 # Consider the following linear system of differential equations: { <mtable columnalign=" William Santiago 2022-05-16 Answered
Consider the following linear system of differential equations:
$\left\{\begin{array}{l}\stackrel{˙}{x}=-4y\\ \stackrel{˙}{y}=x\end{array}$
where $x\left(t\right)$ and $y\left(t\right)$ are unknown real functions.
One can simply verify that the general solution is
$\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)={c}_{1}\left(\begin{array}{c}\mathrm{cos}\left(2t\right)\\ \frac{1}{2}\mathrm{sin}\left(2t\right)\end{array}\right)+{c}_{2}\left(\begin{array}{c}-2\mathrm{sin}\left(2t\right)\\ \mathrm{cos}\left(2t\right)\end{array}\right)$
where ${c}_{1}$ and ${c}_{2}$ are real parameters.
Question: Which is the exponential form of this expression? It should by something like
$\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)={k}_{1}{e}^{i2t}\left(\begin{array}{c}{P}_{11}\\ {P}_{21}\end{array}\right)+{k}_{2}{e}^{-i2t}\left(\begin{array}{c}{P}_{12}\\ {P}_{22}\end{array}\right)$
where ${k}_{1}$ and ${k}_{2}$ should (?) be complex parameters.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it heilaritikermx
I think the key here is to rewrite the linear system as the following matrix equation
$\underset{A\stackrel{\to }{x}=\stackrel{˙}{\stackrel{\to }{x}}}{\underset{⏟}{\left(\begin{array}{cc}0& -4\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\stackrel{˙}{x}\\ \stackrel{˙}{y}\end{array}\right)}}.$
The characteristic equation is ${\lambda }^{2}+4=0,$ so you're correct in using $\mathrm{cos}\left(2t\right)$ and $\mathrm{sin}\left(2t\right).$ Now, to get at a complex exponential solution, you need to find eigenvectors of $A$ with eigenvalues $-2i,2i.$ These are
$\left(\begin{array}{c}2i\\ 1\end{array}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(\begin{array}{c}-2i\\ 1\end{array}\right)$
respectively.
Consequently, you can write
$\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)={c}_{1}\left(\begin{array}{c}2i\\ 1\end{array}\right){e}^{2it}+{c}_{2}\left(\begin{array}{c}-2i\\ 1\end{array}\right){e}^{-2it}.$
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 $A.$.
###### Not exactly what you’re looking for? kwisangqaquqw3
Well, defining $k\equiv {c}_{2}-i{c}_{1}/2$, you get a real

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 ... take the c.c. ... do you see the point?