# Thread: Linear System with Constant Coefficients

1. ## Linear 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

2. Originally Posted by Kim Nu
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
Lets start by writing each equation in operator form

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

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

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

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

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

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

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

so we get

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

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

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

I did this very quickly so check my work.

You should be able to finish from here.

I hope this helps. Good luck.

3. Hi Empty Set,

Thanks a bunch for the response!

One question on this step:

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)

4. Originally Posted by Kim Nu
Hi Empty Set,

Thanks a bunch for the response!

One question on this step:

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)
Remember that $D =\frac{d}{dt}$

So it we distribute the operator we get

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

Now we do what D tells us we take a derivative

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

I hope this clears it up.

Good luck.

5. 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

6. Originally Posted by Kim 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
When you solve for $x_1$ you will get two more constants

$c_3, c_4$

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 $c_1,c_2,c_3,c_4$ and eliminate two of them from your solution. Notice that both solutions only have $c_1,c_2$ 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.

7. 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.

8. ## Here is the rest...

Since you have solved for $x_1$ I will just use my solution

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

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

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

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

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

Collecting like terms are reducing we get

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

so $c_2=-2c_3$

So now we get the solutions

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

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

I hope this helps.

Good luck.

9. You are the master!