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

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) = $\dfrac{5s}{s^2 +4}$

$L(\dfrac{s}{s^2 +a^2})$  = cos at

so,

x = 5  $L^{-1}(\dfrac{s}{s^2 +2^2})$

x = 5 cos 2t