Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

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 In consequence, the second order differential equation has been replaced by a system of two first order differential equations, namely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 $v$.

share|improve this answer
    
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 there....wait...do 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
add comment

$$\dot{x}=v$$

$$\dot{v}=x^3-x^5-v$$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.