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Here's another question from Advanced Engineering Mathematics by Greenberg [Ex 1.2 Q5]:

For what values of the constant $\lambda$ will $y = \exp(\lambda x)$ be a solution of the differential equation? If there are no such $\lambda$'s, state that $y^\prime + 3y = 0$

First of all, I couldn't understand the question because of the thing called $\exp$ in $y = \exp(\lambda x)$. May I know what is $\exp$? I'm confident that if I get to know what $\exp$ is then I can take it from there.

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${\rm exp}\,( x)=e^x$. –  David Mitra Dec 23 '11 at 9:29
    
@DavidMitra Why not use \exp instead of {\rm exp}? –  matt Dec 23 '11 at 9:37
    
@matt Thanks, I didn't realize that was defined already. –  David Mitra Dec 23 '11 at 9:46

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It's an alternative notation for the exponential function: $\exp (x)=e^x$. Some authors use this notation exclusively.

The notation is convenient to use if the argument of the exponential function is "large". For example, contrast $e^{(x-\mu)^2\over 2\sigma^2}$ with $\exp\Bigl({(x-\mu)^2\over 2\sigma^2}\Bigr)$. Some may find the latter notation ugly.

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I would also say that it is also valuable pedagogically: it allows a student to work with it without the preconceived notions of how powers work. $e^x e^y=e^{x+y}$ is "obvious", but proving that for $\exp$ without knowing that it is indeed an exponential function is not too trivial. –  J. M. Dec 23 '11 at 11:53

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