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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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    Possibly related: [Solving wave equation](http://physics.stackexchange.com/questions/46306/solving-wave-equation).2012-12-08
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    Is $y$ a constant? a function of $x$? Is $\psi$ just a function of one variable, $x$? Are you trying to solve for $y$? or for $\psi$?2012-12-08
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    @m0nhawk thats me my physics page was hacked i still dont know why!.2012-12-08
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    @Gerry Myerson just $\psi(x)$2012-12-08
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    @m0nhawk what you think, is it true?2012-12-08
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    Neo, you are a person of few words, far too few for me to understand how your reply to my comment engages with all my concerns.2012-12-09
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    @Gerry Myerson omm, ok just tell me, can i square the wave function $\sqrt {\psi}$?2012-12-09
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    I might be able to answer that if I knew the answers to the questions you have ignored.2012-12-09
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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 $\psi(x)=0$.