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This is from Berkeley Problems in Mathematics, Spring 86. It asks for $\lambda\in \mathbb{R}$, find all solutions of the following two equations:

$\phi(x)=e^{x}+\lambda\int^{x}_{0}e^{x-y}\phi(y)dy; \psi(x)=e^{x}+\lambda\int^{1}_{0}e^{x-y}\psi(y)dy$

My thought is to take the derivative, thus we have $\frac{d}{dx}\phi=e^{x}+\lambda\phi(x)$ because we have $\frac{d}{dx}\lambda \int^{x}_{0}e^{x}/e^{y}\phi(y)dy=\lambda e^{x}/e^{x}\phi(x)=\lambda \phi(x)$. And so is the equation for $\psi$ Thus the difference equation would be $\frac{d}{dx}(\phi-\psi)=\lambda(\phi-\psi)$ and implies $\phi-\psi=Ce^{\lambda x}$ for some $C$. But I do not know how to use this to solve the original equation. The seeming simple equation $\frac{d}{dx}\phi=e^{x}+\lambda\phi(x)$is also not easy to solve. I do not know how to treat the delay term $e^{x}$ or find a special solution for this.

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    I get the differential equation $\phi' = (1+\lambda) \phi$. You differentiated incorrectly, the upper limit of integration is $x$, not a constant.2012-07-22

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The derivative of $\phi(x)=e^{x}+\lambda\int^{x}_{0}e^{x-y}\phi(y) \,dy$ is $\phi'(x) = e^{x}+\lambda \phi(x) + \lambda\int^{x}_{0}e^{x-y}\phi(y)\, dy$, or, more clearly, $\phi' = (1+\lambda) \phi$.

The derivative of $\psi(x)=e^{x}+\lambda\int^{1}_{0}e^{x-y}\psi(y) \, dy$ is $\psi'(x) = e^{x} +\lambda\int^{1}_{0}e^{x-y}\psi(y) \, dy$, o, more clearly, $\psi' = \psi$.

The solutions are $\phi(x) = \phi(0) e^{(1+\lambda) x}$, and $\psi(x) = \psi(0) e^x$.

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    No problem at all.2012-07-22