Dx1 = -x1 - 3e^(-2t)

Dx2 = -2x1 - x2 - 6^(e^-2t)

Find the general solution of the system.

Can a calc guru help walk me through how to solve this system? Thanks,

Kim

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- June 2nd 2008, 12:02 PMKim NuLinear System with Constant Coefficients
Dx1 = -x1 - 3e^(-2t)

Dx2 = -2x1 - x2 - 6^(e^-2t)

Find the general solution of the system.

Can a calc guru help walk me through how to solve this system? Thanks,

Kim - June 2nd 2008, 12:27 PMTheEmptySet
Lets start by writing each equation in operator form

Now multiply equation 2 by and add it 2 equation 1 to get

So

The particular solution is of the form

so we get

So and the solution is

I did this very quickly so check my work.

You should be able to finish from here.

I hope this helps. Good luck. - June 2nd 2008, 01:22 PMKim Nu
Hi Empty Set,

Thanks a bunch for the response!

One question on this step:

http://www.mathhelpforum.com/math-he...cfb8432d-1.gif

How do you make the D on the right hand side dissappear? Shouldn't it be:

3(D + 1)e^(-2t) - 3e^(-2t) = 3e^(-2t) [(D + 1) -1] = (3e^(-2t))(D) - June 2nd 2008, 07:15 PMTheEmptySet
- June 3rd 2008, 07:18 AMKim Nu
Hi Empty Set,

Thanks for your teaching. I reworked the problem and arrived at the same point you did for x2. However the answer in the back of the book for x2 is:

x2 = -2c1te^(-t) + c2e^(-t) + 12e^(-2t)

How do you suppose that -2 snuck its way in front of the c1 for the x2c solution? Thanks,

Kim - June 3rd 2008, 07:28 AMTheEmptySet
When you solve for you will get two more constants

But since each of these systems are only first order and linear you can only have ONE constant per equation.

You will need to plug both of your solutions into the original system and solve for a relationship between and eliminate two of them from your solution. Notice that both solutions only have in them.

I hope this gives you somewhere to start...

I'm leaving for work now, but will try to look back later.

Good luck. - June 3rd 2008, 08:42 AMKim Nu
Thanks Empty Set. I'll look forward to some more input on this one. I have no problem findin the correct answer for x1; its the x2 that is confusing me.

- June 3rd 2008, 09:59 AMTheEmptySetHere is the rest...
Since you have solved for I will just use my solution

Note that this is a different constant then from

Now we know that the constants are related by what I called before. So we plug in and start simplifying.

Collecting like terms are reducing we get

so

So now we get the solutions

and

I hope this helps.

Good luck. - June 3rd 2008, 10:07 AMKim Nu
You are the master!