Discussion
If a transformation \(y = uv\) transforms the given differential equation \(f(x)y'' - 4f'(x)y' + g(x)y = 0\) into the equation of the form \(v'' + h(x)v = 0\), then \(u\) must be
Answer: D
The multiplier is \(u = f^{2}\).
Write the equation as \(y'' + Py' + Qy = 0\) with \(P = -4f'/f\). Substituting \(y = uv\) gives
\[ uv'' + (2u' + Pu)v' + (u'' + Pu' + Qu)v = 0, \]
and the \(v'\) term disappears only if \(2u' + Pu = 0\), i.e. \(u = e^{-\frac{1}{2}\int P\,dx}\). Here
\[ -\frac{1}{2}\int P\,dx = 2\int \frac{f'}{f}\,dx = 2\ln f, \]
so \(u = f^{2}\). The reciprocal \(1/f^{2}\) comes from a sign slip in the exponent, and \(xf\) and \(1/2f\) do not remove the first-derivative term.
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