Answer:
x = 5 cos 2t
Step-by-step explanation:
given equation
x'' + 4x = 0 ; x(0) = 5, x'(0) = 0
L{ x'' + 4 x } = 0
L{x''} + 4 L{x} = 0
s² . L(x) - s . x(0) - x'(0) + 4 L{x} = 0
( s² + 4 ). L(x) - 5 s = 0
L(x) = [tex]\dfrac{5s}{s^2 +4}[/tex]
[tex]L(\dfrac{s}{s^2 +a^2})[/tex] = cos at
so,
x = 5 [tex]L^{-1}(\dfrac{s}{s^2 +2^2})[/tex]
x = 5 cos 2t
