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Find the general solution of x1? = x1 ? 2x2, x2? = 2x1 x2 using the eigenvalue method. do not use complex exponentials in your solution.

User Zig Mandel
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1 Answer

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Guessing on how the system is expressed:


\begin{bmatrix}x_1\\x_2\end{bmatrix}'=\begin{bmatrix}1&-2\\2&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}

The coefficient matrix has eigenvalues
\lambda=1+2i with corresponding eigenvectors
\begin{bmatrix}\pm i&1\end{bmatrix}^\top. This means the characteristic solution is


\begin{bmatrix}x_1\\x_2\end{bmatrix}=Ce^((1+2i)t)\begin{bmatrix}i\\1\end{bmatrix}

\begin{bmatrix}x_1\\x_2\end{bmatrix}=Ce^t(\cos2t+i\sin2t)\begin{bmatrix}i\\1\end{bmatrix}

\begin{bmatrix}x_1\\x_2\end{bmatrix}=C_1\begin{bmatrix}-e^t\sin2t\\e^t\cos2t\end{bmatrix}+C_2\begin{bmatrix}e^t\cos2t\\e^t\sin2t\end{bmatrix}
User Nathan Ostgard
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