Question

Find the general solution of the given system. dx dt = −6x + 4y dy dt = − 7 4 x + 2y

Answer 1

I assume the coefficient on
`x` in the second equation is
`-\frac74`. The coefficient matrix


`\begin{bmatrix}-6&4\\\\-\frac74&2\end{bmatrix}`

has eigenvalues given by


`\begin{vmatrix}-6-\lambda&4\\\\-\frac74&2-\lambda\end{vmatrix}=\lambda^2+4\lambda-5=(\lambda+5)(\lambda-1)=0`

The eigenvalues are thus
`\lambda_1=-5` and
`\lambda_2=1`.

For the eigenvalue
`\lambda_1`, the corresponding eigenvector
`\eta_1` satisfies


`\begin{bmatrix}-1&4\\\\-\frac74&7\end{bmatrix}\begin{bmatrix}\eta_(1,1)\\\eta_(1,2)\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}`

`\implies\eta_(1,1)=4\eta_(1,2)`

so that we can choose
`\eta_1=\begin{bmatrix}4\\1\end{bmatrix}`.

For
`\lamda_2=1`, we have


`\begin{bmatrix}-7&4\\\\-\frac74&1\end{bmatrix}\begin{bmatrix}\eta_(2,1)\\\eta_(2,2)\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}`

`\implies7\eta_(2,1)=4\eta_(2,2)`

and we can choose
`\eta_2=\begin{bmatrix}4\\7\end{bmatrix}` for the corresponding eigenvector.

Then the general solution to the ODE system is given by


`\begin{bmatrix}x(t)\\y(t)\end{bmatrix}=C_1\begin{bmatrix}4\\1\end{bmatrix}e^(-5t)+\C_2\begin{bmatrix}4\\7\end{bmatrix}e^t`