# Integral equation that's cant solve… Need a hand [closed]

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

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

Thanks

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

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. If this question can be reworded to fit the rules in the help center, please edit the question.

• duplicate question asked within the hour: Fredholm equations – Namaste May 29 '13 at 14:52
• Is $sen(x)$ meant to be $\sin(x)$? Also, lol at the duplicate... – Sharkos May 29 '13 at 14:53
• @Sharkos Yes, it's used in another language for $\sin$ – Namaste May 29 '13 at 14:57
• @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". – Calvin Lin May 29 '13 at 15:18
• @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. – 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 $$A=\frac{6\pi\lambda^2}{8\lambda^2-9}\qquad{\rm and}\qquad B=\frac{9\pi\lambda}{2(8\lambda^2-9)}.$$
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...