E1: (2D + 1)x1 + (D^2 - 4)x2 = -7e^(-t)
E2: Dx1 - (D + 2)x2 = -3e^(-t)
Here's how I solved for x1:
1. Multiply E2 by (D - 2) to get:
(Dx1)(D-2) - (D^2 - 4)x2 = -3e^(-t) * (D-2)
2. Simplify the above and then add to E1 to get:
(D^2 + 1)x1 = 2e^(-t)
3. x1c = c1cos (t) + c2sin (t)
4. Then using Ae^(-t), x1p = e^(-t)
5. x1 = c1cos (t) + c2sin (t) + e^(-t)
How do I solve for x2? I'm lost again on this one! Thanks,
Kim
Hey thanks Empty Set,
Now my posts are definitely getting annoying but:
I took your suggestion, which is the correct course of action; I end up with
x2 = c3cos(t) + c4sin(t) + c3e^(-2t) + 2e^(-t)
Then I take the before mentioned solved x1 value and x2 value and plug them into E2: Dx1 - (D + 2)x2 = -3e^(-t)
I work this all out and end up with:
sin (t)[-c1 + c3 -2c4] + cos (t)[c2 - c4 - 2c3) = -8e^(-t)
The answer in the back of the book is:
x1 = 5c1cost(t) + 5c2sin(t) + e^(-t)
x2 = (c1 + 2c2)cost + (-2c1 + c2)sin(t) + c3e^(-2t) + 2e^(-t)
So my answers are close to being correct, I just don't understand how those coefficient were arrived at. Do you have any suggestions? Thanks,
Kim
I think should be this
And your from your 1st post is
So if we put these into we get...
Now we have
We need to solve these for
To get so we want to eliminate so we multiply the 2nd by -2 and add it to the first to get
Now we need to find so we want to dliminate so now we multiply the first by 2 and add it to the 2nd
Now we finally can get the final solution
The answers are equivalent if you replace in both equations you will get the same answer.
Good luck.