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Solve the following equation: $$X'=\pmatrix { 2 & 5 \\ -5 & 8}X$$ $$x(0)=(5,5)^t$$

I already got out the followings:

$$\lambda_{1,2} = 5 \pm 4i$$ $$v_{1} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} - i\cdot \begin{pmatrix} 4 \\ 0 \end{pmatrix}$$ $$v_{1} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} + i\cdot \begin{pmatrix} 4 \\ 0 \end{pmatrix}$$

And so far I have this(Sry its messy): $$\begin{align*}\vec{x}_{1}(t) = (3-4i)e^{5t} \begin{pmatrix}a\cos(4t)+b\sin(4t)\end{pmatrix}\end{align*}$$ $$\begin{align*}\vec{x}_{2}(t) = 5\cdot e^{5t} \begin{pmatrix}c\cos(4t)+d\sin(4t)\end{pmatrix}\end{align*}$$

I dont know if this is right so far... but I also tried to sum up x_1:

My attend to solve x_1

This is the right answer in the end:$$\begin{align*}\vec{x}(t) = e^{5t} \begin{pmatrix}5\cos(4t)+\frac{5}{2}\sin(4t)\\ 5\cos(4t)-\frac{5}{2}\sin(4t)\end{pmatrix}\end{align*}$$

tyx

user577215664
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Darklight66
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  • Please transcribe your work into MathJax. Your scrawls are rather difficult to decipher. You can find a tutorial and quick reference guide here. – amd Feb 24 '18 at 21:21
  • Hallo, sry I dont know MathJax... what do you mean? for me the codes are shown perfectly. – Darklight66 Feb 24 '18 at 22:31
  • Your scribbled handwritten computations are difficult to read. If you want people to spend their own time answering your questions, don’t add to that burden by forcing them to decipher your handwriting. Moreover, images are not searchable nor are they accessible to people using screen readers. If you’re going to participate on this site, you will need to learn MathJax. I gave you a link to a tutorial in my previous comment. – amd Feb 25 '18 at 01:23
  • Anyway, see this question for a way to solve such a system without computing eigenvectors. – amd Feb 25 '18 at 01:25
  • ok thank you... you are right. sry I was in a hurry and this is my first post :D – Darklight66 Feb 25 '18 at 09:51

1 Answers1

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Since we have complex conjugate eigenvalues/eigenvectors, we only need to use one of them.

For $\lambda_1 = 5 + 4 i, v_1 = \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} $, we have

$\begin{align} e^{(5 + 4i)t} \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} \\ = e^{5t}e^{4it} \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix}\end{align} \\ = e^{5t}(\cos 4t + i \sin 4t)\begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} \\ = e^{5t}\begin{pmatrix} 3 \cos 4t + 4 \sin 4t +i(3 \sin 4t - 4 \cos 4t)\\ 5 \cos 4 t + i( 5 \sin 4t )\end{pmatrix}$

We know that the real and imaginary parts are both solutions, so our general solution is

$$\vec{x}(t) = e^{5t}\left(c_1\begin{pmatrix} 3 \cos 4t + 4 \sin 4t \\ 5 \cos 4 t \end{pmatrix} + c_2 \begin{pmatrix} 3 \sin 4t - 4 \cos 4t\\ 5 \sin 4t \end{pmatrix} \right)$$

We now use our initial condition $x(0)=(5,5)^t$

$$c_1\begin{pmatrix} 3 \\ 5 \end{pmatrix} + c_2 \begin{pmatrix} - 4 \\ 0 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}$$

You can use many methods to solve this $2 \times 2$ and it yields $c_1 = 1, c_2 = -\dfrac{1}{2}$ and a final result

$$\begin{align*}\vec{x}(t) = e^{5t} \begin{pmatrix}5\cos 4t+\dfrac{5}{2}\sin 4t\\ 5\cos 4 t-\dfrac{5}{2}\sin 4t\end{pmatrix}\end{align*}$$

Moo
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  • ty u very much :D, u let out something about how we get c1 and c2 after we seperatet real and imaginary parts... but i already knew how. i found a video. https://www.youtube.com/watch?v=TRVS5Wo9LoM&t=156s – Darklight66 Feb 25 '18 at 11:47
  • See these nice notes http://www.math.vt.edu/people/afkhamis/class_home/notes/F08W11b.pdf. – Moo Feb 25 '18 at 12:33