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Given the integral equation: $$\sqrt{f(x)}\int_{0}^{x}f(\tau)d\tau=g(x)$$ with g(x) known function, in what cases and how is it possible to solve it?


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If $g(x)=k_1x^\beta$ then the solution has a simple form $f(x)=\sqrt{(\alpha+1)k_1} x^\alpha$ being $\alpha=\frac{2}{3}(\beta-1)$ with $\beta\ne 1$. – Jon May 31 '12 at 14:08
up vote 3 down vote accepted

Square both sides of the equation : $$F^'(x) \ (F(x))^2=(g(x))^2$$ with $F(x) = \int_{0}^{x}f(\tau)d\tau $.
Now by integration between 0 and t : $$\frac{(F(t))^3}{3} = \int_{0}^{t} (g(x))^2 dx$$ F(t) can now be expressed : $$F(t) =\sqrt[3]{3} \ \left( \int_{0}^{t} (g(x))^2 dx \right)^{1/3}$$ since $F'(t)=f(t)$, by derivation we obtain : $$f(t) = \sqrt[3]{3} \ (g(t))^2 \left(\int_{0}^{t} (g(x))^2 dx \right)^{-2/3}$$

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Let us assume that $g(x)$ is differentiable (in fact $g(x)$ should be also positive and monotonously increasing $f(x)$ has to be positive such that the square root is properly defined).

Take a derivative of your equation and you obtain $$ f(x)^{3/2} + \frac{g(x)\, f'(x)}{2 f(x)} =g'(x). $$ Thus, you have reduced your equation to a differential equation with the initial condition $f(0) = g(0)$.

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