Simple coupled ODE 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.
 A: 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*}
A: Notice, we have $$\frac{dx}{dt}=2x+3y\tag 1$$
$$\frac{dy}{dt}=-3x+2y\tag 2$$


*

*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 
$$u+y\frac{du}{dy}=\frac{2u+3}{-3u+2}$$ $$y\frac{du}{dy}=\frac{2u+3}{-3u+2}-u$$ $$y\frac{du}{dy}=\frac{3(1+u^2)}{2-3u}$$ $$\frac{2-3u}{3(1+u^2)}du=\frac{dy}{y}$$
$$\frac{2}{3}\int \frac{du}{1+u^2}-\frac{1}{2}\int \frac{2udu}{1+u^2}=\int \frac{dy}{y}$$ $$\frac{2}{3}\tan^{-1}(u)-\frac{1}{2}\ln(1+u^2)=\ln y+C_1$$

*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.
