# A second order ODE

1. Oct 22, 2011

### lavinia

d$^{2}$u/ds$^{2}$= cosu[(du/ds)$^{2}$ - k$^{2}$]

2. Oct 22, 2011

### HallsofIvy

"cosu" is cos(u)? Since the independent variable, s, does not appear explicitely in that equation you can use a technique called "quadrature".

Let v= du/ds so that $d^2u/ds^2=dv/ds$ but, by the chain rule, dv/ds= (dv/du)(du/ds)= v(dv/du) so your equation becomes $v dv/du= cos(u)(v^2+ k^2)$

That's a separable first order equation. Once you have solved it for v, integrate to find u.

3. Oct 22, 2011

### jackmell

That lends itself to interpretation:

$$\frac{d^2 u}{ds^2}=\cos(u(v))\biggr|_{v=(u')^2-k^2}$$

that's doable right?

Last edited: Oct 22, 2011
4. Oct 23, 2011

### lavinia

thanks - pretty cool