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Find the general solution:

$\hat y_1'= y_2$

$\hat y_2' = 3y_1 +2y_3$

$\hat y_3' = -y_2$

If I put it in matrix form and get the eigenvalues I got

$\pmatrix{0-\lambda&1&0\\3&0-\lambda&2\\0&-1&0-\lambda}$

but I can't get the eigenvalues since it gives me $0$.

The answer is :

$y_1(t) = -c_1e^{-t} -c_2e^{t} -\frac{2}{3}c_3$,

$y_2(t) = c_1e^{-t}-c_2e^{t}$,

$y_3(t) = c_1e^{-t} +c_2e^{t}+c_3$

How did they get that?

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1 Answer 1

up vote 1 down vote accepted

Let

$$ \textbf{A} = \pmatrix{0 & 1 & 0 \\ 3 & 0 & 2 \\ 0 & -1 & 0} $$ then, expanding on the first row \begin{align} \det(\textbf{A} - \lambda \textbf{I}) = - \lambda \begin{vmatrix}-\lambda & 2 \\ -1 & -\lambda\end{vmatrix}-1 \begin{vmatrix}3 & 2 \\ 0 & -\lambda\end{vmatrix} &= -\lambda^3 - 2 \lambda +3\lambda \\ &= -\lambda(\lambda^2-1) \end{align} hence $$ \lambda_1 = -1,\quad \lambda_2 = 0, \quad \lambda_3 = 1. $$

To calculate the eigenvectors we need to solve $$ (\textbf{A} - \lambda_i \textbf{I})\textbf{v}_i = 0 $$

For $\lambda_1 = -1$ we have $$ (\textbf{A} + \textbf{I})\textbf{v}_1 = \pmatrix{1 & 1 & 0 \\ 3 & 1 & 2 \\ 0 & -1 & 1} \pmatrix{v_{11} \\ v_{21} \\ v_{31} } = \pmatrix{0 \\ 0 \\ 0} $$ and then $\textbf{v}_1 = (-1,1,1)^T$.

For $\lambda_2 = 0$ $$ \textbf{A}\textbf{v}_2 = \pmatrix{0 & 1 & 0 \\ 3 & 0 & 2 \\ 0 & -1 & 0} \pmatrix{v_{12} \\ v_{22} \\ v_{32} } = \pmatrix{0 \\ 0 \\ 0} $$ and $\textbf{v}_2 = \big(-\frac{2}{3}, 0 ,1\big)^T$

Finally, for $\lambda_3 = 1$ $$ (\textbf{A} - \textbf{I})\textbf{v}_3 = \pmatrix{-1 & 1 & 0 \\ 3 & -1 & 2 \\ 0 & -1 & -1} \pmatrix{v_{13} \\ v_{23} \\ v_{33} } = \pmatrix{0 \\ 0 \\ 0} $$ and $\textbf{v}_3 =(-1,-1,1)^T$ and the solution to the system is

$$ \textbf{y} = c_1 \textbf{v}_1 e^{\lambda_1 t} + c_2 \textbf{v}_2 e^{\lambda_2 t} + c_3 \textbf{v}_3 e^{\lambda_3 t} $$

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Thank you! Can you show me how they got the general solution? –  Q.matin Nov 26 '12 at 23:39
    
@Q.matin I've completed the answer. Hope it's clear now. –  Pragabhava Nov 27 '12 at 0:00
    
Thanks a lot!!! –  Q.matin Nov 27 '12 at 0:02
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