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

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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 $y=ax.$