Solving a Fredholm Equation of the second kind I'm trying to solve the Fredholm equation,
$$
\phi(x) = 3 + \lambda \int_{0}^{\pi} \text{cos}(x-s) \, \phi(s) \,ds.
$$
I began by using the method of successive approximations and found the iterated kernels,
$$
\begin{split}
K_1 (x,s) &= \text{cos}(x-s)\\
K_2 (x,s) &= \int_0^\pi \text{cos}(x-t) \, \text{cos}(t-s)\,dt \equiv \frac{\pi}{2}\text{cos}(x-s)\\
K_3(x,s) &= \int_0^\pi \text{cos}(x-t) \, \frac{\pi}{2}\text{cos}(t-s) \,dt \equiv (\frac{\pi}{2})^2 \,\text{cos}(x-s)\\
&\vdots \\
K_{n+1} (x,s) &= (\frac{\pi}{2})^n \,\text{cos}(x-s)\\
&\vdots
\end{split}
\quad\text{ for}\,\, n=0,1,2,\dots.
$$
Noting that the resolvent kernel is given by
$$
R(x,s;\lambda) = \sum_{n=0}^\infty \lambda^n\,K_{n+1}(x,s)
$$
then we find,
$$R(x,s;\lambda) = \frac{\text{cos}(x-s)}{1-\frac{\lambda \pi}{2}}.$$
The equation to be solved is then,
$$
\phi(x) = 3 + \lambda\int_0^\pi R(x,s;\lambda) \, ds
$$
which is solved and found to be,
$$\phi(x) = 3 + \frac{\lambda}{1-\frac{\lambda \pi}{2}} 2 \,\text{sin}(x)$$
I have tried confirming this solution by transforming the initial question to a differential equation but cannot find the same solution as shown above. However, I am certain that the answer above is correct. Can anyone see a flaw in my solution?
 A: There is only a tiny mistake: we have
$$ \phi(x) = 3 + \lambda \int \limits_0^\pi R(x,s;\lambda) \color{red}{3} \, \mathrm {d} s = 3 + \frac{\color{red}{6} \lambda}{1 - \lambda \frac{\pi}{2}} \sin(x) \, .$$
This is the correct solution for all $\lambda \in \mathbb{R} \setminus {\frac{2}{\pi}}$. For $\lambda = \frac{2}{\pi}$ there is no solution, since $\frac{\pi}{2}$ is an eigenvalue of the integral operator. Note, however, that your series only converges for $|\lambda| < \frac{2}{\pi}$.
In order to confirm the above result and to extend its validity to all $\lambda \in \mathbb{R} \setminus {\frac{2}{\pi}}$, we can employ the method of degenerate kernels. Using a trigonometric addition formula we may write the integral equation as
$$ \phi (x) = 3 + \lambda \cos(x) \int \limits_0^\pi \cos(s) \phi(s) \, \mathrm{d} s + \lambda \sin(x) \int \limits_0^\pi \sin(s) \phi(s) \, \mathrm{d} s \, . $$
Therefore, the solution must have the form $\phi(x) = 3 + a \cos(x) + b \sin(x)$ for some $a,b \in \mathbb{R}$. Plugging this result back into the equation we obtain a system of linear equations for $a$ and $b$, which has no solution for $\lambda = \frac{2}{\pi}$ and the unique solution $a = 0$ and $b = \frac{6 \lambda}{1-\lambda \frac{\pi}{2}}$ for $\lambda \in \mathbb{R} \setminus {\frac{2}{\pi}}$ as expected. Alternatively, we can derive a differential equation for $\phi$. Its general solution has the form given above and suitable boundary conditions lead to the same linear system.
