Discussion
If eigenfunctions corresponding to distinct eigenvalues \(\lambda\) of the Sturm-Liouville problem
\[ \begin{aligned} &\frac{d^{2}y}{dx^{2}} - 3\frac{dy}{dx} = \lambda y, \\ &0 \lt x \lt \pi, \end{aligned} \]
\[ y(0) = y(\pi) = 0 \]
are orthogonal with respect to the weight function w(x), then w(x) is
Answer: A
The weight function is \(e^{-3x}\).
Bring the equation to Sturm-Liouville form (p y')' + ... = \(\lambda w y\). Multiply by an integrating factor \(\mu(x)\) chosen so that \(\mu(y'' - 3y') = (\mu y')'\), which needs \(\mu' = -3\mu\), so \(\mu = e^{-3x}\):
\[ \begin{aligned} &\frac{d}{dx}\left(e^{-3x}\frac{dy}{dx}\right) \\ &= \lambda e^{-3x} y \end{aligned} \]
With p = \(e^{-3x}\) and the boundary conditions y(0) = y(\(\pi\)) = 0, eigenfunctions for distinct eigenvalues are orthogonal with weight \(w(x) = e^{-3x}\).
\(e^{3x}\) is the reciprocal, from taking the factor with the wrong sign, and \(e^{\pm 2x}\) does not turn \(y'' - 3y'\) into an exact derivative.
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