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So I have an equation of motion with an additional viscous force shown below:
$ \frac{d^2x}{dt^2} = x^3 - x^5 - \frac{dx}{dt} $

And the question is Rewrite as a system for x(t) and v(t). I don't even understand how to begin this problem. Any ideas?

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Let $\frac{\mathrm{d}x}{\mathrm{d}t}=\dot{x}=v$ and $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=\ddot{x}=\dot{v}$. The original equation can now be written as $\dot{v}=x^3-x^5-v$. In consequence, the second order differential equation has been replaced by a system of two first order differential equations, namely

$$\left\{\begin{array}{l l} \dot{x}=v\\ \dot{v}=x^3-x^5-v \end{array}\right.$$

Note that the derivatives depend only on $x$ and $v$.

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okay I'm with you so far in terms of converting the equation to a first order. But I still don't understand what to do from i solve it like a linear equation $\dot{v} + v = x^3-x^5$ – Richard May 3 '13 at 0:47
This system of equations is non-linear, so it will be difficult to solve it analytically. However, there are plenty of numerical methods for systems of first order differential equations. If you have initial conditions for $x$ and $v$, then you can easily compute the derivatives at that time, which can then be used to find $x$ and $v$ at another time; see Euler method. – Librecoin May 3 '13 at 1:22



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sorry, I'm really stupid and I'm not seeing much. I actually wrote my equation like $a = x^3 - x^5 - v$, kinda the same ...but then what do I do from here – Richard May 3 '13 at 0:35
$a$ is the time derivative of $v$, that's the point. Now you have two first-degree equations rather than a single, second-degree equation. – Ron Gordon May 3 '13 at 0:37
so like I said below, I just solve it like a linear equation? – Richard May 3 '13 at 0:57
@Richard: not really - the equations are coupled together, and non-linear. This is just another way of expressing them, it doesn't really simplify things much except in that it keeps the unknowns to first-order derivatives. – Ron Gordon May 3 '13 at 1:12

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