Difference equation from the square potential step?

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Rearranged slightly, the equation above it is

[tex] \frac{\partial^2 \phi_\epsilon \left( x \right)}{\partial x} = \frac{2m}{\hbar^2} \left( V_\epsilon \left( x \right) - E \right) \phi_\epsilon \left( x \right).[/tex]

What happens when you integrate both sides of this equation from [itex]x = x_1 - \epsilon[/itex] to [itex]x = x_1 + \epsilon[/itex]?