Integrating on both sides - differential equation - intuition

Question

Let's say I know that a curve passes through the point $(x_0,y_0)$ and has a differential form that can be written as : $P(y)dy = Q(x)dx$. I wish to know the equation of the curve. I know I can solve this 2 ways :

1. Taking the antiderivative on both sides and then calculating the integration constant using the initial condition $y(x_0) = y_0$.
2. Doing both at the same time by integrating both sides using the right integration bounds:

\begin{align} \int_{y_0}^{y} P(y') dy' = \int_{x_0}^{x} Q(x') dx'\end{align}

I know that this second method is right because when $x=x_0$ then $y=y_0$ and both integrals equal $0$. I was wondering however, if I could have a more visual intuition for why this equation really is the equation of the curve by thinking of $dx$ and $dy$ as small changes and thinking about these integrals as summations propagating from $(x_0,y_0)$ to $(x,y)$. Any ideas?

Answer 1

Assume that your curve has a parametric representation $$t\mapsto\bigl(x(t),y(t)\bigr)\quad(t\geq0),\qquad\bigl(x(0),y(0)\bigr)=(x_0,y_0)\ .$$ Then the given condition amounts to $$P(y(t))\>\dot y(t)=Q(x(t))\>\dot x\qquad(t\geq0)\ .$$ Integrate this from $t=0$ to an arbitrary upper endpoint $T$ and obtain $$\int_0^T P(y(t))\>\dot y(t)\>dt=\int_0^T Q(x(t))\>\dot x(t)\>dt\qquad(T\geq0)\ .$$ Substituting separately on both sides of the last equation gives $$\int_{y_0}^{y(T)}P(y)\>dy=\int_{x_0}^{x(T)}Q(x)\>dx\ .$$ Do not write the $(T)$ in the end result, and you have your recipe.

Answer 2

"I was wondering however, if I could have a more visual intuition for why this equation really is the equation of the curve by thinking of $dx$ and $dy$ as small changes and thinking about these integrals as summations propagating from $(x_0,y_0)$ to $(x,y)$"

is exactly right. More precisely you should think of them being infinitesimal (instead of small) changes. Mathematicians used to be terrified of infinitesimals (as witnessed in the other answer) but since 1961 students can feel better about them, because Abraham Robinson placed them on solid logical ground.

A technical point: the integral strictly speaking is not exactly an infinite Riemann sum (of infinitesimal terms), but rather the standard part thereof.