Question

Solve the initial value problem
y"+y=0 y(0)=2 y'(0)=0

Answer 1

$y"+y=0,\,\, y(0)=2,\,\, y'(0)=0\\ \textsf{The auxiliary equation is}\\ m^2 + 1 = 0, m^2 = -1\\ m = \pm\sqrt{-1} = \pm j\\ \textsf{The general solution to the}\\ \textsf{differential equation is}\\ y = B\cos{x} + C\sin{x}\\ y(0) = B = 2, \therefore B = 2.\\ y' = -B\sin{x} + C\cos{x}\\ y'(0) = C = 0\\ \therefore \textsf{The particular solution is}\\ y = 2\cos{x}$