Prove that the general solution of

Chapter 8, Problem 26E

(choose chapter or problem)

Prove that the general solution of

\(\mathbf{X}^{\prime}=\left(\begin{array}{rr}

-1 & -1 \\

-1 & 1

\end{array}\right) \mathbf{X}+\left(\begin{array}{l}

1 \\

1

\end{array}\right) t^{2}+\left(\begin{array}{r}

4 \\

-6

\end{array}\right) t+\left(\begin{array}{r}

-1 \\

5

\end{array}\right)

\)

on the interval \((-\infty, \infty)\) is

\(\begin{aligned}

\mathbf{X}=& c_{1}\left(\begin{array}{c}

1 \\

-1-\sqrt{2}

\end{array}\right) e^{\sqrt{2} t}+c_{2}\left(\begin{array}{c}

1 \\

-1+\sqrt{2}

\end{array}\right) e^{-\sqrt{2} t} \\

&+\left(\begin{array}{l}

1 \\

0

\end{array}\right) t^{2}+\left(\begin{array}{r}

-2 \\

4

\end{array}\right) t+\left(\begin{array}{l}

1 \\

0

\end{array}\right) .

\end{aligned}

\)

Text Transcription:

mathbf X^prime=({array} rr-1 & -1 -1 & 1 {array}) mathbf X+({array} l 1 1 {array}) t^2+({array} r 4 -6 {array}) t+ ({array} r -1 5 {array})

(-infty, infty)

mathbf X=& c_1 ({array} c 1 -1-sqrt 2 {array}) e^sqrt 2 t+c_2 ({array} c 1 -1+sqrt 2 {array}) e^-sqrt 2 t &+ ({array} l 1 0 {array}) t^2+ ({array} r-2 4 {array} ) t+ ({array} l1 0 {array})