Question

Simultaneous ODE
I have $y_{1} ' = \frac{5}{2} y_{1} - \frac{3}{2} y_{2}$ and $y_{2} ' = -\frac{3}{2} y_{1} + \frac{5}{2} y_{2}$
How do I proceed to find y_{1} and y_{2}?

Answer 1

$\mathbf y'=\mathbf A\mathbf y\iff\begin{bmatrix}{y_1}'\\{y_2}'\end{bmatrix}=\begin{bmatrix}\frac52&-\frac32\\-\frac32&\frac52\end{bmatrix}\begin{bmatrix}y_1\\y_2\end{bmatrix}$

Find the eigensystem corresponding to the coefficient matrix.

$\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}\frac52-\lambda&-\frac32\\-\frac32&\frac52-\lambda\end{vmatrix}=0$
$\left(\dfrac52-\lambda\right)^2-\left(-\dfrac32\right)^2=0$
$\lambda^2-5\lambda+4=0$
$(\lambda-1)(\lambda-4)=0$
$\implies \lambda_1=1,\lambda_2=4$

For $\lambda_1=1$, the associated eigenvector satisfies

$(\mathbf A-\mathbf I)\mathbf v_1=\mathbf 0\iff\begin{bmatrix}\frac32&-\frac32\\-\frac32&\frac32\end{bmatrix}\begin{bmatrix}v_{11}\\v_{12}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies v_{11}-v_{12}=0\implies\mathbf v_1=\begin{bmatrix}1\\1\end{bmatrix}$

For $\lambda_2=4$, we have

$(\mathbf A-4\mathbf I)\mathbf v_2=\mathbf 0\iff\begin{bmatrix}-\frac32&-\frac32\\-\frac32&-\frac32\end{bmatrix}\begin{bmatrix}v_{21}\\v_{22}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies v_{21}+v_{22}=0\implies\mathbf v_2=\begin{bmatrix}1\\-1\end{bmatrix}$

The general solution for the ODE system is then

$\mathbf y=C_1e^{\lambda_1t}\mathbf v_1+C_2e^{\lambda_2t}\mathbf v_2$
$\iff\begin{bmatrix}y_1\\y_2\end{bmatrix}=C_1e^t\begin{bmatrix}1\\1\end{bmatrix}+C_2e^{4t}\begin{bmatrix}1\\-1\end{bmatrix}$
$\implies\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}C_1e^t+C_2e^{4t}\\C_1e^t-C_2e^{4t}\end{bmatrix}$