

The second - component is (2t+12 - 12\cos t-6\sin t)

Answer:

(x(t)=\begin{bmatrix}4t + 28\cos t-28\2t+12 - 12\cos t-6\sin t\end{bmatrix})