x ˙ 1 = − x 1 − x 2 x ˙ 2 = x 1 − x 2
Here x 1 and x 2 are functions of time t . What happens to the solutions x 1 ( t ) and x 2 ( t ) (subject to any initial conditions) after a very long time?
Bonus: Attempt to derive a general relation between x 1 and x 2 which is independent of time.
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Solving the given two differential equations, we get d t 2 d 2 x 2 + 2 d t d x 2 + 2 x 2 = 0 and a similar equation for x 1 . So x 2 decays exponentially with time and so also x 1 .
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The system has the form:
x ˙ = A x
The solutions therefore take the form:
x = c 1 e λ 1 t v 1 + c 2 e λ 2 t v 2
In the above equation, the λ values are the eigenvalues of the matrix A , and the v vectors are the eigenvectors.
As it turns out, the eigenvalues of the matrix are complex, with negative real parts. So the solutions are oscillatory, but their envelopes tend toward zero.