has the general solution
Given x(0) = -2/3 and y(0) = 1/3, solve
So I thought we just need to add a term to our previous solution such that it differentiates to gives 2e^(2t) for x and 3e^(2t) so we don't really need to worry about the other stuff. I did it in a rather hack way:
Differentiate and set equal to the dx/dt and dy/dt expressions above and you get
C = -2/(3q) and
pC + 2qC + 3 = 0
Which, upon substitution, leads to
2p = 5q
So if we define p = 5, we get q = 2 and hence C = -1/3. Now if we use the initial conditions to find A and B we get A = 0 and B = 1. In short:
(you can verify by differentiation that this does satisfy the above equations)
Now my questions
1) Is there any elegant way of getting values for C, p or q? My first instinct was to guess p = 3 and q = 2 but that was wrong. Was C = -2/(3q) predictable?
2) Why are there an infinite number of solutions? I choose p = 5, but p = 10 would be perfectly fine also surely? What initial condition is missing?
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- Thread Starter
- 18-11-2008 04:21
- 18-11-2008 18:08
This C in your particular integral is not necessary.
you should be able to work out why?
This is why use have a degree of freedom in choices of C,p,q