Originally Posted by
Whitewolfblue x' = -3x + 2y,
y' = -3x + 4y,
I eliminated the y's first in order to solve for x, I'm sure I have to eliminate x's next.
I just realized what I did, I crossed my notes with the question I was working on.
So looking at things I should have gotten:
5.5x" + .5x' - 3x = 0 (that is if my new math is correct).
What do I need to use the x(0) = 0 with? I need to find Yc again, correct?
-WBB
-EDIT-
I got x(t) I think, not sure which one it is. In any case, when I eliminate y from the equations I get the equation:
.5x" - .5x' - 3x = 0
Thus producing Ae^(-2t) + Be^(3t). Mr F says: Now substitute x = 0 and t = 0 to get B = -A.
But, when I eliminate x from the equations I get:
-(1/3)y" + (1/3)y' - y = 0
I can't get the zero values for that equation because it never crosses the x axis. Did I do something wrong, or is there a step that I need to take in order to solve it?