Final answer:
To solve the initial-value problem, find the general solution of the differential equation, and use the initial conditions to find the specific solution.
Step-by-step explanation:
To solve the initial-value problem, we need to find the general solution of the differential equation and then use the given initial conditions to find the specific solution. The differential equation is y'' + 36y = 0.
The characteristic equation is r^2 + 36 = 0. Solving this equation gives us complex roots r = ±6i. Therefore, the general solution is y = c1*cos(6x) + c2*sin(6x). To find the specific solution, we use the initial conditions y(π/6) = 0 and y'(π/6) = 0.
Plugging in the values, we get 0 = c1*cos(π/6) + c2*sin(π/6) and 0 = -6c1*sin(π/6) + 6c2*cos(π/6). From the first equation, we can determine that c2 = -c1*tan(π/6). Substituting this into the second equation and solving for c1, we get c1 = 0. Therefore, the specific solution is y = 0.