Final answer:
To solve the initial value problem z''(x) + z(x) = 6e^-5x with z(0) = 0 and z'(0) = 0, we combine a particular solution of the non-homogeneous equation with the general solution of the homogeneous equation and apply the initial conditions to find the constants.
Step-by-step explanation:
The student is asked to solve an initial value problem for a second-order differential equation z''(x) + z(x) = 6e-5x, with initial conditions z(0) = 0, and z'(0) = 0. To solve it, we look for a particular solution to the non-homogeneous equation and a general solution to the associated homogeneous equation z''(x) + z(x) = 0. The solution to the homogeneous equation is zh(x) = C1cos(x) + C2sin(x), where C1 and C2 are constants to be determined by the initial conditions.
For the particular solution, we make a guess zp(x) = Ae-5x and find that A must satisfy A = -6/24 = -1/4 to fulfill the non-homogeneous part. So, the particular solution is zp(x) = -1/4e-5x.
Combining both parts, the general solution is z(x) = zh(x) + zp(x) = C1cos(x) + C2sin(x) - 1/4e-5x. Using the initial conditions, we get z(0) = C1 - 1/4 = 0, which means C1 = 1/4. To find C2, we differentiate the general solution to get z'(x) = -C1sin(x) + C2cos(x) + 5/4e-5x and use the initial condition z'(0) = C2 + 5/4 = 0, which gives C2 = -5/4.
The final solution is therefore z(x) = 1/4cos(x) - 5/4sin(x) - 1/4e-5x.