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ecMathGeek · #5 May 28th, 2007, 03:40 PM

Quote:

Originally Posted by ThePerfectHacker

Let $f$ be continously differenciable on $\mathbb{R}$ . Solve the integral equation:

$[f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt$

I would much rather deal with a differential equation. Hopefully this works.

$\frac{d}{dx}[f(x)]^2=\frac{d}{dx}\int_0^x [f(t)]^2+[f'(t)]^2dt$

$2[f(x)]f'(x) = [f(x)]^2+[f'(x)]^2$

$[f(x)]^2-2 [f(x) \cdot f'(x)] + [f'(x)]^2 = 0$

$[f(x) - f'(x)]^2=0$

$f(x)-f'(x)=0$
$f(x)=f'(x)$
$f(x) = Ae^x$

Am I right?