Posted by David on Saturday, July 27, 2013 at 4:33pm.
Solve the DE by using the Laplace transform:
x''+x=cos(3t), x(0)=x'(0)=0

calculus  Steve, Saturday, July 27, 2013 at 6:03pm
L{x"} = s^2*F(s)  s*f(0)  f'(0)
= s^2*F(s)
so,
s^2 F(s) + F(s) = s/(s^2+9)
F(s) = s/((s^2+9)(s^2+1))
= 1/8 (s/(s^2+1)  s/(s^2+9))
f(t) = 1/8 (cos(t)  cos(3t))
check:
x' = 1/8 (3sin(3t)sin(t))
x" = 1/8 (9cos(3t)cos(t))
x'+x = 1/8 (8cos(3t)) = cos(3t)