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Ok. So the problem is:

Show that the damped equation

$\ddot{x} + k\dot{x} + (\gamma + \beta cos(t))x = 0$

can be transformed into a Mathieu equation by the change of variable $x = ze^{\mu t}$ for a suitable choice for $\mu$

Second attempt at solution:

We have:

$x=ze^{\mu t}$

$\dot{x}=\mu z e^{\mu t} + \dot{z}e^{\mu t}$

$\ddot{x} = \mu^2 ze^{\mu t}+2\mu \dot{z}e^{\mu t} + \ddot{z} + e^{\mu t}$

Inserting this into our equation gives, after some changing around:

$\ddot{z}e^{\mu t} + (2\mu + k)\dot{z}e^{\mu t} + (\mu^2 + k\mu + \gamma + \beta cos(t))ze^{\mu t} = 0$

We then choose $\mu$ so that $2\mu + k = 0$. By doing this, and dividing both sides with $e^{\mu t}$ we obtain:

$\ddot{z} + (\mu^2 + k\mu + \gamma + \beta cos(t))z = 0$

And now we just perform a simple change of variable, as shown in my last post, and we are done!

Does it look OK now? :)

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Note that the change of variables you are performing says $x(t) = z(t) \cdot \mathrm{e}^{\mu t}$, thus $\dot{x}(t) = \mu z(t) \cdot \mathrm{e}^{\mu t} + \dot{z}(t) \cdot \mathrm{e}^{\mu t}$. – Sasha Mar 25 '12 at 18:58
Oh yeah, you're right. Thanks. For some reason I thought of this as a constant. Think I need to get back to the drawing board for a while. Suddenly this got more complicated! – Kristian Mar 25 '12 at 19:11
Hi. I've tried a new approach to the problem. Would be very grateful if you could look at it again. – Kristian Mar 25 '12 at 19:35

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