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I have $$f'(x) - f(x)^2=0$$ Looks simple enough but I can't work it out. $f(x)=0$ is obviously a solution but, if it's the only solution, how do I show that?

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Clearly, as you have stated in the question $f(x) = 0$ is a solution. If $f(x) \neq 0$, $$\dfrac{df}{dx} = f^2 \implies \dfrac{df}{f^2} = dx \implies -\dfrac1f = x - c \implies f = \dfrac1{c-x}$$

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Thanks! I hadn't tried separation of variables. :) –  Korgan Rivera Oct 27 '12 at 19:23

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