Differentiating the function two times with respect to
t: 
y = cos(k x t) 
y' = -k x sin(k x t) 
y'' = -(k^2) x cos(k x t) 
Plug this back into the Differential equation : 
4*(k^2)*cos(k*t) = -9*cos(k*t) 
4*k^2 = 9 
k^2 = 9/4
k = +9/4 and k = -9/4 
k = +sqrt(9/4) and l = -sqrt(9/4)
k = 1.5 and I = -1.5