# 2nd order ODE including a quadratic

• Dec 30th 2010, 11:06 PM
smolloy
2nd order ODE including a quadratic
Hi all,
I am trying to solve for the rocking frequency of a semi-circular object (imagine a cylinder cut down the z axis so that the cross section is semi-circular), and I have got stuck right at the end.

I figured out the location of the centre of mass so that I could write down the torque experienced by the object when sitting at any angle, and am now faced by a differential equation that looks substantially trickier than I imagined.

Note that I converted sines and cosines to the following:
sin(x) ---> x
cos(x) ---> 1 - 0.5x^2

After all this work I am left with an equation of the form: x'' = f(x)
Where f(x) contains a linear and quadratic term.

Can anyone give me any clues as to how I might attack this equation?

Thanks!
• Dec 31st 2010, 02:20 AM
HallsofIvy
Since the independent variable (which I will call "t") does not appear in that equation there is a standard method of reducing the order of the equation (often called "quadrature").

Let y= x'. Then x''= y'= $\displaystyle \frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}= y\frac{dy}{dx}$.

So the second order equation for x as a function of t converts to a first order equation for y as a function of x:
$\displaystyle y\frac{dy}{dx}= f(x)$
$\displaystyle \int y dy= \int f(x)dx$

$\displaystyle \frac{1}{2}y^2= \int f(x)dx+ C$
(That $\displaystyle y^2$ is the reason for the name "quadrature".)

$\displaystyle y= \frac{dx}{dt}= \pm\sqrt{2\left(\int f(x)dx+ C\right)}$

$\displaystyle \int \frac{dx}{\sqrt{2\left(\int f(x)dx+ C\right)}}= t+ D$

How difficult those integrals are to do depends strongly on the function f(x).
• Dec 31st 2010, 03:36 AM
smolloy
Thanks HallsofIvy. That's a nice trick I will have to remember.

Since f(x) is polynomial in x (with only linear and quadratic terms), the first integral should be quite simple to do. The second looks pretty horrible though :( How to do an integral of the inverse of a third order polynomial?! Sounds like another variable substitution might help. Perhaps I should make life easy for myself, and remove the quadratic term by approximating cos(x)--->1.....

Is there a name for differential equations of the form: x''=f(x) ?