# Asking solutions for the integral equations

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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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 7:22
There is typo as you can check here:math.berkeley.edu/sites/default/files/pages/Spring86.pdf But my derivative may be wrong as you said. –  user32240 Jul 22 '12 at 7:25
I double checked, and see no problem with the derivative. Yes it is an uninteresting problem to me... –  user32240 Jul 22 '12 at 7:34
The upper limit on the integral with the $\psi$ equation should be $1$, not $x$. –  copper.hat Jul 22 '12 at 7:59
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 8:03

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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I am confused, the equation is the same, why the derivative is different? –  user32240 Jul 22 '12 at 8:10
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 8:12
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. –  user32240 Jul 22 '12 at 8:13
I see..I see. The Leibniz formula. –  user32240 Jul 22 '12 at 8:15
Thank you! Sorry to trouble you. –  user32240 Jul 22 '12 at 8:15
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