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Is there a way to calculate $\psi$ which is a function of $x$ out of this differential equation:

$$ \frac{d^2 \psi}{d x^2} = x^2 \psi $$

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up vote 2 down vote accepted

$\dfrac{d^2\psi}{dx^2}=x^2\psi$

$\dfrac{d^2\psi}{dx^2}-x^2\psi=0$

Note that this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf.

Let $\psi=e^{-\frac{x^2}{2}}y$ ,

Then $\dfrac{d\psi}{dx}=e^{-\frac{x^2}{2}}\dfrac{dy}{dx}-xe^{-\frac{x^2}{2}}y$

$\dfrac{d^2\psi}{dx^2}=e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}-xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}+\left(x^2e^{-\frac{x^2}{2}}-e^{-\frac{x^2}{2}}\right)y=e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-2xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}+(x^2-1)e^{-\frac{x^2}{2}}y$

$\therefore e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-2xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}+(x^2-1)e^{-\frac{x^2}{2}}y-x^2e^{-\frac{x^2}{2}}y=0$

$e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-2xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}-e^{-\frac{x^2}{2}}y=0$

$\dfrac{d^2y}{dx^2}-2x\dfrac{dy}{dx}-y=0$

You can apply the procedure in Help on solving an apparently simple differential equation to get $y=c_1\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt}}{\sqrt{t}}dt+c_2\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt}}{\sqrt{t}}dt$

$\therefore\psi=c_1e^{-\frac{x^2}{2}}\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt}}{\sqrt{t}}dt+c_2e^{-\frac{x^2}{2}}\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt}}{\sqrt{t}}dt$

$\psi=c_1\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt-\frac{x^2}{2}}}{\sqrt{t}}dt+c_2\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt-\frac{x^2}{2}}}{\sqrt{t}}dt$

$\psi=C_1\int_0^\infty e^{-\frac{t^2}{4}+xt-\frac{x^2}{2}}~d(\sqrt{t})+C_2\int_0^\infty e^{-\frac{t^2}{4}-xt-\frac{x^2}{2}}~d(\sqrt{t})$

$\psi=C_1\int_0^\infty e^{-\frac{t^4}{4}+xt^2-\frac{x^2}{2}}~dt+C_2\int_0^\infty e^{-\frac{t^4}{4}-xt^2-\frac{x^2}{2}}~dt$

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