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How to solve this wave equation:

$ay^2\psi+by\psi+\frac {d^2\psi}{dx^2}=0$ here $c=\frac {d^2\psi}{dx^2}$ where, the equation is a quadratic equation (a univariate polynomial equation of the second degree). example: $ay^2+by+c=0,$

$y=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$

the problem is that here there are two wave functions after $y^2$ and $y$:

$ay^2\psi+by\psi+c=0,$

thats why i dont know how to solve it.

Remark: Is this solution true?

$(y+\frac {b}{2a})^2\psi=\frac {(b^2\psi-4ac)}{4a^2},$ to solve this i take the wave function under the radical. $y\sqrt {\psi}=\frac{-b\sqrt {\psi}\pm\sqrt{b^2\psi-4ac}}{2a}$

isn't it wrong to take take the wave function under the radical$\sqrt {\psi}$ ?

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    i didnt ingnored anything or anyone!!! Is$y$a constant? i dont know!, but$y$is not a function. Is $\psi$ just a function of one variable,$x$? yes im trying to solve for y2012-12-09

2 Answers 2

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sn't it wrong to take take the wave function under the radical? no, it is not wrong. for example: $\psi=A\sin kx$ $\sqrt {\psi}=\sqrt {A\sin kx}$

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If you are trying to solve $a\psi y^2+b\psi y+{d^2\psi\over dx^2}=0$ for $y$, where $a,b$ are constants, $a\ne0$, and $\psi$ is a function of $x$, then I can't see any reason for not using the quadratic formula, getting $y={-b\psi\pm\sqrt{b^2\psi^2-4a\psi\psi''}\over2a\psi}$ where I have written $\psi''$ for ${d^2\psi\over dx^2}$. One must avoid those $x$ for which $\psi(x)=0$.