TITLE: Damped Harmonic Oscillator $2y''+8y'+8y=0$
QUESTION [0 upvotes]: Underdamped Harmonic Oscillator $2y''+8y'+8y=0$
I am given mass $m=2$, damping coefficient $b=8$ and spring constant $k=8$
I first need to change this into a first order system so I get:
$\frac{dy}{dt} = v$ and $\frac{dv}{dt} = -4y - 4v$ so if I change this into matrix form:
$$ \frac{dY}{dt} = \Bigg(\begin{array}{cc} 0 & 1 \\ -4 & -4 \end{array} \Bigg) Y$$
My question now is it fine to solve this by using the eigenvalues and eigenvectors? I only ask because we previously solved it differently in class.

REPLY [1 votes]: I am not sure if it is easier to use this approach.  You could just say the characteristic equation is
$2r^2 + 8r + 8 = 0\\
2(r+2)^2 = 0\\
y = C_1  e^{-2t} + C_2 t e^{-2t}\\
$
Anyway, if you want to keep going with the approach you have chosen.
$det(A - \lambda I) = 0\\
\lambda^2 - trace(A) \lambda + det(A) = 0\\
\lambda^2 +4 \lambda + 4 = 0\\ 
(\lambda + 2)^2 = 0\\
$
Doesn't that look familiar.
$(A - \lambda I)v = \begin{pmatrix} 2&1\\-4&-2\end{pmatrix}v = 0\\
v = \begin{pmatrix} 1\\-2\end{pmatrix}$
we have one eigenvector.  When we have duplicated eigenvectors we look for some $w$ such that
$(A - \lambda I)w = v\\
w = \begin{pmatrix} 0\\1\end{pmatrix}$
will suffice.
$Y = C_1 v e^{-2t} +  C_2(v t e^{-2t} + w e^{-2t})\\
Y(0) = C_1 v+C_2 w = \begin{pmatrix} C_1\\-2C_1 + C_2\end{pmatrix} =\begin{pmatrix} y(0)\\y'(0)\end{pmatrix}\\
C_1 = y(0), C_2 = y'(0) - 2y(0)\\
y(t) = y(0) e^{-2t} - 2y(0) t e^{-2t} + y'(0) t e^{-2t}$
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