Any help with this question would be great Thank you Solve t

Any help with this question would be great! Thank you!

Solve the given initial-value problem. y\" + 4y = 0, y(0) = 3, y\'(0) = -3 y(x) = 3 cos(2x) - 1/3 sin(2x)

Solution

y\'\'+4y=0

y\'\'=-2^2y

General solution is

y=A cos(2x)+B sin(2x)

y(0)=A=3

y\'(0)=2B=-3 hence, B=-3/2

y(x)=3 cos(2x)-3/2 sin(2x)