- #1

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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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- Thread starter lavinia
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- #1

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

- #2

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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.

- #3

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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:

- #4

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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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