So i have solved a system of linear ODEs problem given the matrix

$\begin{bmatrix}-1 & -2\\5 & -3\end{bmatrix}$ and initial conditions

y $(0)$= $\begin{bmatrix}2\\1\end{bmatrix}$

but dont know exactly how to find the real and imaginary solutions to the following solutions I calculated:

$y_1$ = $\begin{bmatrix} \frac{1}{5} + \frac{3i}{5}\\1\end{bmatrix} e^ {-2+3i}$ , and

$y_2$ = $\begin{bmatrix} \frac{1}{5} - \frac{3i}{5}\\1\end{bmatrix} e^ {-2-3i}$

From here, i tried applying Euler's formula $e^{i \theta}$ = $cos \theta + isin \theta$ to obtain:

$Re \begin{bmatrix} -2cos \theta \\ 2sin \theta \end{bmatrix}$ and $Im \begin{bmatrix} 3sin \theta \\ 3 cos \theta \end{bmatrix}$

Resulting in general solution $y_{general} = C_1 \begin{bmatrix} -2cos \theta \\ 2sin \theta \end{bmatrix} + C_2 \begin{bmatrix} 3sin \theta \\ 3 cos \theta \end{bmatrix}$

Im not sure about the way to obtain the real and imaginary parts of the solution but this was my attempt. Any help is greatly appreciated!


We have

$e^{\lambda_1 t}v_1 = e^{(-2+3i)t}\begin{bmatrix}1+3i\\5\end{bmatrix} = e^{-2t}e^{3it}\begin{bmatrix}1+3i\\5\end{bmatrix} = e^{-2t}(\cos 3t + i \sin 3t)\begin{bmatrix}1+3i\\5\end{bmatrix} = \begin{bmatrix} e^{-2t}(\cos 3t -3 \sin 3 t) + i (3 \cos 3t +\sin 3 t)\\ e^{-2t}(5\cos3 t + 5i\sin 3t) \end{bmatrix} $

Because the real and imaginary parts are both independent solutions, we can now write

$$X(t) = \begin{bmatrix}x(t)\\ y(t) \end{bmatrix} = c_1 e^{-2t}\begin{bmatrix}\cos 3t -3 \sin 3 t \\ 5\cos 3t \end{bmatrix} + c_2 e^{-2t}\begin{bmatrix}3 \cos 3t+\sin 3t\\ 5\sin 3t \end{bmatrix}$$

Use the ICs to find $c_1$ and $c_2$.

The final solution is

$$X(t) = \begin{bmatrix}x(t)\\ y(t) \end{bmatrix} = e^{-2t}\begin{bmatrix}2\cos 3t \\ \cos 3t + 3 \sin 3t \end{bmatrix} $$

  • $\begingroup$ Thank you so much!! $\endgroup$ – lohboys May 16 at 22:18

Hint: $\tilde{y_{1}}=\frac{y_{1}+y_{2}}{2}, \tilde{y_{2}}=\frac{y_{1}-y_{2}}{2i}$

  • $\begingroup$ i've tried working it out using this hint to no success. I don't obtain anything which looks remotely like a possible answer :'(. Could you perhaps go through using this hint? $\endgroup$ – lohboys May 16 at 1:39

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