# Homogenous system of differential equation

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?

• You need variable $t$ also in the matrix ($\beta$ should be $\beta t$) – Jean Marie Jan 14 at 12:12
• @JeanMarie Right, I'll edit it right away! – C. Cristi Jan 14 at 12:13
• You can find the answer in any textbook about systems of differential equations. – Jean Marie Jan 14 at 12:14
• You need to apply your matrix to a vector of initial conditions. See the following MIT document : ocw.mit.edu/courses/mathematics/… – Jean Marie Jan 14 at 12:28
• 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? – Christoph Jan 14 at 15:38

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 $$\cos\theta = \frac12\left(e^{i\theta}+e^{-i\theta}\right) \\ \sin\theta = \frac1{2i}\left(e^{i\theta}-e^{-i\theta}\right).$$