Solving second order ODE with d²u/ds² = cos u[(du/ds)² - k²]

lavinia · Oct 22, 2011

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

HallsofIvy · Oct 22, 2011

"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 [itex]d^2u/ds^2=dv/ds[/itex] but, by the chain rule, dv/ds= (dv/du)(du/ds)= v(dv/du) so your equation becomes [itex]v dv/du= cos(u)(v^2+ k^2)[/itex]

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

jackmell · Oct 22, 2011

lavinia said:

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

That lends itself to interpretation:

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

that's doable right?

lavinia · Oct 23, 2011

HallsofIvy said:

"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 [itex]d^2u/ds^2=dv/ds[/itex] but, by the chain rule, dv/ds= (dv/du)(du/ds)= v(dv/du) so your equation becomes [itex]v dv/du= cos(u)(v^2+ k^2)[/itex]

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

thanks - pretty cool

Solving second order ODE with d²u/ds² = cos u[(du/ds)² - k²]

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Solving second order ODE with d²u/ds² = cos u[(du/ds)² - k²]

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