Solving a Non-Linear Equation: Need Help!

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SUMMARY

The discussion focuses on solving the non-linear Emden-Fowler equation represented as E/m = xd²x/dt² with initial conditions x=a and dx/dt(a)=0. The solution involves multiplying by x' and integrating to derive the equation ½x'² = (E/m) ln{x} + C. A change of variable is introduced, leading to the formulation x'' = A e^(-m x'²/2E), where A is a constant. The final solution is expressed as x(t) = i√(2E/m) ∫erf⁻¹(αt + β) dt, utilizing the inverse error function.

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  • Understanding of non-linear differential equations, specifically the Emden-Fowler equation.
  • Familiarity with integration techniques and the error function (erf).
  • Knowledge of initial value problems in differential equations.
  • Proficiency in mathematical software tools like Maple for symbolic computation.
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  • Study the properties and applications of the Emden-Fowler equation in mathematical physics.
  • Learn advanced integration techniques involving special functions, particularly the error function.
  • Explore numerical methods for solving non-linear differential equations.
  • Investigate the use of Maple for solving complex differential equations and symbolic manipulation.
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Mathematicians, physicists, and engineers dealing with non-linear dynamics, as well as students and researchers interested in advanced differential equations and their solutions.

CPL.Luke
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hey does anybody have any idea how to solve this equation?

E/m= xd^2x/dt^2with initial conditions x=a and dx/dt(a)=0

its non-linear and so I don't have any idea what to do with it, and maple won't give me an answer.
 
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Yeah, it's a tricky Emden-Fowler equation. The only thing I can think of is to multiply by x' and integrate to get

[tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{x} + C[/tex]

Now, swap for x from the original equation, i.e.,

[tex]x = \frac{E}{m x^{\prime \prime}}[/tex]

such that

[tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{\frac{E}{m x^{\prime \prime}}} + C[/tex]

rearrange to get

[tex]x^{\prime \prime} = A e^{-\frac{m x^{\prime 2}}{2 E}}[/tex]

where A is a constant. Now make the change of variable

[tex]x^{\prime} = i \sqrt{\frac{2 E}{m}} z[/tex]

to get

[tex]i \sqrt{\frac{2E}{m}} z^{\prime} e^{-z^2} = A[/tex]

but of course

[tex]\frac{ d }{dt} erf{(z)} = \sqrt{\frac{2}{\pi}} z^{\prime} e^{-z^2}[/tex]

so integrating gives you

[tex]erf{(z)} = \alpha t + \beta[/tex]

where alpha and beta are constants that may or may not be complex

So there's a solution of the form

[tex]x(t) = i\sqrt{\frac{2 E}{m}} \int{erf^{-1}{(\alpha t + \beta)} dt}[/tex]

Note: the inverse erf function is integrable -- see this Wolfram page.
 
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