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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- Jun 2nd 2008, 11:02 AMKim 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 - Jun 2nd 2008, 11:27 AMTheEmptySet
Lets start by writing each equation in operator form

$\displaystyle E_1: \\\ (D+1)x_1=-3e^{-2t}$

$\displaystyle E_2: \\\ (D+1)x_2+2x_1=-6e^{-2t}$

Now multiply equation 2 by $\displaystyle -\frac{1}{2}(D+1)$ and add it 2 equation 1 to get

$\displaystyle -\frac{1}{2}(D+1)^2x_2=3(D+1)e^{-2t}-3e^{-2t} \iff -\frac{1}{2}(D+1)^2x_2=-6e^{-2t} $

$\displaystyle (D+1)^2x_2=12e^{-2t}$

So $\displaystyle x_{2c}=c_1e^{-t}+c_2te^{-t}$

The particular solution is of the form $\displaystyle x_{2p}=Ae^{-2t}$

so we get

$\displaystyle 4Ae^{-2t}+2(-2)Ae^{-2t}+Ae^{-2t}=12e^{-2t} \iff A=12$

So $\displaystyle x_{2p}=12e^{-2t}$ and the solution is

$\displaystyle x_2= c_1e^{-t}+c_2te^{-t}+12e^{-2t}$

I did this very quickly so check my work.

You should be able to finish from here.

I hope this helps. Good luck. - Jun 2nd 2008, 12:22 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) - Jun 2nd 2008, 06:15 PMTheEmptySet
Remember that $\displaystyle D =\frac{d}{dt}$

So it we distribute the operator we get

$\displaystyle 3(D+1)e^{-2t}-3e^{-2t}=3[De^{-2t}+e^{-2t}]-3e^{-2t}$

Now we do what D tells us we take a derivative

$\displaystyle 3[De^{-2t}+e^{-2t}]-3e^{-2t}=3[-2e^{-2t}+e^{-2t}]-3e^{-2t}=-6e^{-2t}+3e^{-2t}-3e^{-2t}=-6e^{-2t}$

I hope this clears it up.

Good luck. - Jun 3rd 2008, 06:18 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 - Jun 3rd 2008, 06:28 AMTheEmptySet
When you solve for $\displaystyle x_1$ you will get two more constants

$\displaystyle c_3, c_4$

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 $\displaystyle c_1,c_2,c_3,c_4$ and eliminate two of them from your solution. Notice that both solutions only have $\displaystyle c_1,c_2$ 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. - Jun 3rd 2008, 07:42 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.

- Jun 3rd 2008, 08:59 AMTheEmptySetHere is the rest...
Since you have solved for $\displaystyle x_1$ I will just use my solution

$\displaystyle x_1=c_3e^{-t}+3e^{-2t}$

Note that this is a different constant then from $\displaystyle x_2$

Now we know that the constants $\displaystyle c_1,c_2,c_3$ are related by what I called $\displaystyle E_2$ before. So we plug in $\displaystyle x_1,x_2$ and start simplifying.

$\displaystyle (D+1)x_2+2x_1=-6e^{-2t}$

$\displaystyle -c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-24e^{-2t}+c_1e^{-t}+c_2te^{-t}+12e^{-2t}+2c_3e^{-t}+6e^{-2t}=-6e^{-2t}$

Collecting like terms are reducing we get

$\displaystyle (-c_1+c_2+c_1+2c_3)e^{-t}=0 \iff (c_2+2c_3)e^{-t}=0$

so $\displaystyle c_2=-2c_3$

So now we get the solutions

$\displaystyle x_1=c_3e^{-t}+3e^{-2t}$ and

$\displaystyle x_2=c_1e^{-t}\underbrace{-2c_3te^{-t}}_{c_2=-2c_3}+12e^{-2t}$

I hope this helps.

Good luck. - Jun 3rd 2008, 09:07 AMKim Nu
You are the master!