[SOLVED]System of linear ODEs check

dwsmith

Well-known member
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
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.

Last edited: