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

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Discussion Overview

The discussion centers around solving a second-order ordinary differential equation (ODE) of the form d²u/ds² = cos(u)[(du/ds)² - k²]. Participants explore techniques for solving this equation, including the use of quadrature and the transformation of variables.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the original second-order ODE and raises the question of the interpretation of "cosu".
  • Post 2 suggests using the substitution v = du/ds to transform the equation into a separable first-order equation, indicating that the independent variable s does not appear explicitly.
  • Post 3 reiterates the original equation and proposes an interpretation involving the substitution of v, questioning the feasibility of the approach.
  • Post 4 echoes the earlier suggestion of using quadrature and the substitution method, affirming the process and expressing appreciation for the technique.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using quadrature and the substitution of variables, but there is no consensus on the interpretation of "cosu" or the overall feasibility of the proposed methods.

Contextual Notes

The discussion does not clarify certain assumptions regarding the nature of the function u or the conditions under which the proposed methods apply. The dependence on the interpretation of "cosu" remains unresolved.

lavinia
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d[itex]^{2}[/itex]u/ds[itex]^{2}[/itex]= cosu[(du/ds)[itex]^{2}[/itex] - k[itex]^{2}[/itex]]
 
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"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.
 
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?
 
Last edited:
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
 

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