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# Simultaneous Differential Equations watch

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1. Note that

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

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?
2. 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

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Updated: November 18, 2008
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