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.

$\dfrac{dx}{dt}=2\dfrac{\sqrt{2g(\sin c- \sin x)}}{\sqrt{l}}$. $g$, $c$, and $l$ are all constants. Is there a way to solve for $x$ in terms of $t$ here? Once I did separation of variables and plugged in the integral into wolframalpha I got a pretty horrendous integral on the side with $x$. I was wondering if it could perhaps be simplified especially when you solve for $x$ in terms of $t$. Thanks

share|improve this question
1  
Nonlinear, inhomogeneous, implicit ODE? Maybe it's doable, but my gut feeling says NO. –  Kaster May 2 '13 at 2:14
1  
Hold on, are you trying to solve pendulum problem? There's no closed form for that in elementary functions. –  Kaster May 2 '13 at 2:20
    
Yup lol how did you know? Did you try it once yourself too? –  Ovi May 2 '13 at 2:21
    
I know because the ODE looks familiar. I learned being in high school that this ODE cannot be solved in elementary functions. You can use elliptic integral though. Check here for more info. –  Kaster May 2 '13 at 2:24
2  
Ordinary Differential Equation. That's the thing you're trying to solve. –  Kaster May 2 '13 at 2:26

1 Answer 1

up vote 1 down vote accepted

If it's OK for you to use the Taylor series approximation of $\sin(x)$ as $\sin(x) \approx x$ (for small x), then you can rewrite your equation as

$\dfrac{dx}{dt}=2\dfrac{\sqrt{A - 2gx}}{\sqrt{l}}$

where $A = 2g\sin(c)$.

You then have a more straightforward separable equation with the solution

$x(t) = \frac{Al - g^2(c_1 +2t)^2}{2gl}$.

This is called the "small angle approximation."

share|improve this answer
    
Thanks, I am aware of the pendulum equation for small angles but I was trying to find this for a physics problem involving a lever falling from above the horizontal. However, from the comments I found out that this does not have a closed form in elementary functions, and it is beyond my purposes to get a really messy answer to this question. –  Ovi May 2 '13 at 23:32

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.