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Mariah's avatar

Can you help me understand why my answer is wrong (system of differential equations, solved via eigenvectors)?

Asked by Mariah (19187 points ) October 27th, 2012

This is homework, so please do not directly give me any answers!

I have a system of differential equations which must be solved using eigenvectors (I know there are other ways, but I have to use eigenvectors):
dx/dt = 4x + y
dy/dt = 4y

So my matrix A is:
[4 1]
[0 4]

Fluther formatting is making this hard, but just pretend that is one set of vertical square brackets enclosing all four numbers.

I get a repeated root of 4 for the eigenvalue. I’m confident this part is right.

Then I have to find the eigenvector, and I think this part might be where I am screwing up, although I can’t figure out how.

[4 1] * [x] = 4 * [x]
[0 4]...[y].........[y]

This translates to the two equations:

4x + y = 4x
4y = 4y

The latter equation gives us no information, but 4x’s in the first cancel and you get y=0, x is free. So for my eigenvector I choose:

[1]
[0]

The formula then to find x(t) and y(t) is:

[x(t)] = (at + b) * e^(eigenvalue*t) * eigenvector
[y(t)]

where a and b are constants.

So we have x(t) = (at + b) * e^(4t) and y(t) = 0.

This very well parallels a similar example problem the prof did in class, but the thing is, I know this answer isn’t right. I know because when I try to check my answer by differentiating x with respect to t, my answer is not equivalent to 4x + y.

dx/dt = a*e^(4t) + 4(at+b)*e^(4t)
4x + y = 4(at+b)*e^(4t)

I can see that y should equal a*e^(4t) to make this equation true (and this would be consistent with the equation for dy/dt as well), but I don’t know how that can be when I get 0 in the y entry of the eigenvector.

So greatly appreciate it if anyone is able to help!

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10 Answers

Mariah's avatar

Ah, I think I figured it out, kind of. It seems the formula the professor gave us was wrong? I have a solution, but not via eigenvectors. Would still appreciate help if anyone can offer it.

Sunny2's avatar

I wish I could help, but I can’t begin to conceive of the problems you face. I so admire that you are smart enough to face them. Surely we have somebody who can speak and think in this language to help you. Good luck.

lillycoyote's avatar

I have no idea if this helps because I don’t understand it at all but and I don’t know if you have shown every step… my mother was a mathematician and a Associate Professor of statistics and she once told me that she always had to remind her students, and herself, that when you are doing higher math problems and you don’t arrive at the right answer that the problems if very often not that you haven’t done the “higher math” parts correctly, but that somewhere along the line you added 2+2 and got 5, if you know what I mean. I guess what she was trying to say is that you can rack your brain checking the complex parts, convinced you’ve done one the complex, hard parts wrong, when it’s quite possible that it could be something very simple, that you are overlooking, something like minor computational error. If that is possible I would check for something like that. I just don’t know.

Anyway, I don’t know if that advice helps, but it’s all I’ve got.

Good luck.

Mariah's avatar

Asked my TA this question today and he was stumped. He’s gonna talk to the prof and report back.

BonusQuestion's avatar

When you have a double root for characteristic polynomial and a single eigenvector your solutions are not what you have here:

[x(t)] = (at + b) * e^(eigenvalue*t) * eigenvector
[y(t)]

The solutions are in form (a v1 + b(t v1 + v2)) e^( c t)

Where v1 is the eigenvector and c is the only eigenvalue and v2 is the solution to (cI – A) v2 = v1.

Mariah's avatar

@BonusQuestion Thank you…yeah, that formula that you say is wrong was given to us by the professor during lecture. I was already pretty convinced from my own analysis that that formula was wrong but it helps to hear it from someone else. I’ll try your method.

BonusQuestion's avatar

Yes, the formula is wrong. He probably mistook it with solutions to linear differential equations. If you have a zero c with multiplicity for characteristic polynomial of a linear DE then solutions corresponding to that zero are e^(ct), te^(ct), etc.

Mariah's avatar

Prof admitted in class today that he gave us a wrong formula and taught us the right method. Thanks for all the help!

lillycoyote's avatar

I’m just glad that you found out what was wrong with the whole business. It was probably quite maddening! When you believe, when you are almost certain, that you’re doing everything right, and it still, just keeps coming out wrong. :-)

Mariah's avatar

@lillycoyote It is maddening! The funny thing is that it’s becoming a bit of a trend; seems like every time I stoop to Fluther for help with my homework it turns out that somebody made a mistake, and it wasn’t me.

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