Integral equation that's cant solve... Need a hand

Question

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

Answer 1

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 $$A=\frac{6\pi\lambda^2}{8\lambda^2-9}\qquad{\rm and}\qquad B=\frac{9\pi\lambda}{2(8\lambda^2-9)}.$$

Answer 2

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...