Question

Solve this system of Differential Equations:

Answer 1

\[x ^{'} = \left[\begin{matrix}3 & -2 \\ 4 & -1\end{matrix}\right]x\]

Answer 2

Yielding to this solution:

Answer 3

It's a bit complicated... do you know change of basis matrices?

Answer 4

i've had a crack at this and i reckon your eigenvalues are \(\lambda_1 = 1 + 2i\) and \(\lambda_2 = 1 - 2i\)

the corresponding e-vectors are

\[\vec v_1 = \left(\begin{matrix} { 1 \over 2} ( 1 + i) \\ 1\end{matrix}\right)\]

\[\vec v_2 = \left(\begin{matrix} { 1 \over 2} ( 1 - i) \\ 1\end{matrix}\right)\]

from there, it is slog :-(

so we need to plug this into the idea that the independent solutions are of form \(\Large \vec x_{1,2} = \vec v_{1,2} e ^ {\lambda_{1,2} t}\)

i am happy to plough on with this if you so wish.

Answer 5

Err. Change of basis shouldn't be needed for this! @IrishBoy123 you should just do it on a piece of paper, it that's easier for you!
My problem is that whenever I try to expand one vector (because you just need to consider one, right?), and then Euler it, I arrive at a solution not given by the book!

Answer 6

Sorry, change of basis is how I think of it conceptually. :P Hmmm, can you show your working @sh3lsh please, and then we can see where you can go wrong?

Answer 7

If the solution spans the same space as your textbook solutiom, I call it good enough.

Answer 8

Hi, here's a bit of my work.
Sorry if I'm skipping a few of my steps. My professor expect us to do these questions with little time, so I'm rushing a tad.

Answer 9

I want to expand this and separate this into the real part and imaginary part and call it done. However, these weren't the solutions.

Answer 10

Oh right. I believe what you've done is fine. It's just that they've grouped the solutions differently. See how no imaginary number is mentioned in their solution? Well it's because they've group the solutions into the real and imaginary parts and then "changed the constants", absorbing the imaginary number i into their constant... lol, not explained well, but kinda get what I mean?
tl;dr, they grouped the solutions so that the imaginary number i is absorbed into the constants.

Answer 11

Hmm.
But there's still a floating negative.
If I grouped the reals I would obtain
\[\left(\begin{matrix}\cos2t \\ \cos2t-\sin2t\end{matrix}\right)\]

Answer 12

So you've converted the \(c_1\) vector into trig, now do the same with \(c_2\) and combine ALL the real parts and the imaginary parts... I hope that will give you the required answer :P

Answer 13

Err. I found out my problem. My eigenvector calculations were wrong. Here's the solution all!
(sorry if it's messy!)