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Given: $$\vec x'(t) = \begin{bmatrix} 4&-1\\ 13&0 \end{bmatrix} \vec x(t) $$

Evaluating to find eigenvalues: $$ (4-\lambda)(-\lambda)+13=0 $$$$ (\lambda-2)^2=-9$$ $$\lambda=2\pm3i$$

Finding the eigenvector for the eigenvalue $\lambda = 2+3i$: $$ \left[ \begin{array}{cc|c} 2-3i&-1&0\\ 13&-2-3i&0 \end{array} \right] $$ Since, for a 2x2, the rows must be complex scalar multiples of each other (here: R2 = (2 + 3i)R1 ), then picking row 2:

\begin{align*} \begin{cases} v_1=\frac{2+3i}{13}v_2 \\ v_2=free \end{cases} \end{align*}

$$\vec v= v_2 \begin{bmatrix} \frac{2+3i}{13}\\ 1 \end{bmatrix} = v_2\begin{bmatrix} 2+3i\\ 13 \end{bmatrix}$$ The solution takes the form of: $$e^{\lambda_1t}\vec v_1 = e^{(2+3i)t}\begin{bmatrix} 2+3i\\ 13 \end{bmatrix} $$ However, this is not a real-valued vector solution, since the eigenvector has an imaginary part. How should I proceed?

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  • $\begingroup$ Can you edit the entries of $A$ matrix? Second element should be -1 and not 11. $\endgroup$ – Maxtron Sep 11 '18 at 20:40
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    $\begingroup$ See this answer for a method that doesn’t require computing any eigenvectors. $\endgroup$ – amd Sep 11 '18 at 20:54
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You may split this solution into real and imaginary parts and $$e^{\lambda_1t}\vec v_1 = e^{(2+3i)t}\begin{bmatrix} 2+3i\\ 13 \end{bmatrix}=U +iV$$

Each one of those $U$ and$V$ is a real solution to your system.

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  • $\begingroup$ I have attempted to do so in one of the solutions posted; have I made any errors? $\endgroup$ – Essan Verne Sep 11 '18 at 21:24
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    $\begingroup$ I have checked your work and it is correct. Please add $e^{2t}$ to your second solution as well. $\endgroup$ – Mohammad Riazi-Kermani Sep 11 '18 at 22:07
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Use Euler's identity to split the solution into its real and imaginary parts:

$$e^{(2+3i)t}=e^{2t+i3t}=e^{2t}*e^{i3t}$$

$$e^{i3t} = \text{cos}(3t)+i\text{sin}(3t)$$ Therefore: $$e^{\lambda_1t}\vec v_1 = e^{2t}*e^{i3t}\begin{bmatrix} 2+3i\\ 13 \end{bmatrix} $$

$$= e^{2t}*[\text{cos}(3t)+i\text{sin}(3t)]*\left(\begin{bmatrix} 2\\ 13 \end{bmatrix}+i\begin{bmatrix} 3\\ 0 \end{bmatrix}\right) $$

Foiling: $$e^{2t}\left(\begin{bmatrix} 2\text{cos}(3t)\\ 13\text{cos}(3t) \end{bmatrix}+i\begin{bmatrix} 3\text{cos}(3t)\\ 0 \end{bmatrix}+i\begin{bmatrix} 2\text{sin}(3t)\\ 13\text{sin(3t)} \end{bmatrix}+i^2\begin{bmatrix} 3\text{sin}(3t)\\ 0 \end{bmatrix}\right)$$

$$\underbrace{e^{2t}\begin{bmatrix} 2\text{cos}(3t)-3\text{sin}(3t)\\ 13\text{cos}(3t) \end{bmatrix}}_{{\text{Real Part}}}+i\underbrace{e^{2t}\begin{bmatrix} 2\text{sin}(3t)+3\text{cos}(3t)\\ 13\text{sin}(3t) \end{bmatrix}}_{{\text{Imaginary Part}}}$$

The solution takes the form: $$\vec x(t)=C_1(\text{Real Part}) + C_2(\text{Imaginary Part})$$ $$\vec x(t)=C_1e^{2t}\begin{bmatrix} 2\text{cos}(3t)-3\text{sin}(3t)\\ 13\text{cos}(3t) \end{bmatrix}+C_2e^{2t}\begin{bmatrix} 2\text{sin}(3t)+3\text{cos}(3t)\\ 13\text{sin}(3t) \end{bmatrix}$$

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  • $\begingroup$ Looks good. Please add $e^{2t}$ to your second solution as well. $\endgroup$ – Mohammad Riazi-Kermani Sep 11 '18 at 22:05
  • $\begingroup$ Is there a theorem behind "solution takes the form:"? I realize it's a standard form, but it's still not obvious to me how an imaginary part of a solution projects itself onto the real plane by multiplication by a constant C_2. $\endgroup$ – Ben Ogorek Dec 31 '18 at 17:19
  • $\begingroup$ After help from @KittyL in comment section here, I think I understand. Above, the particular solution corresponding to one eigenvector was found. The other eig solution is the complex conjugate. Applying the superposition principle, functions of the form $d_1 soln1 + d_2 soln2$ are also solutions, leading to a form $(d_1 + d_2) * real part + (d_1 - d_2) i * imag part$. You can choose $d_1$ and $d_2$ such that $i$ is gone. $\endgroup$ – Ben Ogorek Jan 1 '19 at 21:45
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Your matrix is actually similar to one of the form $\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}$ with transition matrix $\begin{bmatrix} 2&3\\ 13&0 \end{bmatrix}$ given respectively by the eigenvalues' real and imaginary parts and the transition is given (in columns) by real and imaginary parts of the first eigenvector. The solution to the equation will then be given by $\vec x(t)=e^{2t}\begin{bmatrix} cos(3t)& sin(3t)\\ -sin(3t)&cos(3t) \end{bmatrix}.\vec x(0)$

To give a proof of this fact, apply Taylor expansion and definition of exponential of a matrix. Regards

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