Exponential form of general integral of a linear system of ODE

Question

Consider the following linear system of differential equations: $$ \begin{cases} \dot{x}=-4y \\ \dot{y}=x \end{cases} $$ where $x(t)$ and $y(t)$ are unknown real functions.

One can simply verify that the general solution is $$ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 \begin{pmatrix} \cos(2t) \\ \frac{1}{2}\sin(2t) \end{pmatrix} +c_2 \begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} $$ where $c_1$ and $c_2$ are real parameters.

Question: Which is the exponential form of this expression? It should by something like $$ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = k_1 e^{i2t} \begin{pmatrix} P_{11}\\ P_{21} \end{pmatrix} +k_2 e^{-i2t} \begin{pmatrix} P_{12} \\ P_{22} \end{pmatrix} $$ where $k_1$ and $k_2$ should (?) be complex parameters.

Answer 1

I think the key here is to rewrite the linear system as the following matrix equation

$$\underbrace{\begin{pmatrix} 0& -4\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} \dot{x}\\ \dot{y} \end{pmatrix}}_{A\vec{x}=\dot{\vec{x}}}.$$

The characteristic equation is $\lambda^2+4=0,$ so you're correct in using $\cos(2t)$ and $\sin(2t).$ Now, to get at a complex exponential solution, you need to find eigenvectors of $A$ with eigenvalues $-2i,2i.$ These are $$\begin{pmatrix} 2i\\ 1 \end{pmatrix}\quad \text{and}\quad \begin{pmatrix} -2i\\ 1 \end{pmatrix}$$ respectively.

Consequently, you can write $$\begin{pmatrix} x(t)\\ y(t) \end{pmatrix}=c _1\begin{pmatrix} 2i\\ 1 \end{pmatrix}e^{2it}+c_2\begin{pmatrix} -2i\\ 1 \end{pmatrix}e^{-2it}.$$ It is worth noticing that the eigenvectors, are attached to their associated eigenfunctions, and that the eigenvalue of the derivative operator is identical to that of the eigenvalue under the action of $A.$

Answer 2

Well, defining $k\equiv c_2-ic_1/2$, you get a real $$ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = k ~e^{i2t} \begin{pmatrix} i\\ 1/2 \end{pmatrix} +k^* ~e^{-i2t} \begin{pmatrix} -i \\ 1/2 \end{pmatrix}. $$

You did not specify anything about the form of the Ps, but you have sufficient freedom to recast these into something of your choice.

Note the problem simplifies to a triviality if you defined $z\equiv x+i2y$, so $\dot{z}= i2~z$... take the c.c. ... do you see the point?