# Non-linear ode

1. Mar 13, 2007

### CPL.Luke

hey does anybody have any idea how to solve this equation?

E/m= xd^2x/dt^2

with initial conditions x=a and dx/dt(a)=0

its non-linear and so I don't have any idea what to do with it, and maple won't give me an answer.

Last edited: Mar 13, 2007
2. Mar 13, 2007

### Matthew Rodman

Yeah, it's a tricky Emden-Fowler equation. The only thing I can think of is to multiply by x' and integrate to get

$$\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{x} + C$$

Now, swap for x from the original equation, i.e.,

$$x = \frac{E}{m x^{\prime \prime}}$$

such that

$$\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{\frac{E}{m x^{\prime \prime}}} + C$$

rearrange to get

$$x^{\prime \prime} = A e^{-\frac{m x^{\prime 2}}{2 E}}$$

where A is a constant. Now make the change of variable

$$x^{\prime} = i \sqrt{\frac{2 E}{m}} z$$

to get

$$i \sqrt{\frac{2E}{m}} z^{\prime} e^{-z^2} = A$$

but of course

$$\frac{ d }{dt} erf{(z)} = \sqrt{\frac{2}{\pi}} z^{\prime} e^{-z^2}$$

so integrating gives you

$$erf{(z)} = \alpha t + \beta$$

where alpha and beta are constants that may or may not be complex

So there's a solution of the form

$$x(t) = i\sqrt{\frac{2 E}{m}} \int{erf^{-1}{(\alpha t + \beta)} dt}$$

Note: the inverse erf function is integrable -- see this Wolfram page.

Last edited: Mar 13, 2007