Final answer:
The given differential equation y'' - 36y = 0 can be solved by assuming the solution is of the form y(x) = e^(rx), where r is a constant. The general solution is y(x) = c1e^(6x) + c2e^(-6x), and to find the particular solution that satisfies the boundary conditions y(0) = 0 and y(b) = 0, we substitute them into the general solution. After solving for the constants, we find that the particular solution is y(x) = 0.
Step-by-step explanation:
The given equation is a second order linear homogeneous differential equation. To solve it, we assume the solution is of the form y(x) = e^(rx), where r is a constant. Substituting this into the equation, we get the characteristic equation r^2 - 36 = 0, which can be factored as (r - 6)(r + 6) = 0. So r = 6 or r = -6.
The general solution is y(x) = c1e^(6x) + c2e^(-6x), where c1 and c2 are constants. To find the particular solution that satisfies the boundary conditions, we substitute y(0) = 0 and y(b) = 0 into the general solution.
If we substitute y(0) = 0 into the general solution, we get 0 = c1 + c2. If we substitute y(b) = 0 into the general solution, we get 0 = c1e^(6b) + c2e^(-6b).
Since e^(-6b) is never equal to 0 for any real value of b, we must have c2 = 0. Substituting this into the equation 0 = c1 + c2, we get 0 = c1. Therefore, the particular solution that satisfies the boundary conditions is y(x) = 0.