for distinct roots (k1, k2) of the equation k^2 + 2Bk + w^2 show that x(t) = Ae^(k1t) + Be^(k2t) is a solution of the following differential equation: (d^2)x/dt^2 + 2B(dx/dt) + (w^2)x = 0

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- Feb 5th 2009, 01:49 PMcraftyshowing differential equation solution
for distinct roots (k1, k2) of the equation k^2 + 2Bk + w^2 show that x(t) = Ae^(k1t) + Be^(k2t) is a solution of the following differential equation: (d^2)x/dt^2 + 2B(dx/dt) + (w^2)x = 0

- Feb 5th 2009, 02:10 PMRincewind
All you need to do is take the general solution and sub it into the d.e. EG

Substituting into the d.e.

Now collecting like terms

Now since and are roots of the terms on the LHS are both zero therefore the d.e. is satisfied. - Feb 5th 2009, 02:31 PMcrafty
thanks for that i was going about it in such a difficult way, this was much simpler than i had anticipated...

- Feb 5th 2009, 02:35 PMcrafty
theres another part to this question which i still can't seem to get use the initial conditions x(0) = 1 and x'(0) = 0 to determine the values of A and B, describe how the solution behaves as t -> infinity

- Feb 5th 2009, 04:17 PMRincewind
What you need to do here is substitute in the boundary conditions to determine A and B so...

and

You can solve these simultaneous as 2 equations with 2 unknowns (A and B)

This problem is a bit tricky since throughout the B in the original equation is different from the B in the general solution.

Regarding the general behavior it depends on the sign of and . If we assume the B in the original equation is real and positive and the w is real and less than B then the roots are

which are both definitely negative and in that case the exponential terms go to zero as t goes to infinity.

Hope this helps. - Feb 5th 2009, 06:17 PMcrafty
and how do i solve these simultaneous equations

- Feb 5th 2009, 07:03 PMRincewind

Step 1:

Step 2:

Step 3:

Step 4: Substitute into

- Feb 5th 2009, 07:25 PMcrafty
by the way for that part about infinity i forgot to mention that W>B and that means the roots are complex, so what happens then??? sorry about forgetting to mention this earlier

- Feb 5th 2009, 10:20 PMRincewind
OK so if W > B and B > 0 then you have the roots equal to

and

where

Then your solution becomes

So when t goes to infinity the exponential term sends the function to zero. The other terms just cause the function to wiggle around but the overall trend is .

If B < 0 then the function x goes to infinity.