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Find two linearly independent solutions to the pair of coupled ODEs

$\frac{dx}{dt} = 2x + 3y$,

$\frac{dy}{ dt} =-3x+2y$.

I figured the eigenvalues/eigenvectors of the corresponding matrix to be $$e^{(2+3i)t} \quad\text{with}\quad \left[\begin{array}{r} i \\ 1 \end{array}\right] \quad\text{and} \quad e^{(2-3i)t} \quad\text{with}\quad\left[\begin{array}{r} -i \\ 1 \end{array}\right].$$ But apparently the solutions are $$\left[\begin{array}{r} x \\ y \end{array}\right] =e^{2t} \left[\begin{array}{r} \cos3t \\ -\sin3t \end{array}\right] ,\left[\begin{array}{r} x \\ y \end{array}\right] =e^{2t} \left[\begin{array}{r} \sin3t \\ \cos3t \end{array}\right].$$

Even after transforming the exponential eigenvalues to trigonometric functions I still don't get the desired results. I checked the eigenvectors/values with a computer and they seem to be correct. Please help.

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    $\begingroup$ You need to find real solutions. And it's standard trick: $\cos z = (e^{iz}+e^{-iz})/2$ $\endgroup$ – Michael Galuza Aug 16 '15 at 9:10
  • $\begingroup$ Why just real ? $\endgroup$ – user117498 Aug 16 '15 at 9:15
  • $\begingroup$ Form of solution say us about it;) Anyway, this solutions are the same, because $c_1e^{a+bt}+c_2e^{a-bt}=e^{at}[(c_1+c_2)\cos bt + i(c_1-c_2)\sin bt]$ $\endgroup$ – Michael Galuza Aug 16 '15 at 9:18
  • $\begingroup$ Sorry, i don't get how that brings me to the desired result $\endgroup$ – user117498 Aug 16 '15 at 9:25
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For $\lambda = 2 + 3i$, you should get an eigenvector of: $$ \vec v = \begin{bmatrix} -i \\ 1 \end{bmatrix} $$ The two linearly independent (real) solutions are the real and imaginary parts of $e^{\lambda t}\vec v$, which can be found by applying Euler's formula: \begin{align*} e^{(2 + 3i)t}\begin{bmatrix} -i \\ 1 \end{bmatrix} &= e^{2t}(\cos 3t + i\sin 3t)\begin{bmatrix} -i \\ 1 \end{bmatrix} \\ &= e^{2t}\begin{bmatrix} -i\cos 3t + \sin 3t \\ \cos 3t + i\sin 3t \end{bmatrix} \\ &= \underbrace{e^{2t}\begin{bmatrix} \sin 3t \\ \cos 3t \end{bmatrix}}_{\text{real part}} + i \cdot \underbrace{e^{2t}\begin{bmatrix} -\cos 3t \\ \sin 3t \end{bmatrix}}_{\text{imaginary part}} \end{align*}

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Notice, we have $$\frac{dx}{dt}=2x+3y\tag 1$$ $$\frac{dy}{dt}=-3x+2y\tag 2$$

  1. Diving (1) by (2), we get $$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{2x+3y}{-3x+2y}$$ $$\frac{dx}{dy}=\frac{2\frac{x}{y}+3}{-3\frac{x}{y}+2}$$ Now, let $x=uy\implies \frac{dx}{dy}=u+y\frac{du}{dy}$, setting the values we get $$u+y\frac{du}{dy}=\frac{2u+3}{-3u+2}$$ $$y\frac{du}{dy}=\frac{2u+3}{-3u+2}-u$$ $$y\frac{du}{dy}=\frac{3(1+u^2)}{2-3u}$$ $$\frac{2-3u}{3(1+u^2)}du=\frac{dy}{y}$$ $$\frac{2}{3}\int \frac{du}{1+u^2}-\frac{1}{2}\int \frac{2udu}{1+u^2}=\int \frac{dy}{y}$$ $$\frac{2}{3}\tan^{-1}(u)-\frac{1}{2}\ln(1+u^2)=\ln y+C_1$$

  2. Diving (2) by (1), we get $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3x+2y}{ 2x+3y}$$ $$\frac{dy}{dx}=\frac{-3+2\frac{y}{x}}{2+3\frac{y}{x}}$$

    Now, let $y=ux\implies \frac{dy}{dx}=u+x\frac{du}{dx}$, setting the values we get $$u+x\frac{du}{dx}=\frac{2+3u}{-3+2u}$$ $$x\frac{du}{dx}=\frac{2+3u}{-3+2u}-u$$ $$x\frac{du}{dx}=\frac{2+6u-2u^2}{-3+2u}$$ $$\frac{-3+2u}{2+6u-u^2}du=\frac{dx}{x}$$ I hope you can solve further.

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