Question

Please help! I need to find the solution to the system x' = 6y, y' = -5x, with the initial conditions being x(0) = 1, y(0)= 5. I believe I have to do eigenvalues, but my eqns aren't coming out right.

Answer 1

What eigenvalues have you found?

Answer 2

I think I have found lambda equals +/- sqrt(30), based on my matrix: [0- lambda, 6; -5, 0-lambda]

Answer 3

+- sqrt(30) i
They are imaginary.

Answer 4

OH. my goodness. Math error! Thanks!

Answer 5

But---

Answer 6

and hence the eigenvectors will also be complex. To deal with this, use the same trick you use when the roots of the characteristic equation are imaginary: take sums to find trig functions cos and sin

Answer 7

I want to double check my method: okay, so, I find the values, then the vectors, and then I set up an eqn that looks some thing like x(t) = c1v1e^(eigenvector)t...?

Answer 8

oh not, I replace the e^(eigenvector) with (cost +isint)

Answer 9

call the eigenvalues l1 and l2, with corresponding eigenvectors v1 and v2. Then the general solution is
c1.v1.e^(l1.t) + c2.v2.e^(l2.t)
where c1 and c2 are constants.

Answer 10

but what happens to the (cost + isint)?

Answer 11

Yes, change basis and have two new eigenfunctions:
1/2 . ( e^(l1.t) + e^(l2.t) ) = cos(l1.t)
1/2i . ( e^(l1.t) - e^(-l2.t) ) = sin(l1.t)

remember that l1 = -l2

Answer 12

never tried these before but i did just watch a video by arthur mattuck tho.

x' = 6y +0x
y' = 0y - 5y

would our matrix be:

6 0
0 -5

or am i going astray there?

Answer 13

@amistre: your coefficients are in the wrong place

Answer 14

x' = 0x + 6y
etc.

Answer 15

we have to start with the x part?

Answer 16

The column vector of the functions x and y have to be in the same order on both sides of the equation

Answer 17

yep. And so now I finally have my vector [5; -30i]

Answer 18

nope, i think it should be [5, 30i] for complex conjugates

Answer 19

nvm keep it the way is was

Answer 20

this is wrong...

Answer 21

help :/ I don't know what I am doing any more :C I know that my eigen values are +/- 30i, and when I try to find the eigenvectors, I find -30i*a + 5*b = 0 and -5*a - 30i*b = 0

Answer 22

I don't thing the vector v = [5, -30i] is correct; and this affects my eqns for x and y

Answer 23

\[\begin{vmatrix}6&0\\0&-5\end{vmatrix}\to \lambda^2+\lambda-30=0;\ \lambda={-6,5} \]

\[\begin{vmatrix}0&6\\-5&0\end{vmatrix}\to \lambda^2+ 30=0;\ \lambda={\pm i\sqrt{30}} \]

so i spose it does matter which column is where

Answer 24

Yeah, the x values are far left and the y values are far right. So your bottom one is correct. Sorry, I didn't put sqrt() around my 30

Answer 25

I am making my vectors again

Answer 26

so it should be v = [-sqrt(30)i; 6]

Answer 27

hello, there anyone else who can help?