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## Homework Statement

Solving the following differential equation with the given boundary conditions:

[tex]\hbar^2 \frac{d^2}{dx^2}\psi (x) = 2mE\psi (x), \ \ \ \ \ \forall \ \hbar^2,\ m,\ E > 0 [/tex]

[tex]\psi(a) = \psi(-a) = 0 [/tex]

## Homework Equations

## The Attempt at a Solution

[tex]\hbar^2 \frac{d^2}{dx^2}\psi (x) = 2mE\psi (x) [/tex]

[tex]\frac{d^2}{dx^2}\psi (x) = \frac{2mE}{\hbar^2}\psi (x) [/tex]

[tex]\frac{d^2}{dx^2}\psi (x) - \frac{2mE}{\hbar^2}\psi (x) = 0 [/tex]

Let

[tex]\frac{2mE}{\hbar^2} = \rho[/tex]

Then

[tex]\frac{d^2}{dx^2}\psi (x) - \rho \psi (x) = 0 [/tex]

The characteristic polynomial for this ODE shall be

[tex]r^2 - \rho = 0 [/tex]

[tex]r^2 = \rho ,\ r_1 = -\sqrt{\rho},\ r_2 = \sqrt{\rho} [/tex]

Therefore

[tex]\psi (x) = c_1e^{r_1 x} + c_2e^{r_2 x} [/tex]

[tex]\psi (x) = c_1^{-\sqrt{\rho} x} + c_2e^{\sqrt{\rho} x} [/tex]

By using the given conditions the only possible solution I can get is

[tex]c_1 = c_2 = 0,\ \psi (x) = 0 [/tex]

As I get to something like

[tex]e^{4 \sqrt{\rho} a} = 1 [/tex]

And since

[tex]\rho,\ a > 0[/tex]

So

[tex]\psi(x) = 0 [/tex]

seems to be the only solution.

Is this it?

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