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[SOLVED] System of linear ODEs check

dwsmith

Well-known member
Feb 1, 2012
1,673
Just checking a solution.

$y' = \begin{pmatrix}4 & -1\\ 2 & 1\end{pmatrix}y $
$$
\lambda^2 - 5\lambda + 6 = (\lambda - 3)(\lambda - 2) = 0.
$$
So the eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 2$.
To find the eigenvectors, we must solve $(4 - \lambda)y_1 - y_2 = 0\iff y_2 = (4 - \lambda)y_1$.
Then
$$
y = \begin{pmatrix}1\\ 4 - \lambda\end{pmatrix}.
$$
Now $\mathbf{y_1} = e^{3t}\begin{pmatrix}1\\ 1\end{pmatrix}$ and $\mathbf{y_2} = e^{2t}\begin{pmatrix}1\\ 2\end{pmatrix}$.
Thus, the solution is
$$
y = Ae^{3t}\begin{pmatrix}1\\ 1\end{pmatrix} + Be^{2t}\begin{pmatrix}1\\ 2\end{pmatrix}.
$$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Just checking a solution.

$y' = \begin{pmatrix}4 & -1\\ 2 & 1\end{pmatrix}y $
$$
\lambda^2 - 5\lambda + 6 = (\lambda - 3)(\lambda - 2) = 0.
$$
So the eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 2$.
To find the eigenvectors, we must solve $(4 - \lambda)y_1 - y_2 = 0\iff y_2 = (4 - \lambda)y_1$.
Then
$$
y = \begin{pmatrix}1\\ 4 - \lambda\end{pmatrix}.
$$
Now $\mathbf{y_1} = e^{3t}\begin{pmatrix}1\\ 1\end{pmatrix}$ and $\mathbf{y_2} = e^{2t}\begin{pmatrix}1\\ 2\end{pmatrix}$.
Thus, the solution is
$$
y = Ae^{3t}\begin{pmatrix}1\\ 1\end{pmatrix} + Be^{2t}\begin{pmatrix}1\\ 2\end{pmatrix}.
$$
Yes it's correct. (Yes)
 
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