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

Our final problem comes close to the end of Lovelace’s correspondence course with De Morgan. By early November 1841, she had progressed to the subject of second-order differential equations. One of the examples with which she had trouble was on page 156 of De Morgan’s Calculus. Given the nonhomogeneous equation

\[\frac{d^2 u}{d\theta^2 }+u=\cos⁡\theta\]

De Morgan gave its general solution as

\[u=C \sin ⁡\theta + C' \cos⁡ \theta + \sin⁡\theta \int cos^2 \theta d \theta - \frac{1}{2}  \cos⁡ \theta \int \sin ⁡2\theta  d\theta,\] which, since

\[ \int \cos^2 \theta d\theta = \frac{1}{2} \theta + \frac{1}{4} \sin 2 ⁡  \theta \, \,   \mbox{ and  } \,\, \int \sin ⁡2\theta  d\theta = -\frac{1}{2} \cos ⁡2\theta \] resulted in

\[u=C \sin\theta  + C' \cos⁡ \theta +\frac{1}{2}\theta \sin⁡ \theta + \frac{1}{4}  \sin ⁡2 \theta  \sin⁡ \theta+  \frac{1}{4} \cos⁡\theta  \cos ⁡2\theta.\] Using double angle formulae, he converted this into

\[u=C \sin⁡ \theta + C'  \cos⁡ \theta + \frac{1}{2}\theta \sin \theta+ \frac{1}{4}  \cos⁡ \theta,\] which he then expressed in its final form as

\[u=C \sin⁡ \theta+C'  \cos⁡ \theta + \frac{1}{2} \sin⁡ \theta,\] along with a challenge: ‘Explain this step?’ (See Figure 12).

Figure 12. Problem from page 156 of De Morgan’s Differential and Integral Calculus.

The trouble was, as Lovelace remarked in a letter to De Morgan, ‘I cannot “explain this step”.’ She noted that ‘in the previous line, we have:   \[\begin{array}{r}  (1) \ldots u = C \sin⁡  \theta +C'  \cos⁡ \theta + \frac{1}{2}\theta  \sin⁡ \theta + \frac{1}{4} \cos \theta ⁡\mbox{ (quite clear)} \end{array}\]   \[\begin{array}{rcl}  (2) \ldots  \mbox{ And }  u &=&\cos⁡ \theta -\frac{d^2 u}{d \theta ^2 } \mbox{ (by hypothesis)} \\                                                             &=& \frac{1}{4}  \cos⁡ \theta+\left (\frac{3}{4} \cos⁡ \theta-\frac{d^2 u}{d\theta^2} \right ) \end{array}\]

whence one may conclude that \[C \sin⁡\theta+C'  \cos⁡ \theta + \frac{1}{2}\theta\sin⁡ \theta=\frac{3}{4}\cos⁡ \theta-\frac{d^2 u}{d\theta^2} \] But how \(u=C \sin⁡  \theta+C'  \cos⁡  \theta+\sin⁡\theta\cdot\frac{1}{2} \theta\) is to be deduced I do not discover.’ [LB 170, 4 Nov. [1841], ff. 132v-133r]

Not deterred, she tried again:

By subtracting \(\frac{1}{4}\cos⁡ \theta\)  from both sides of (1), we get \[u-\frac{1}{4} \cos⁡ \theta= C \sin⁡ \theta+C' \cos⁡ \theta+ \frac{1}{2} \theta \sin⁡ \theta\] But unless \(\frac{1}{4} \cos⁡ \theta=0\), (which would only be the case I conceive if \(\theta=\pi/2\)), I do not see how to derive the equation \(\ldots\) [LB 170, 4 Nov. [1841], f. 133r].

De Morgan’s final solution is certainly correct. So can you explain why the \(\frac{1}{4} \cos⁡ \theta\) mysteriously disappears?

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