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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