Note that there is the infinity eigenvalue, $\lambda = 6/5\pi$, so by the Fredholm Alternative, we know that there is a non-trivial solution to $\mathcal{L}y = 0$, and thus no unique solution to $\mathcal{L}y = f(x)$ for any $f$. The real question, then, is whether there are none or infinitely many solutions.

**Addition for Trinity revisions (May 3, 2013)**: Remember that if we have the degenerate Fredholm integral equation, with
\[
h(x) y(x) + \int_a^b K(x,t) u(t) \, \de{t} = f(x),
\]

then separated using $K(x,t) = \sum_j \alpha_j(x) \beta_j(t)$,
\[
h(x) y(x) + \sum_{j=1}^\infty c_j \alpha_j(x) = f(x),
\]

where $c_j = \langle \beta_j, y\rangle$. Thus, we can always write the solution as (assuming $h \neq 0$)
\[
y(x) = \frac{f(x)}{h(x)} + \frac{1}{h(x)} \sum_{j=1}^\infty d_j \alpha_j(x),
\]

for constants $d_j$. *End of addition (May 3, 2013)*

Consider now the forcing (i) given by $f(x) = 2\sin(2\pi x - \pi/6)$. The solution must take the form of $y = \sum c_j a_j(x) + (5\pi/6) f(x)$. It is then somewhat easier to write $f$ as
\[
f(x) = 2\sin(2\pi x - \pi/6) = \sqrt{3} \sin(2\pi x) - \cos(2\pi x),
\]

so that we may write the general ansatz as $y(x) = D_1 \sin (2\pi x) + D_2 \cos(2\pi x)$. Once substituted into $\mathcal{L}y$ we must end up with the identical LHS as \eqref{myLy}. The equation to be solved is then
\[
\frac{2}{5\pi} D_1 \sin(2\pi x) = \sqrt{3} \sin(2\pi x) - \cos(2\pi x),
\]

which can never be satisfied due to the cosine term. Therefore, there are no solutions to the forced equation.

Consider now the forcing (ii) given by $f(x) = \sin(2\pi x)$. Again, the general solution for which we try is the same as above, except now, when it is plugged into the equation, we get
\[
\frac{2}{5\pi} D_1 \sin(2\pi x) = \sin(2\pi x),
\]

so we see that $D_1 = 5\pi/2$ and $D_2$ is left free. The general solution is then
\[
y(x) = \frac{5\pi}{2} \sin(2\pi x) + D_2 \cos(2\pi x).
\]

**How does the general solvability condition work?**

Let us remind ourselves how the solvability condition comes into play. Imagine that we have some general problem to be solved, $\mathcal{L}y = f(x)$. We've also found the eigenvectors, $y_k$ and $w_k$ of the original and adjoint problem. We then seek the coefficients of an eigenfunction expansion of the solution, $y = \sum c_k y_k$ by using the standard trick:
\begin{align*}
\langle w_k, \mathcal{L}y \rangle &= \langle w_k f(x) \rangle \\
\text{BCs} + \langle \mathcal{L}^*w_k, \sum c_j y_j \rangle &= \langle w_k f(x) \rangle \\
\text{BCs} + \lambda_k c_k \langle \mathcal{L}^*w_k, y_k \rangle &= \langle w_k f(x) \rangle
\end{align*}

except of course, when $\lambda_k = 0$, then the above does not yield a way of obtaining the corresponding $c_k$. However, in order for it to be consistent, we need $\langle w_k, f\rangle = 0$ (assuming the BCs of the adjoint problem are zero), which is the *solvability* condition.

Coming back to this problem \eqref{fie4a}, we define the usual inner product,
\[
\langle w, \mathcal{L}y\rangle = \int_0^1 w \mathcal{L}y \, \de{x} = \int_0^1 w(x) \left[\tfrac{6}{5\pi} y(x) + \int_0^1 K(x,t) y(t) \, \de{t}\right] \de{x}.
\]

Switching the order gives
\begin{align*}
\langle w, \mathcal{L}y\rangle &= \int_0^1 y(t) \left[Aw(t) + \int_0^1 K(x,t) w(x) \, \de{x}\right] \de{t} \\
&= \int_0^1 y(x) \left[Aw(x) + \int_0^1 K(t,x) w(t) \, \de{t}\right] \de{x} \\
&= \langle \mathcal{L}^* w, y \rangle.
\end{align*}

so as we should know, the adjoint problem is simply the original problem with the kernel $K(t,x) = \sin(2\pi t - 3\pi x)$. We need to know the eigenfunction, $w_0$, that corresponds to the zero eigenvalue. This time around, the ansatz is
\[
w = C_1 \sin(3\pi x) + C_2 \cos(3\pi x).
\]

We can verify that
\begin{align*}
\mathcal{L}^*w &= \tfrac{6}{5\pi} \Bigl[ C_1 \sin(3\pi x) + C_2\cos(3\pi x)\Bigr] \\
& \quad - \tfrac{2}{5\pi} \Bigl[ 3C_1 \sin(3\pi x) + 2C_2\cos(3\pi x)\Bigr] \\
&= \tfrac{2}{5\pi} C_2 \cos(3\pi x).
\end{align*}

In other words, when solving $\mathcal{L}^* w = 0$ for the zero eigenvalue, we get the eigenfunction of $w_0(x) = \sin(3\pi x)$ and therefore the solvability condition is
\[
\langle w_0, f \rangle = \int_0^1 f(x) \sin(3\pi x)\, \de{x} = 0.
\]