6

In know that in a rigorous setting $dy \over dx$ isn't a fraction but is instead defined as the function: $\frac{d}{dx}y$ . Treating it as a fraction is technically considered an abuse of notation that is sometimes done as a shortcut for solving problems or explaining concepts. However, I have been trying to help another person with their intro to differential equations class and am confused on what the solution to the following problem is in a rigorous setting. Here is the problem (and solution):
$$ dy - (y-1)^2 \space dx = 0 $$ Such that (given in their notes):
$$ p(y) \space dy = g(x) \space dx $$ Then (according to their solution): $$ dy = (y-1)^2 \space dx \rightarrow \frac{1}{(y-1)^2} \space dy = dx$$ Which in turn implies that: $$\int (y-1)^{-2} \space dy = \int dx = \frac{-1}{y-1} = x + c $$ $$\rightarrow y = 1 - \frac{1}{x+c}$$ I' ve heard before that the $dy$ and $dx$ in integrals can be thought of a "signaling" which variable to integrate, but here you can see them being abused as symbolic variables that are manipulated as if they were algebraic symbols.

Not only that, but they are being manipulated as if there was an additive inverse $-dx$ for the $dx$, otherwise the first step of the solution wouldn't be possible. My friend is confused on what the professor is doing, and I can't in good faith right now explain to him why this is allowable, because this solution feels like a heavy abuse of notation, and I don't know what is happening conceptually that allows the solution.

What allows you to treat $dx$ and $dy$ as real-valued elements of a fraction in this solution? In a rigorous setting, what is really going on that allows you to solve this problem for $y$?

Nate
  • 778

1 Answers1

4

Rigorously you apply the chain rule which in integration is the substitution rule: $$ y'(x)=f(y(x))\implies \int_a^b dx = \int_a^b \frac{y'(x)}{f(y(x))}dx=\int_{y(a)}^{y(b)}\frac1{f(u)}du, $$ the expression $y(x)$ gets replaces by a new integration variable $u$.

Lutz Lehmann
  • 126,666