Question

This is the first part of a four-part problem. Let P = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{x}_1(t) = \begin{bmatrix} 2e^{2t} - 4e^{-t} \\ 3e^{2t} - 10e^{-t} \end{bmatrix}, \vec{x}_2(t) = \begin{bmatrix} 4e^{2t} + 2e^{-t} \\ 6e^{2t} + 5e^{-t} \end{bmatrix} a. Show that $\vec{x}_1(t)$ is a solution to the system $\vec{x}' = P\vec{x}$ by evaluating derivatives and the matrix product Enter your answers in terms of the variable $t$. $\vec{x}_1'(t) = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{x}_1(t)$ $\begin{bmatrix} \\ \\ \end{bmatrix} = \begin{bmatrix} \\ \\ \end{bmatrix}$ b. Show that $\vec{x}_2(t)$ is a solution to the system $\vec{x}' = P\vec{x}$ by evaluating derivatives and the matrix product Enter your answers in terms of the variable $t$. $\vec{x}_2'(t) = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{x}_2(t)$ $\begin{bmatrix} \\ \\ \end{bmatrix} = \begin{bmatrix} \\ \\ \end{bmatrix}$

          This is the first part of a four-part problem.
Let
P = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix}
\vec{x}_1(t) = \begin{bmatrix} 2e^{2t} - 4e^{-t} \\ 3e^{2t} - 10e^{-t} \end{bmatrix}, \vec{x}_2(t) = \begin{bmatrix} 4e^{2t} + 2e^{-t} \\ 6e^{2t} + 5e^{-t} \end{bmatrix}
a. Show that $\vec{x}_1(t)$ is a solution to the system $\vec{x}' = P\vec{x}$ by evaluating derivatives and the matrix product
Enter your answers in terms of the variable $t$.
$\vec{x}_1'(t) = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{x}_1(t)$
$\begin{bmatrix} \\ \\ \end{bmatrix} = \begin{bmatrix} \\ \\ \end{bmatrix}$
b. Show that $\vec{x}_2(t)$ is a solution to the system $\vec{x}' = P\vec{x}$ by evaluating derivatives and the matrix product
Enter your answers in terms of the variable $t$.
$\vec{x}_2'(t) = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{x}_2(t)$
$\begin{bmatrix} \\ \\ \end{bmatrix} = \begin{bmatrix} \\ \\ \end{bmatrix}$

This is the first part of a four-part problem.
Let
P =
< b m a t r i x >

x⃗1(t) =
< b m a t r i x >
, x⃗2(t) =
< b m a t r i x >

a. Show that x⃗1(t) is a solution to the system x⃗' = Px⃗ by evaluating derivatives and the matrix product
Enter your answers in terms of the variable t.
x⃗1'(t) =
< b m a t r i x >
x⃗1(t)
< b m a t r i x >
=
< b m a t r i x >
b. Show that x⃗2(t) is a solution to the system x⃗' = Px⃗ by evaluating derivatives and the matrix product
Enter your answers in terms of the variable t.
x⃗2'(t) =
< b m a t r i x >
x⃗2(t)
< b m a t r i x >
=
< b m a t r i x >

Added by Phillip A.

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