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y = acos kx + bsinx
show that ( d^2y/dx^2 ) + k^2y= 0
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Sagot :

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).

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