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Let the following homogenous system of differential equation:

$$\left\{\begin{array}{l}x'=ax+by\\y'=cx+dy\end{array}\right.$$

Let $A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $A$ is not diagonalizable and has complex roots conjugate: $\alpha\pm i\beta$

How do I arrive at the fundamental Matrix of solutions:

$$e^{\alpha t}\begin{pmatrix}\cos\beta t&\sin\beta t\\\sin\beta t&\cos\beta t\end{pmatrix}$$

Why this particular matrix?

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  • $\begingroup$ You need variable $t$ also in the matrix ($\beta$ should be $\beta t$) $\endgroup$ – Jean Marie Jan 14 at 12:12
  • $\begingroup$ @JeanMarie Right, I'll edit it right away! $\endgroup$ – C. Cristi Jan 14 at 12:13
  • $\begingroup$ You can find the answer in any textbook about systems of differential equations. $\endgroup$ – Jean Marie Jan 14 at 12:14
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    $\begingroup$ You need to apply your matrix to a vector of initial conditions. See the following MIT document : ocw.mit.edu/courses/mathematics/… $\endgroup$ – Jean Marie Jan 14 at 12:28
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    $\begingroup$ I see, thanks! I believe there must also be a minus sign somewhere in your fundamental matrix, can you please check again with your source? $\endgroup$ – Christoph Jan 14 at 15:38
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Pretend for the moment that you’re working with complex matrices. The coefficient matrix has two distinct eigenvalues and you can find corresponding complex eigenvectors $v_1$ and $v_2$ using the usual method, but working in the complex domain. Plunging ahead, you then have $$e^{tA} = \begin{bmatrix}v_1\\v_2\end{bmatrix} \begin{bmatrix}e^{(\alpha+i\beta)t}&0\\0&e^{(\alpha-i\beta)t}\end{bmatrix} \begin{bmatrix}v_1\\v_2\end{bmatrix}^{-1}.$$ Multiply this out and simplify using the identities $$\cos\theta = \frac12\left(e^{i\theta}+e^{-i\theta}\right) \\ \sin\theta = \frac1{2i}\left(e^{i\theta}-e^{-i\theta}\right).$$

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