# Finding two linearly independent solutions to a system of ODEs

Find two linearly independent solutions to the pair of coupled ODES $$\frac{dx}{dt}=2x+3y$$ $$\frac{dy}{dt}=-3x+2y$$

My attempt:

Consider the matrix $\ A=\begin{bmatrix} 2 & 3 \\ -3 & 2 \\ \end{bmatrix}$, which has eigenvalues $2-3i, 2+3i$ with corresponding eigenvectors $\vec{v_1}=\begin{pmatrix} 1 \\ -i \\ \end{pmatrix}$ and $\vec{v_2}=\begin{pmatrix} 1 \\ i \\ \end{pmatrix}$ respectively. Hence the general solution is $$\vec{x}(t)=Ae^{(2-3i)t}\vec{v_1}+Be^{(2+3i)t}\vec{v_2} \ \ \ \ A,B\in\mathbb{R}$$ Where $\ \vec{x_1}(t)=e^{(2-3i)t}\vec{v_1} \$ and $\ \vec{x_2}(t)= e^{(2+3i)t}\vec{v_2}\$ are two linearly independent solutions.

Are these in fact independent? The solutions I have wish to use Euler's formula to write $$\vec{x_1}(t)=e^{2t}\begin{pmatrix} \cos(3t) \\ -\sin(3t) \\ \end{pmatrix} \ \ \vec{x_2}(t)=e^{2t}\begin{pmatrix} \cos(3t) \\ \sin(3t) \\ \end{pmatrix}$$ But this seems unnecessary to me.

Theoretically, $$\vec{x_1}(t)=e^{(2-3i)t}\vec{v_1}$$
and $$\vec{x_2}(t)= e^{(2+3i)t}\vec{v_2}$$ are solutions and they are linearly independent.
• Quick question, how does $$e^{2t}(\cos(3t)+i\sin(3t))\begin{pmatrix} 1 \\ i \\ \end{pmatrix}=e^{2t}\begin{pmatrix} \cos(3t) \\ \sin(3t) \\ \end{pmatrix}?$$ How do you obtain the RHS? – user557493 Aug 30 '18 at 2:50