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

Author(s):
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 $y=ax.$

Adrian Rice (Randolph-Macon College), "Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 7," Convergence (September 2021)