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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.

J. M. ain't a mathematician
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Bombyx mori
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  • Double-check your first derivative; it is incorrect. Also, is there a typo in the question? The two equations seem to be identical copies, making the problem very uninteresting. – Erick Wong Jul 22 '12 at 07:22
  • There is typo as you can check here:http://math.berkeley.edu/sites/default/files/pages/Spring86.pdf But my derivative may be wrong as you said. – Bombyx mori Jul 22 '12 at 07:25
  • I double checked, and see no problem with the derivative. Yes it is an uninteresting problem to me... – Bombyx mori Jul 22 '12 at 07:34
  • The upper limit on the integral with the $\psi$ equation should be $1$, not $x$. – copper.hat Jul 22 '12 at 07:59
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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. – copper.hat Jul 22 '12 at 08:03

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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 $\psi(x) = \psi(0) e^x$.

copper.hat
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  • I am confused, the equation is the same, why the derivative is different? – Bombyx mori Jul 22 '12 at 08:10
  • Sorry, I made the mistake of copying your equation text. The upper limit of integration on the $\psi$ integral is $1$, not $x$. I have fixed it. Look at the reference you provided more carefully. – copper.hat Jul 22 '12 at 08:12
  • Sorry, I made the mistake of copying down the wrong problem. I want to ask why we had the extra $\lambda\int^{x}_{0}e^{x-y}dy$ term in the derivative, it feels not needed. – Bombyx mori Jul 22 '12 at 08:13
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    I see..I see. The Leibniz formula. – Bombyx mori Jul 22 '12 at 08:15
  • Thank you! Sorry to trouble you. – Bombyx mori Jul 22 '12 at 08:15
  • No problem at all. – copper.hat Jul 22 '12 at 08:15
  • @copper.hat Could you explain what garantees that you can differentiate the equations? The problem does not say anything about their differentiability and I believe you would have to prove that first, no? – TeX Apprentice Aug 19 '23 at 11:01
  • @TeXApprentice Since $\phi, \psi$ appear under an integral sign, there is an implicit assumption that they are integrable. It follows from the fixed point equations that they are continuous and again it follows that they are differentiable. – copper.hat Aug 19 '23 at 16:16