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I'm having problem obtaining the solution of the homogeneous Fredholm Integral Equation of the 2nd kind, with separable kernel. I always get a zero if I use the normal method i was taught for the nonhomogeneous type. I have an example: $$y(x) = \lambda\int_{-1}^1(x + z)y(z)dz$$

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certainly, your method of solving gives the general form $y(x)=c_1x+c_2$

Then, bringing it back into the Fredholm Integral Equation leads to the condition : $$ y(x)=c_1x+c_2=2\lambda c_2x+\frac{2}{3}\lambda c_1$$ which implies $c_1=2\lambda c_2$ and $c_2=\frac{2}{3}\lambda c_1$

$ (\frac{4}{3}\lambda^2 -1 ) c_1=0$

So, they are two cases :

If $\lambda=\pm \frac{\sqrt3}{2}$ , then $c_2=\frac{2}{3}\lambda c_1=$any value, and the solutions are $y(x)=2 \lambda c_2 x + c_2$.

If $\lambda \neq \pm \frac{\sqrt3}{2}$ , then $c_1=c_2=0$ and the solution is $y(x)=0$

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