Question

Find the general solution to 2 y^{\,\prime\prime} + 32 y = 0. give your answer as y = . . . \ . in your answer, use c_1 and c_2 to denote arbitrary constants and x the independent variable. enter c_1 as c1 and c_2 as c2

Answer 1

We want to solve
2y'' + 32y = 0

That is,
y'' + 16y = 0

Solve the indicial equation.
r² + 16 = 0
r² = -16
r = 4i, -4i which yields the functions sin(4x), cos(4x).

Answer:
The general solution is
y(x) = c₁cos(4x) + c₂sin(4x)