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$mx''(t) = -kx(t) - cx'(t)$

Express the equation of motion as a linear system of first-order differential equations. Use c=4, m=1, k=3

Here is the correct answer with some work:

The anwser is

$x_1 = x_2$

$x_2 = (-k/m) x_1 (-c/m)x_1$

Eigenvalues: -3, -1

General solution is : $\begin{bmatrix}x_1\\x_2\end{bmatrix} = c_1\begin{bmatrix}1\\-3\end{bmatrix}$e^(-3t)$ + c_2\begin{bmatrix}1\\-1\end{bmatrix}$e^(-t)

Then, using the initial condition $x_1(0), x_2(0) = (15,0)$

We get $c_1 = -7.5 , c_2 = 22.5$

How are $c_1, c_2$ being found?

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Just solve the system $$AC=X$$ Where: $$A=\left[ \begin{align} 1 && 1 \\ -3 && -1 \end{align} \right] $$ $$C=\left[ \begin{align} c_1 \\ c_2 \end{align} \right]$$ $$X(0)=\left[ \begin{align} 15 \\ 0 \end{align} \right]$$

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