Question

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

Answer 1

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

Answer 2

The positive value of k that satisfies the differential equation is k = 12.

To find the value of k that satisfies the equation, we differentiate the function y = sin(kt) twice to obtain y" and we insert it into the differential equation y" + 144y = 0.

So, y' = dy/dt

= dsin(kt)/dt

= kcos(kt)

y" = dy'/dt

y" = dkcos(kt)/dt

y" = -k²sin(kt)

So, substituting y and y" into the differential equation, we have

y" + 144y = 0

-k²sin(kt) + 144sin(kt) = 0

-k²sin(kt) = -144sin(kt)

k² = 144

k = ±√144

k = ±12

Since we require a positive value, k = 12

So, the positive value of k that satisfies the differential equation is k = 12.

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