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The question is to find the solution for $$\frac{dx}{dt}=2x+5y$$ $$\frac{dy}{dt}=-x-2y$$ What I have done is to find the eigenvalues and eigenvectors for matrix $$\begin{bmatrix} \ 2 & 5 \\ \ -1 & -2 \end{bmatrix}$$ And solve for eigenvectors $$\lambda_1=i, x_1=[2+i,-1];\lambda_2=-i,x_2=[i-2,1]$$ However my solution for $(x,y)$ is $$x(t)=C_1(2\cos t-\sin t)+iC_2(\cos t+2\sin t)$$ $$y(t)=C_1(-\cos t)+iC_2(-\sin t)$$No matter what I try, I cannot reach the answer given which the question asks me to show such that, for constant $\alpha$ and $\beta$, $$x(t)=\alpha (5\cos t)+\beta(5\sin t)$$ $$y(t)=\alpha(-2\cos t-\sin t)+\beta(\cos t -2\sin t)$$ I don't know what steps am I missing or what linear combinations should I looking for to reach the given solution.

Thanks in advance

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The ODE system can be written as

$$ \frac{{\rm d}u(t)}{{\rm d}t} = A u(t) \tag{1} $$

with

$$ A = \left(\begin{array}{cc} 2 & 5 \\ -1 & -2\end{array}\right) $$

and $u(t) = (x(t), y (t))^T$. The formal solution of Eq. (1) is

$$ u(t) = e^{At}u(0) \tag{2} $$

In order to calculate $e^{At}$ we solve the eigensystem $A = U\Lambda U^{-1}$ and find

$$ U = \left(\begin{array}{cc} -2-i & -2 + i \\ 1 & 1\end{array}\right) ~~~\mbox{and}~~~ \Lambda = \left(\begin{array}{cc} i & 0 \\ 0 & -i\end{array}\right) $$

Therefore Eq. (2) becomes

$$ u(t) = U\left(\begin{array}{cc} e^{it} & 0 \\ 0 & e^{-it}\end{array}\right) U^{-1}\left(\begin{array}{c} x_0 \\ y_0\end{array}\right) = \left(\begin{array}{c} x_0\cos t + (2x_0 + 5y_0)\sin t \\ y_0\cos t - (x_0 + 2y_0)\sin t\end{array}\right) \tag{3} $$

where $x_0 = x(0)$ and $y_0 = y(0)$. Now, from Eq (3) define $\alpha$ and $\beta$ such that $5\alpha = x_0$ and $5\beta = (2x_0 + 5y_0)$, solving for $x_0$ and $y_0$ we find

$$ x_0 = 5\alpha ~~~\mbox{and}~~~ y_0 = \beta - 2\alpha $$

With this Eq. (3) becomes

$$ \left(\begin{array}{c} x(t) \\ y(t)\end{array}\right) = \left(\begin{array}{c} \alpha(5\cos t) + \beta(5\sin t) \\ \alpha(-2\cos t - \sin t) +\beta (\cos t - 2\sin t)\end{array}\right) $$

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