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I'm reading An Introduction to Ordinary Differential Equations by Agarwal and O'Regan. On page 28, I have the expression

$$y\left(x\right)=c\exp\left(-\int^x{p\left(t\right)\, dt} \right)$$

which is equation 5.4.

My problem is the missing lower limit of integration. I don't understand what the notation means. The context of the equation is solving the homogenous equation


This leads to


The text says that by integrating both sides, we get the expression that is puzzling me. Any advice on how to interpret this notation would be appreciated.

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The book really writes that? Put any constant you want as the lower limit. You should check (using Fundamental Theorem of Calculus) to make sure that solution works. – Matt Mar 30 '11 at 4:17
I understand now based on the answers that the missing lower limit means it doesn't matter. Is that a standard way of staying that the limit doesn't matter? – Henry B. Mar 30 '11 at 4:34
I have seen this notation before in old books on analysis (maybe Whittaker and Watson) where there are other curiosities like using $(-)^n$ for $(-1)^n$ or Shew instead of Show. – KCd Mar 30 '11 at 9:16
To add to @KCd's comment: sometimes you'll see formulae with $(-)^n$ in old handbooks. For "shew", these are probably the same books that use words like "inflexion"... – J. M. Apr 7 '11 at 8:38
up vote 2 down vote accepted

The point is that the lower limit doesn't matter: If you have $y\left(x\right)=c\exp\left(-\int_a^x{p\left(t\right)\, dt} \right)$ and $z\left(x\right)=c\exp\left(-\int_b^x{p\left(t\right)\, dt} \right)$ they are equal [(as long as $p(t)$ is integrable over (a,b)] to within a multiplicative constant, which get absorbed into the constant of integration.

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When you wrote $y'$ mean to write another $y$? Also, the constant is multiplicative. – Mariano Suárez-Alvarez Mar 30 '11 at 4:19
@Mariano: You are right. Prime can be another or derivative-I should have chosen a different symbol. I changed $y'$ to $z$ to try to eliminate confusion (but I know this is impossible). Additive vs. multiplicative fixed. Thanks. – Ross Millikan Mar 30 '11 at 4:42

The lower limit of integration is where the arbitrary constant $c$ comes from. It doesn't really matter what lower limit you use, as $c\exp\left(-\int_a^x{p\left(t\right)\, dt} \right) = c\exp\left(-\int_b^x{p\left(t\right)\, dt} - \int_a^b{p(t)\,dt}\right) = c\left(\int_a^b{p(t)\,dt}\right)^{-1}\exp\left(-\int_b^x{p\left(t\right)\, dt} \right)$ and the definite integral from $a$ to $b$ gets absorbed in the constant $c$.

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