Determine the General solution to the differential Equation system X'=AX

A=

0 -4

4 0

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- May 9th 2009, 09:43 PMRaidanSystems of differential equations...
**Determine the General solution to the differential Equation system X'=AX**

A=

0 -4

4 0

- May 10th 2009, 05:32 AMJester
- May 10th 2009, 06:19 AMHallsofIvy
For a simple problem like this, danny arrigo's suggestion is best, but you can do as a matrix problem.

Find the eigenvalues of . It should be easy to see that they are 4i and -4i. Since those are distinct, the matrix can be diagonalize (over the complex numbers).

With eigenvalue 4i, we have which gives the two equations -4y= 4ix and 4x= 4iy. Those both reduce to y=ix so one eigenvector is <1, i>.

With eigenvalue -4i, we have which gives the two equations -4y= -4ix and 4x= -4iy. Those both reduce to y=-ix so another, independent, eigenvector is <1, i>.

If we use those as columns in matrix P, then and

Then it is easy to see that .

If we multiply, on the left, by the constant matrix , we have .

Let so that and substitue PY for X in the equation: so our differential equation becomes .

The general solution to that, just like any y'= ay, is and because that is a diagonal matrix it is easy to show that we get

Now we use to write .

And, then use the fact that , to change into sine and cosine, getting danny aririgo's answer.

Whew! Well, I said that, for a simple problem like this danny arrigo's solution was best! - May 10th 2009, 06:36 AMRaidan
**Ahh.. thanx guys..I got it now..(Nod)**