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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



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


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) = (1x^4)/2
Hope this helps.



