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

Adrian Rice (Randolph-Macon College)

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).

From page 156 of De Morgan's calculus textbook.

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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