Question

Simultaneous ODE

I have
`y_(1) ' = (5)/(2) y_(1) - (3)/(2) y_(2)` and
`y_(2) ' = -(3)/(2) y_(1) + (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(\frac52-\lambda\right)^2-\left(-\frac32\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}`