# Solving Polynomial wave equation of the second degree

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. – m0nhawk Dec 8 '12 at 21:38
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 $\psi$? – Gerry Myerson Dec 8 '12 at 21:38
@m0nhawk thats me my physics page was hacked i still dont know why!. – Neo Dec 8 '12 at 21:42
@Gerry Myerson just $\psi(x)$ – Neo Dec 8 '12 at 21:43
@m0nhawk what you think, is it true? – Neo Dec 8 '12 at 22:59

sn't it wrong to take take the wave function under the radical? no, it is not wrong. for example: $\psi=A\sin kx$ $\sqrt {\psi}=\sqrt {A\sin kx}$
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 $\psi(x)=0$.