Use the Laplace transform to solve differential equation

dx/dt + 7x = 5cos (2t)

Initial conditions: x (0) = 4, x' (0) = - 4

Assume forcing functions are 0 prior to t = 0 - ... ?

Using Laplace transform we have:L (x') + 7L (x) = 5L (cos (2t)) sL (x) - x (0) + 7L (x) = 5s / (s^2+4) (s+7) L (x) - 4 = 5s / (s^2+4) (s+7) L (x) = (5s - 4s^2 - 16) / (s^2+4)

=> L (x) = - (4s^2 - 5s + 16) / (s^2+4) (s+7)

now the boring part, using partial fractions we separate 1 / (s^2+4) (s+7) that is: (7-s) / [53 (s^2+4) ] + 1/53 (s+7). So:

L (x) = (1/53) [ (-28s^2+4s^3-4s^2+35s-5s^2+5s) / (s^2+4) + (-4s^2+5s-16) / (s+7) ]L (x) = (1/53) [ (4s^3 - 37s^2 + 40s) / (s^2+4) + (-4s^2+5s-16) / (s+7) ]

denoting T: = L^ (-1) and x = (4/53) T (s^3 / (s^2+4)) - (37/53) T (s^2 / (s^2+4)) + (40/53) T (s^2+4) - (4/53) T (s^2/s+7) + (5/53) T (s/s+7) - (16/53) T (1/s+7)