First Order Nonlinear Differential Equation

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SUMMARY

The discussion focuses on solving the first-order nonlinear differential equation given by (dy/dx)^2 = ((ay^4)/2) - (a+1)y^2 + 1, with initial conditions y(0)=0 and y'(0)=1, where 'a' is constrained within the interval [0,1]. Participants highlight that a direct approach leads to complex results, specifically an elliptic integral, which complicates the solution process. The conversation emphasizes the need for alternative methods to simplify the resolution of this equation.

PREREQUISITES
  • Understanding of first-order nonlinear differential equations
  • Familiarity with elliptic integrals
  • Basic knowledge of initial value problems
  • Proficiency in mathematical notation and terminology
NEXT STEPS
  • Research methods for solving first-order nonlinear differential equations
  • Explore techniques for simplifying elliptic integrals
  • Study initial value problem-solving strategies in differential equations
  • Learn about numerical methods for approximating solutions to complex ODEs
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Mathematicians, students studying differential equations, and researchers dealing with nonlinear dynamics will benefit from this discussion.

frank1234
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Hi, I need help solving this ode, when I try to solve it I end up with a big crazy answer and I believe it should be simpler.

(dy/dx)^2=((ay^4)/2)-(a+1)y^2+1

y(0)=0, y'(0)=1 and a is within [0,1]
 
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frank1234 said:
Hi, I need help solving this ode, when I try to solve it I end up with a big crazy answer and I believe it should be simpler.

(dy/dx)^2=((ay^4)/2)-(a+1)y^2+1

y(0)=0, y'(0)=1 and a is within [0,1]

I'm not surprised because that is the natural consequence of ...

http://mathhelpboards.com/differential-equations-17/need-help-solving-2nd-order-nonlinear-differential-equation-12222.html#post58240

Unfortunately the 'direct attack' to the ODE...

$\displaystyle y^{\ '} = \sqrt{a\ y^{4} - (1 + a) y^{2} + 1}\ (1)$

... leads to an elliptic integral, a rather indigestible dish:( ... but never say never again!...

By the way, welcome to MHB! :)...

Kind regards

$\chi$ $\sigma$
 

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