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Keith_Richards [23]
4 years ago
5

For what positive values of k does the function y=sin(kt) satisfy the differential equation y''+144y=0 ?

Mathematics
1 answer:
lina2011 [118]4 years ago
7 0

Answer:

The positive value of k that the function y = sin(kt) satisfies the differential equation y'' + 144y = 0 is +12

Step-by-step explanation:

To determine the positive values of k that the function y = sin(kt) satisfy the differential equation y''+144y=0.

First, we will determine y''.

From y = sin(kt)

y' = \frac{d}{dt}(y)

y' = \frac{d}{dt}(sin(kt))\\

y' = kcos(kt)

Now for y''

y'' = \frac{d}{dt}(y')

y'' = \frac{d}{dt}(kcos(kt))

y'' = -k^{2}sin(kt)

Hence, the equation y'' + 144y = becomes

-k^{2}sin(kt) + 144(sin(kt)) = 0

(144 - k^{2})(sin(kt)) = 0

(144 - k^{2})= 0

∴ k^{2} = 144\\

k = ±\sqrt{144}\\

k = ± 12

∴ k = +12 or -12

Hence, the positive value of k that the function y = sin(kt) satisfies the differential equation y'' + 144y = 0 is +12

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