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

Author(s):
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 $u=C \sin⁡\theta+C'' \cos⁡\theta+ \frac{1}{2} \theta \sin⁡\theta.$