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Topic: solution of the diff eqn, (y')^2 = (y^4) + A, where A < 0 ?
Replies: 4   Last Post: Jun 25, 2011 2:56 PM

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achille

Posts: 575
Registered: 2/10/09
Re: solution of the diff eqn, (y')^2 = (y^4) + A, where A < 0 ?
Posted: Jun 25, 2011 5:23 AM
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On Jun 25, 4:09 pm, hanrahan...@yahoo.co.uk wrote:
> I'd be grateful for some help with this differential equation:
>
> (y')^2 = (y^4) + A, where A is a negative constant
>
> Clearly if A could be 0, then (1/x) is a solution. But we are given
> that A < 0.
>
> Apparently y must have two zeros.
>
> Thanks!
>
> Michael


Let A = -a^4, the ODE can be rewritten into the form:

dt = dy / sqrt(y^4 - a^4)

In general, if an ODE has the form dt = dy / sqrt(P(y)) where
P is an quartic polynomial, then y(t) can be expressed in terms
of Jacobi elliptic functions. For you case, the general solution
is:

a/cn( sqrt(2)(a t + const), 1/sqrt(2) )

For more details, look at Wiki's entry on Jacobi elliptic functions

http://en.wikipedia.org/wiki/Jacobi%27s_elliptic_functions

For your own benefit, you should verify for your y(t), if you define
x(t) such that y(t) = a / x(sqrt(2)at), then x(t) satisfies the ODE
for cn(t,1/sqrt(2)):

x'^2 = (1 - x^2)( 1 - (1/2) + (1/2) x^2) = (1-x^4)/2

Hope this helps.



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