- #1
jostpuur
- 2,116
- 19
What are solutions to
[tex]
\psi''(x) = (a_0 + a_1 x)\psi(x)
[/tex]
?
First idea I've had was that I could try some kind of perturbation with respect to the [itex]a_1[/itex] variable, so that
[tex]
\psi(x) = A_1e^{\sqrt{a_0}x} + A_2e^{-\sqrt{a_0}x} + \psi_1(x)
[/tex]
would be an attempt. But I couldn't find anything useful for [itex]\psi_1[/itex] when [itex]a_1\neq 0[/itex], so it got stuck there.
Second idea was to use Fourier transform to transform the problem into this form
[tex]
\phi'(x) = (b_0 + b_1x^2)\phi(x)
[/tex]
Unfortunately the solutions
[tex]
\phi(x) = Be^{b_0x + \frac{1}{3}b_1x^3}
[/tex]
don't have converging Fourier transforms, so that doesn't help either.
[tex]
\psi''(x) = (a_0 + a_1 x)\psi(x)
[/tex]
?
First idea I've had was that I could try some kind of perturbation with respect to the [itex]a_1[/itex] variable, so that
[tex]
\psi(x) = A_1e^{\sqrt{a_0}x} + A_2e^{-\sqrt{a_0}x} + \psi_1(x)
[/tex]
would be an attempt. But I couldn't find anything useful for [itex]\psi_1[/itex] when [itex]a_1\neq 0[/itex], so it got stuck there.
Second idea was to use Fourier transform to transform the problem into this form
[tex]
\phi'(x) = (b_0 + b_1x^2)\phi(x)
[/tex]
Unfortunately the solutions
[tex]
\phi(x) = Be^{b_0x + \frac{1}{3}b_1x^3}
[/tex]
don't have converging Fourier transforms, so that doesn't help either.