Help me solve this integral equation, I'm having some troubles... I need to use the Fredholm method for second kind integral equations.

$$\phi(x)= \sin(x)+ \lambda \int_{0}^{\pi}\cos(2x+y)\phi (y)dy$$



closed as not constructive by Pedro Tamaroff, Amzoti, Start wearing purple, TMM, Micah May 29 '13 at 15:32

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  • 2
    $\begingroup$ duplicate question asked within the hour: Fredholm equations $\endgroup$ – Namaste May 29 '13 at 14:52
  • $\begingroup$ Is $sen(x)$ meant to be $\sin(x)$? Also, lol at the duplicate... $\endgroup$ – Sharkos May 29 '13 at 14:53
  • $\begingroup$ @Sharkos Yes, it's used in another language for $\sin$ $\endgroup$ – Namaste May 29 '13 at 14:57
  • $\begingroup$ @amWhy Since the other question has no answer, should this be closed as duplicate? I noticed that only 1 vote is cast, under "not constructive". $\endgroup$ – Calvin Lin May 29 '13 at 15:18
  • $\begingroup$ @Calvin Lin: I think the other question shows a LOT of effort that this post lacks. I posted the duplicate comment long before the answers. I wish those who answered here would have taken the time to answer the question that showed effort, and not this post, if only answering one of the two. $\endgroup$ – Namaste May 29 '13 at 15:21

I don't know the method you mention, but notice that you can differentiate twice to get $$\phi''(x)=-\sin x-4\lambda\int_0^\pi dy\ \phi(y)\cos(2x+y),$$ so that $$\phi''(x)+4\phi(x)=3\sin x.$$ This is easily solved as $$\phi(x)=\sin x+A\cos(2x)+B\sin(2x)$$ with constants of integration $A$ and $B$. Now plug $\phi$ back in to the integral and solve for $A$ and $B$ in terms of $\lambda$. I get $$A=\frac{6\pi\lambda^2}{8\lambda^2-9}\qquad{\rm and}\qquad B=\frac{9\pi\lambda}{2(8\lambda^2-9)}.$$

  • $\begingroup$ Not this method but thanks anyway :) $\endgroup$ – Deiota May 29 '13 at 15:31

Hint: expand out $\cos(2x+y)$, and you see that the right side will always be a linear combination of $\sin(x)$, $\cos(2x)$ and $\sin(2x)$. So the left side...


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