Suppose $$y''+a^2 y=f(x), y(0)=y(1), y'(0)=y'(1).$$
By integration by parts and some evaluations we end up with $$ y = \int_o^x (x-z)f(z)dz - a^2 \int_0^x (x-z)y(z)dz + c x + d. $$
But I am not able to eliminate the constants $c$ and $d$ using the boundary conditions. We have been given that final solution turns out to be $$ y(x) = \int_0^1 \frac{\cos(a(0.5-|x-z|))}{\sin(0.5a)} f(z) dz. $$
Any steps of hints are appreciated!
Recall that for some linear operator $L$, the Green's function $G(x,s)$ is the function satisfying $LG(x,s)=\delta(x-s)$, and that $y(x)=\int_0^1f(s)G(x,s)ds$. For $Ly=y''+a^2y$, we have that $LG(x,s)=\frac{\partial^2 G(x,s)}{\partial x^2} + a^2G(x,s)=\delta(x-s)$.
We first obtain the solution $$ G(x,s)= \begin{cases} c_1(s)\cos(ax)+c_2(s)\sin(ax), \quad x<s \\ \\ d_1(s)\cos(ax)+d_2(s)\sin(ax), \quad x>s. \end{cases} $$
Although the constants are not explicitly functions of $s$, they are implicitly determined by the $s$ parameter of the Green's function and where the solutions are valid.
Applying the boundary conditions yields first that $$c_1(s)=d_1(s)\cos(a)+d_2(s)\sin(a)$$ $$c_2(s)=d_2(s)\cos(a)-d_1(s)\sin(a)$$ Finally, we apply continuity and the Green's function criterion. Continuity first requires that $$d_2(s)=d_1(s)\frac{\cos(0.5a)-1}{\sin(0.5a)}$$ The Green's function criterion requires a jump discontinuity of magnitude 1 at $x=s$ such that $$\frac{\partial G(x-s)}{\partial x}|_{x=s+}-\frac{\partial G(x-s)}{\partial x}|_{x=s-}=1$$ and from this we can reconstitute $$ G(x,s)= \begin{cases} \frac{\cos(0.5a+a(s-x))}{\sin(0.5a)}, \quad x<s \\ \frac{\cos(0.5a+a(s-x))}{\sin(0.5a)}, \quad x>s. \end{cases} $$
With the Green's function at hand we may now write the solution to be $$y(x)=\int_0^1f(s)G(x,s)ds=\int_0^1f(s)\frac{\cos(0.5a-a|x-s|)}{\sin(0.5a)}ds$$ as required.