Let then .

In the first case, , we get and so . Now we use the boundary conditions. Since . So either is identically zero or . In the former case if is identically zero it would mean is a trivial solution. Thus, and so . But under the condition we can only get the trivial solution. But of course the trivial solution does not solve this Dirichlet problem and so we can reject the trivial solution.

In the second case, , we get and so . The condition . To have a non-trivial solution we must have . Therefore, . The condition . To have a non-trivial solution for we must have with . Thus, . At this point we can assume by letting it get absorbed into the constants. Thus, so far we established the solutions need to have form . These are our eigenfunctions. If are arbitray numbers for we see that is a solution to this equation (without the last condition yet). We are told that the solution statys bounded. The only way that can happen is if the (why is that?). We would like to satisfy the condition therefore we want for . The sad case here is that finite sums do not do, we need to now think of infinite sums of sines. By Fourier analysis (assuming I did not make mistakes) . This gives us a solution

You think about the final case.