Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 9

Since  \(y^2=2ax+x^2=(2a+x)x\) in Lovelace's approach to Problem 9,  using the product rule we have   \[ydy=(x+a)dx    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,            [2]\] so that \[(x+a)dy+ydy=(x+a)(dx+dy)\] or \[(x+a+y)dy=(x+a)(dx+da+dy)\] and thus \[\frac{dy}{x+a} = \frac{d(x+a+y)}{x+a+y}.\] Now since, by [2], \[\frac{dy}{x+a}=\frac{dx}{y}\] this means that \[\frac{dx}{y}=\frac{d(x+a+y)}{x+a+y}.\] So, recalling that \(y^2=2ax+x^2\), we have \[\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(x+a+\sqrt{2ax+x^2}) = \log\left (\frac {x}{2} +\frac{a}{2} + \frac{\sqrt{2ax+x^2}}{2} \right )+ \log ⁡2\] which is Lovelace’s result.

Now recall that De Morgan had obtained \[\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(x+a+\sqrt{2ax+x^2}) + \log ⁡2.\] Strictly speaking, his result should read \[\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(2x+2a+2\sqrt{2ax+x^2}) +C\] and Lovelace’s result should be \[\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(x + a +\sqrt{2ax+x^2}) +C'.\] Thus, letting Lovelace’s constant of integration \(C'=C+\log ⁡2\) reveals that the two results are equivalent and also explains De Morgan’s remark that the \(\log 2\), ‘being a Constant, matters nothing.’

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