# Homework Help: System of differential equations

1. Aug 10, 2010

### MaxManus

1. The problem statement, all variables and given/known data

Find the general solution of

$$\dot{x}$$ = x + e^{2t}p [/tex]
$$\dot{p} = 2e^{-2t}x - p$$

2. Relevant equations

3. The attempt at a solution
I know how to solve systems with constant coefficients using eigenvalues and eigenvectors, but what should I do in this case?

2. Aug 10, 2010

### jackmell

Hello Max. Here's a challenging way: try it with power series. I did for the IVP:

$$x'=x+e^{2t}p,\quad x(0)=1$$

$$p'=2e^{-2t}x-p,\quad p(0)=1$$

and obtained the plot below which agrees nicely with the numerical results cus' I actually plotted them both and they're right on top of one another. Also,I'm no expert so may be an easier way to solve this. :)

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Last edited: Aug 10, 2010
3. Aug 12, 2010

### MaxManus

Thanks, but we have not learned power series, I looked up the method, but I can't solve the system. Did you find the solution?
The book says
$$x = Ae^{(1+\sqrt{2})t} + Be^{(1-\sqrt{2})t}$$
$$p = A\sqrt(2)e^{(\sqrt{2}-1)t} -B\sqrt{2}e^{(-\sqrt{2}-1)t}$$

Last edited: Aug 12, 2010
4. Aug 12, 2010

### jackmell

Well not that solution but a purist would say the power series is the solution although not in closed form like the above. I don't know how to arrive at your solution.

5. Aug 12, 2010

### Dick

Find the eigenvalues and eigenvectors of [[1,exp(2t)],[2*exp(-2t),-1]]. The eigenvectors will be functions of t, but that shouldn't stop you from treating this the same way you would if the matrix were numerical.

6. Aug 12, 2010

### jackmell

Ok. I see now. Thank you. I assume Max got it too. Personally I think Max should work out all the details in nice pretty latex and post it but I don't want to bug him about it. :)

7. Aug 15, 2010

### MaxManus

Eigenvalues
$$(1-\lambda)(-1-\lambda) - e^{2t}2e^{-2t} = 0$$
$$\lambda = \pm- \sqrt{3}$$

for $$\lambda = \sqrt(3)$$

$$(1-\sqrt{3})x + e^{2t}y = 0$$

$$v_1 = \begin{array}{c} 1\\ (\sqrt{3}-1})e^{-2t}\\$$

for $$\lambda = -\sqrt{3}$$
$$(1+ \sqrt{3})x + e^{2t}y = 0$$
$$v_2 = \begin{array}{c} 1\\ (-\sqrt{3}-1})e^{-2t}\\$$

$$x = Ae^{\sqrt{3t}} + Be^{-\sqrt{3t}$$
$$p = A(\sqrt{3}-1)e^{(-2+\sqrt{3})t} - B(\sqrt{3}+1)e^{(-2-\sqrt{3})t}$$

Not sure what I did wrong, but tried calculating the eigenvalue several times, but that is not important. Thanks again to both of you for all the help.

8. Aug 15, 2010

### HallsofIvy

A somewhat more primitive way: differentiate the first equation again to get
$$x''= x'+ 2e^{2t}p+ e^{2t}p'$$
From the second equation,
$$p'= 2e^{-2t}x- p$$
Putting that in,
$$x''= x'+ 2e^{2t}p+ 2x- e^{2t}p= x' + 2x$$

Solve
$$x''- x' - 2x= 0$$
for x(t).
Again, from the first equation,
$$e^{2t}p= x'- x$$
so that
$$p= e^{-2t}(x'- x)$$.

9. Aug 15, 2010

### ehild

Your solution is correct, the book is wrong. Try to plug in back into the original equations

10. Aug 15, 2010

### ehild

A term e2tp is missing.

ehild