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 $-\dfrac74$. 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

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