May 28th, 2007, 03:22 PM
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 | Grand Panjandrum | | Join Date: Nov 2005 Location: South of England
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Quote:
Originally Posted by ThePerfectHacker  Let  be continously differenciable on  . Solve the integral equation: ![[f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt](http://www.mathhelpforum.com/math-help/latex2/img/0c50ccce995ceac784c469416148960b-1.gif "[f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt") | Nice.
RonL
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