Benedict Irwin edited General Solution.tex  over 9 years ago

Commit id: 335b896cb7c9c1c0053da31bc78b7352bc9c8820

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In this circumstance we have \begin{equation}  \frac{d\psi(x)}{dx}=G(x)\psi(x)-\frac{kV'(x)\psi(x)}{\alpha-V(x)} \frac{d\psi(x)}{dx}=-V(x)G(x)\psi(x)-\frac{kV'(x)\psi(x)}{\alpha-V(x)}  \end{equation}  "RECHECK WORKING FROM THIS POINT MAY HAVE MISSED A FACTOR OF V(x)"  and then \begin{equation}  \frac{d^2\psi(x)}{dx^2}=\bigg( G' + G^2 - \frac{kGV'}{\alpha-V} +