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
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.
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 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.
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.