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My question is to find the function $f(t)$ such that $$\frac{df}{dt} = -2f(t)\int_{0}^{t}f(s)\, ds$$ with $f(0) = 1$.

My idea is to divide both sides by $-2f$ and differentiate both sides, and then let $g = \frac{df}{dt}$ and consider $g$ as a function of $f$ which would reduce the order of the differential equation, but this doesn't seem to be working. Is this the right way? Is there another way to do this?

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up vote 3 down vote accepted

I don't know if your way works or not, but here is how I would do it. Define $F(t)=\int_0^t f(s)\,ds$. The ODE becomes $F''(t)=-2F'(t)F(t)=-\frac{d}{dt}(F(t)^2)$. Integrate both sides, get $F'(t)=-F(t)^2+c$. $F(0) = 0$ and $F'(0) = f(0) = 1$, so $c=1$, and $\frac{dF}{dt}=1-F^2$. This is a first-order separable ODE that can be solved by separation of variables.

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This is much simpler. Thanks! –  298378 Jan 26 '12 at 22:23
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