the eigenvalue problem
Pluck a guitar string and it does not vibrate at just any frequency — it rings only at certain special tones, a fundamental and its overtones. Blow across a bottle and only particular pitches resonate. Nature, when confined between two fixed ends, allows only a discrete menu of permitted patterns. The eigenvalue problem is the mathematics behind this universal fact: a parameter in a boundary value problem is allowed only at a special set of values, and at each of those values a special shape appears.
The model problem is y'' + lambda y = 0 on 0 <= x <= L with y(0) = 0 and y(L) = 0, where lambda is an adjustable number. For most lambda the only solution is y = 0. But solve the equation in general: if lambda > 0 the solutions are combinations of sin(sqrt(lambda) x) and cos(sqrt(lambda) x). The condition y(0) = 0 kills the cosine, leaving y = c sin(sqrt(lambda) x). The condition y(L) = 0 then demands sin(sqrt(lambda) L) = 0, which forces sqrt(lambda) L = n pi for a whole number n. So a non-trivial solution exists only when lambda = (n pi / L)^2 for n = 1, 2, 3, and so on. These permitted lambda values are the eigenvalues, and the matching shapes sin(n pi x / L) are the eigenfunctions.
The name and idea come straight from linear algebra, where A x = lambda x asks for the special vectors a matrix merely stretches. Here the matrix is replaced by the differential operator 'take y'' (with these boundary conditions)', the vector by a function, and the special values lambda are again the ones for which a nonzero 'eigenvector' (now an eigenfunction) exists. This problem is the engine of vibration analysis, quantum energy levels, heat-conduction modes, and the separation-of-variables solution of every classical partial differential equation.
For y'' + lambda y = 0 with y(0) = 0, y(pi) = 0, the eigenvalues are lambda_n = n^2 for n = 1, 2, 3, ... with eigenfunctions y_n = sin(n x). Check: y_2 = sin(2x) gives y'' = -4 sin(2x) = -4 y, matching lambda = 4 = 2^2, and sin(2·0) = sin(2·pi) = 0.
Only the discrete values lambda_n = n^2 permit a nonzero solution; each carries its own eigenfunction sin(n x).
Sign matters: for y'' + lambda y = 0 with these zero-at-both-ends conditions, only positive lambda gives non-trivial solutions — lambda <= 0 forces y = 0. Always check the lambda < 0 and lambda = 0 cases explicitly rather than assuming sines appear.