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

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 $$y=ax.$$

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