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CaptainBlack · #11 May 30th, 2007, 10:18 PM

Quote:

Originally Posted by topsquark

Okay, I oopsied. But still:

$f(x) = Ae^x$

Thus
$f(t) = Ae^t$

$\frac{df}{dt} = Ae^t$

Thus
$\int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2$

$= \int_0^x dt \, (A^2e^{2t} + A^2e^{2t} )$

$= \int_0^x dt \, 2A^2e^{2t}$

$= 2A^2 \int_0^x dt \, e^{2t}$

$= 2A^2 \frac{1}{2}e^{2t}|_0^x$

$= A^2(e^{2x} - 1)$

And
$[ f(x) ] ^2 = A^2e^{2x}$

So
$[ f(x) ] ^2 \neq \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2$

How can this then be the solution of the integral equation? I'm missing something here...

-Dan

So you have shown that if $f(t) = Ae^t$ , then:

$A^2e^{2x}$ $= A^2(e^{2x} - 1)$

RonL