Linear system of D.E's

x' = -3x + 2y,

y' = -3x + 4y,

Initial conditions

x(0) = 0;

y(0) = 2;

Using these equations, I derived x" - 4x' + 3x = 0

And then found Yc = Ae^x + Be^(3x).

Now what should I do?

-WWB

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- April 18th 2009, 10:31 PMWhitewolfblueParticular solution of linear systems
Linear system of D.E's

x' = -3x + 2y,

y' = -3x + 4y,

Initial conditions

x(0) = 0;

y(0) = 2;

Using these equations, I derived x" - 4x' + 3x = 0

And then found Yc = Ae^x + Be^(3x).

Now what should I do?

-WWB - April 19th 2009, 12:05 AMmr fantastic
Please show how you got x" - 4x' + 3x = 0 (I get something different). Then you use x(0) = 0 to eliminate one of the arbitrary constants.

Then you substitute x = x(t) into the first DE and solve for y. Then use y(0) = 2 to eliminate the other arbitrary constant. - April 19th 2009, 01:01 PMWhitewolfblueHow I got the equation
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).

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? - April 19th 2009, 09:44 PMmr fantastic