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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

$$y^\prime+p\left(x\right)y=0.$$

This leads to

$$\frac{y^\prime}{y}+p\left(x\right)=0.$$

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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1  
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
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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

2 Answers 2

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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