Linear ode with variable coefficient

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Discussion Overview

The discussion revolves around solving a linear ordinary differential equation (ODE) with variable coefficients, specifically the equation p''(x) - 2(1-3x^2)p'(x) = 0. Participants explore methods for finding explicit solutions and discuss the challenges associated with the non-constant coefficients.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using a substitution q(x) = p'(x) to transform the original equation into a first-order linear ODE, which can be solved using an integrating factor.
  • Another participant provides the integrating factor and derives a solution for q(x), but notes that the resulting function does not have a "nice" antiderivative, leading to difficulties in finding p(x) explicitly.
  • A different participant argues that an explicit solution is available through an integral expression, although they acknowledge that it cannot be solved in closed form.
  • There is a mention of the possibility of using the Frobenius method to find a polynomial solution, but it is suggested that this would yield similar results to the previous approaches.
  • Participants express differing views on the necessity of an explicit solution, with some indicating that the derived expressions are sufficiently explicit despite not being closed-form solutions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of an explicit solution or the nature of the solutions derived. There are competing views on whether the existing expressions are satisfactory and whether alternative methods like the Frobenius method would provide different insights.

Contextual Notes

There are unresolved issues regarding the closed-form solvability of the integral mentioned, and the discussion reflects varying interpretations of what constitutes an explicit solution in this context.

leinadrc
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Does anyone have any guidance on how to solve a problem of thes form

p''(x) - 2(1-3x^2)p'(x) = 0


It looks very similar to a characteristic equation, but the coefficients aren't constant
 
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Sort of, let q(x) = p'(x) so that q'(x) = p''(x). The "new" equation is

q'(x) - 2(1 - 3x^2)q(x) = 0,

which can be solved with an integrating factor.

In this case, the integrating factor is e^(2x(x^2 - 1)).

e^(2x(x^2 - 1))q'(x) + e^(2x(x^2 - 1))(-2)(1 - 3x^2)q(x) = 0
d/dx( e^(2x(x^2 - 1)) * q(x) ) = 0
e^(2x(x^2 - 1)) * q(x) = C
q(x) = Ce^(-2x(x^2 - 1))

So p'(x) = Ce^(-2x(x^2 - 1)). But that function doesn't seem to have a "nice" antiderivative, so I don't know how to continue.
 
so what?? do you need an explicit solution?? you have it

[tex]p(x)=p(x_0)+p'(x_0)e^{2(x_0^3-x_0)}\int_{x_0}^{x}e^{2(\xi-\xi^3)}d\xi[/tex]

you may have a polinomial solution using Frobenius method, but youll get the same thing (as expected).
 
Last edited:
I did do that and I did get p'(x)=c exp(2*(x-x^3)) also...yeah I do need an explicit solution...and this is also where I got stuck because I couldn't see how to solve for p explicitly
 
it is solved explicitly, you have the expresion right there...

the integral above can't be solved in a closed form, but it is as explicit as a trig function
 

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