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In college classes, we solve differential equations. However, I have never seen a book give a rigorous definition of what a differential equation actually is. Can someone give me a rigorous definition of a differential equation? I want a definition in terms of set theory.

user107952
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    https://math.stackexchange.com/questions/33153/definition-of-a-differential-equation?rq=1 – Enrico M. Dec 09 '21 at 19:35
  • Maybe the reason the term “differential equation” is sometimes not defined precisely is that we can avoid using the term entirely if we just make statements like this, for example: If $t_0$ and $y_0$ are real numbers then there exists a differentiable function $y:\mathbb R \to \mathbb R$ such that $y’(t) = y(t)$ for all $t \in \mathbb R$ and $y(t_0) = y_0$. – littleO Dec 09 '21 at 19:40
  • Differential equations can be many things, partial DE, ordinary DE, delay DE, fractional DE, ... The usual theory is developed for the "ordinary" case, and I think that leading up to the Picard-Lindelöf theorem also an exact definition is given what an ODE is. – Lutz Lehmann Dec 09 '21 at 20:01
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    The basic idea is that an $n$th order ODE is an “equation of the form” $f(t,y(t),y’(t),\ldots,y^{(n)}(t)) =0$, where $f:\mathbb R^{n+2} \to \mathbb R$ is continuous. – littleO Dec 09 '21 at 20:04
  • If you want to be very precise, you might object: “What is an equation?” In response, I might redefine an $n$th order ODE to be the function $f$ itself, and define a solution of this ODE to be a function $y:I \to \mathbb R$ such that: 1) $y$ is $n$ times differentiable; 2) if $t \in I$ then the point $(t,y(t), y’(t),\ldots,y^{(n)}(t))$ is in the domain of $f$; 3) $ f(t,y(t), \ldots, y^{(n)}(t))=0$ for all $t \in I$. Here $I$ is an interval in $\mathbb R$. – littleO Dec 09 '21 at 20:04

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