Answer to Question #152183 in Differential Equations for Mayuri

Question

The solution of D.E(D^4-16)y=0 is

Answer 1

$y^{(iv)} - 16y = 0\\ \textsf{The auxiliary equation is}\,\,\,m^4 - 16 = 0.\\ (m^2 - 4)(m^2 + 4) = 0\\ m^2 = 4, m^2 = -4\\ m = \pm 2, \pm 2i.\\ \textsf{Recall that if the solution to the}\\ \textsf{auxiliary equation is of the form}\,\,\, \alpha + i\beta, \\ \textsf{the solution is of the form} \\ y = e^{\alpha x}\left(A\cos(\beta x) + B\sin(\beta x)\right).\\ \therefore y = Ae^{2x} + Be^{-2x} + C\cos(2x) + D\sin(2x).\,\,\,\textsf{is a solution to the ODE}$