That one's nasty. WolframAlpha gives parabolic cylinder functions. Mathematica gives hypergeometric functions. In any case, a series solution is, I think, the way to get where you need to go.
That one's nasty. WolframAlpha gives parabolic cylinder functions. Mathematica gives hypergeometric functions. In any case, a series solution is, I think, the way to get where you need to go.
If we write the ODE as...
(1)
... it is clear that the general solution is...
(2)
... where...
(3)
... is an even function and...
(4)
... is an odd function. The and are found directly from (1) imposing initial conditions ...
(5)
(6)
(7)
(8)
Of course You can proceed in the computation... in next post we will try to find a general expression for the ...
Kind regards
In the previous post it has been demonstrated that the second order linear ODE...
(1)
... has general solution...
(2)
... where...
(3)
... is an even function and...
(4)
... is an odd function. In general the computation od the derivatives of y(*) from (1) can be done recursively as...
(5)
... and from (5), setting , we can derive the recursive relations...
(6)
(7)
Deriving explicit solutions of the recursive relations (6) and (7) seems not very confortable... the recursive computation of the and is however very easy. The figure shows the values of and in computed using 25 terms of (3) and (4)...
Kind regards