Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 7

Adrian Rice (Randolph-Macon College)

In trying to solve the differential equation in Problem 7, \[\frac{dy}{dx}=\frac{y}{x},\] Lovelace’s reasoning was that, since the derivative \(dy/dx\) is a limit, and since a limit is a ‘constant \& fixed thing’, this must imply that the ratio \(y/x\) is also a constant. Calling this constant \(a\) would mean that \[\frac{y}{x}=a\] or \[y=ax.\]

But of course the flaw in this argument is that limits of functions are not in general equal to constants. Indeed, taking the standard definition of the derivative as \[f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] we can see that, while the quantity \(h\) approaches zero, the variable \(x\) is free to assume any value whatsoever—quite the opposite of a constant.

The simplest method for solving the equation correctly would be to separate the variables and integrate, giving \[\int \frac{dy}{y} =\int \frac {dx}{x}\] or \[ \log ⁡ y = \log ⁡x+c,\] which, letting \(a=e^c\), is \[y=ax.\]

Return to Main Solutions Page.
Continue to Solution to Problem 8.