Answer:
Step-by-step explanation:
To show that (d^2y/dx^2) + k^2y = 0 for the given equation y = acos(kx) + bsin(x), we need to find the first and second derivatives of y with respect to x.
Given: y = acos(kx) + bsin(x)
1. Find the first derivative of y with respect to x:
(dy/dx) = -aksin(kx) + bcos(x)
2. Find the second derivative of y with respect to x:
(d^2y/dx^2) = -ak^2cos(kx) - bsin(x)
Now, substitute y, (dy/dx), and (d^2y/dx^2) into the expression (d^2y/dx^2) + k^2y:
(-ak^2cos(kx) - bsin(x)) + k^2(acos(kx) + bsin(x))
Simplify the expression:
-ak^2cos(kx) - bsin(x) + ak^2cos(kx) + bksin(x)
The terms ak^2cos(kx) and -ak^2cos(kx) cancel out, and the terms bsin(x) and bksin(x) also cancel out.
Therefore, the expression simplifies to:0
Thus, we have shown that (d^2y/dx^2) + k^2y = 0 for the given equation y = acos(kx) + bsin(x).
