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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.

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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.

The only problem is that they are complex solutions and we need real solutions.

That is why we combine them to get real solutions.

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  • $\begingroup$ 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? $\endgroup$ – user557493 Aug 30 '18 at 2:50

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