Question

Use Laplace transforms to solve the initial value problem, then give the value of x(?).

x'' + 4x = 0; x(0) = 5, x'(0) = 0

Answer 1

x = 5 cos 2t

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) =
`(5s)/(s^2 +4)`

`L((s)/(s^2 +a^2))` = cos at

so,

x = 5
`L^(-1)((s)/(s^2 +2^2))`

x = 5 cos 2t