Solving 2nd Order ODE: du/ds & k^2

[lavinia]

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

 

[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.

 

[jackmell]

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

That lends itself to interpretation:

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

that's doable right?

 

[lavinia]

"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.

thanks - pretty cool