Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 10

Author(s): 
Adrian Rice (Randolph-Macon College)

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 \[u=C \sin⁡\theta+C'' \cos⁡\theta+ \frac{1}{2} \theta \sin⁡\theta.\]

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