To reconcile the solutions to Problem 10 given by Lovelace and De Morgan, how can the solution \[u=C \sin\theta+C' \cos\theta+\frac{1}{2} \theta \sin \theta+ \frac{1}{4} \cos\theta\] be converted into its final form \[u=C \sin\theta+C' \cos\theta+\frac{1}{2} \theta \sin\theta ?\] As Lovelace suggested, substituting \(\theta=\frac{\pi}{2}\) would obviously eliminate the \(\frac{1}{4} \cos\theta\) term, but that clearly would not be correct as the other terms would also be changed. Why then does the \(\frac{1}{4} \cos\theta\) disappear?
The answer lies in the fact that the two \(\cos\theta\) terms both have constant coefficients, namely, \(C'\) and \(\frac{1}{4}\). They can thus be combined into a single term \[\left(C'+\frac{1}{4}\right) \cos\theta\] or \[C'' \cos\theta\] where \(C''=C'+\frac{1}{4}\). Thus De Morgan’s final answer, while not wrong, is slightly misleading, since in light of the above discussion it would be better expressed as \[u=C \sin\theta+C'' \cos\theta+ \frac{1}{2} \theta \sin\theta.\]