Use the Laplace tables to find the inverse Laplace transform of 18/((s+3)(s^2+9)) .

Answer 1

18/[(s + 3)(s² + 9)] = a/(s + 3) + (bs + c)/(s² + 9);

18 = a(s² + 9) + (bs + c)(s + 3); a + b = 0; 3b + c = 0; 3c = 18; c = 6, b = - 2; a = 2;

18/[(s + 3)(s² + 9)] = 2/(s + 3) - 2(s - 3)/(s² + 9)

L^-1(18/[(s + 3)(s² + 9)] = L^-1( 2/(s + 3) - 2(s - 3)/(s² + 9)) = 2e^-3t - 2cos(3t) + 2sin(3t)