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