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I'm working through Pollard & Tenenbaum's "Ordinary Differential Equations".

In their treatment of first-order differential equations they write that for an equation of form

$$P(x)dx + Q(y)dy = 0$$

its solution is given by

$$\int P(x)dx + \int Q(y)dy = c $$

In other words, they take indefinite integrals to obtain the solution.

Later in the part on first-order equations, they arrive at exact differential equations, and use definite integrals in their proof of the exact differential equation's solution, to arrive at

$$\int_{x_0}^x P(x, y)dx + \int_{y_0}^{y} Q(x_0, y)dy = c$$

as a solution of

$$P(x, y)dx + Q(x, y)dy = 0$$

Where $P$ and $Q$ are partial derivatives of a multivariable function $f$.

When do I use the definite integral $\int_{x_0}^x f(x)dx$ and when can I use $\int f(x)dx$, taking just the antiderivative and being done with it?

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A rule of thumb would be to use indefinite integrals when there is just a differential equation, and use definite integrals when it's an initial value problem. If the initial condition is $y(x_0)=y_0$, then the lower limit on the definite integral should be $x_0$.

But the rule of thumb isn't hard and fast. Sometimes, even without an initial condition, it's convenient to have the arbitrary constant explicit. The indefinite integral $\int f(x) \; dx$ has an implicit, but invisible, arbitrary constant. Sometimes, to do a derivation, one needs that constant explicitly written in the equation. Either one writes $\int f(x) \; dx+c$ (which is a bit reduntant, but correct) or fixes it with a definite integral $_{x_0}^x \int f(t) \; dt +c.$

Mostly, try to be clear for your reader.

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  • $\begingroup$ So for all intents and purposes, the indefinite integral of F and the definite integral of F on a domain <x0, x> are equivalent? Do I need to prove this (through the FTC I reckon) or is it mathematically sufficient to note that derivation takes both options back to F(x)? $\endgroup$ – Heatherfield Jul 24 '17 at 14:07
  • $\begingroup$ Well, for "almost" all intents and purposes. As long as you pay attention to that arbitrary constant (it's either $+c$ or $F(x_0)$) you should be fine. $\endgroup$ – B. Goddard Jul 24 '17 at 14:18
  • $\begingroup$ Yeah I was trying to recreate the proof of the theorem I started the question with, replacing $\int_{x_0}^x$ with $\int + c$ and it didn't give me a meaningful result. But I'm content with knowing that taking a definite integral can solve a differential equation as well was taking the set of antiderivatives. $\endgroup$ – Heatherfield Jul 24 '17 at 15:56

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