Linear ode with variable coefficient

[leinadrc]

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

 

[Muzza]

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.

 

[ReyChiquito]

so what?? do you need an explicit solution?? you have it

$$p(x)=p(x_0)+p'(x_0)e^{2(x_0^3-x_0)}\int_{x_0}^{x}e^{2(\xi-\xi^3)}d\xi$$

you may have a polinomial solution using Frobenius method, but youll get the same thing (as expected).

 

[leinadrc]

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

 

[ReyChiquito]

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