System of differential equations

[MaxManus]

Homework Statement

Find the general solution of

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

Homework Equations

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?

 

[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. :)

 

[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}

 

[jackmell]

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}

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.

 

[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.

 

[jackmell]

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.

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. :)

 

[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.

 

[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).

 

[ehild]

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.

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

 

[ehild]

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

A term e^2tp is missing.

ehild