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# Show that $x_{1}(t) \text { and } x_{2}(t)$ are solutions to the initial-value problem $d \mathbf{x} / d t=A \mathbf{x} \text { with } \mathbf{x}(0)=\mathbf{x}_{0}.$$A=\left[\begin{array}{rr} {1} & {-1} \\ {0} & {1} \end{array}\right] \quad \mathbf{x}_{0}=\left[\begin{array}{l} {2} \\ {1} \end{array}\right]$$x_{1}(t)=2 e^{t}-t e^{t}, x_{2}(t)=e^{t}$

Solution

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Denote:

$x=\begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}=\begin{bmatrix} 2e^{t}-te^{t}\\ e^{t} \end{bmatrix}\tag{1}$

To prove the functions $x_{1}(t)$ and $x_{2}(t)$ are solutions to the initial value problem:

$\frac{dx}{dt}=Ax,$

where

$A=\begin{bmatrix} 1&-1\\ 0&1 \end{bmatrix},\quad\text{and}\quad x_{0}=\begin{bmatrix} 2\\ 1 \end{bmatrix}$

we will find $\frac{dx}{dt}$, $Ax$, and verify that they are equal. After that we will compute

$x(0)=\begin{bmatrix} x_{1}(0)\\ x_{2}(0) \end{bmatrix}$

to verify that the initial condition is satisfied as well.

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