Respuesta :
Answer:
-10/3, 10/3
Step-by-step explanation:
(In this answer I will use y' to denote the derivative of y with respect to t. You shouldn't normally do this because y' normally means the derivative of y with respect to x but I'll be a bit messy for this case)
First calculate the derivatives:
[tex]y=\cos(kt) \Rightarrow y'=-k\sin(kt) \Rightarrow y'' = -k^2\cos(kt)[/tex].
Then plug the derivtes y'' and y into the equation:
[tex]-9k^2\cos(kt) = -100\cos(kt)[/tex]
Solve the equation for k:
[tex]100\cos(kt) - 9k^2\cos(kt) = 0 \\\\\Rightarrow \cos(kt)(100-9k^2) = 0[/tex]
So then we have that [tex]y=\cos(kt)[/tex] satisfies the differential equation when [tex]\cos(kt) = 0[/tex] or when [tex]100-9k^2=0[/tex] (or both). The solutions to these equations are:
[tex]\left \{ {{\cos(kt)=0 \Rightarrow k=\frac{n\pi}{2t}} \atop {100-9k^2 = 0 \Rightarrow k= \pm \sqrt{\frac{100}{9}}=\pm \frac{10}{3}}} \right.[/tex]
I understand that looks a bit complicated and I doubt you would have to give your answers in terms of t so if it asks for a separated list of answers I would go for:
k = -10/3, 10/3.
The values are [tex]k = \pm \frac{10}{3}[/tex].
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- To find the values of k, we have to replace the derivatives into the equation.
The function is:
[tex]y = \cos{kt}[/tex]
The derivatives are:
[tex]y^{\prime}(t) = -k\sin{kt}[/tex]
[tex]y^{\prime\prime}(t) = -k^2\cos{kt}[/tex]
The equation is:
[tex]9y^{\prime\prime} = -100y[/tex]
Replacing:
[tex]-9k^2\cos{kt} = -100\cos{kt}[/tex]
[tex]9k^2 = 100[/tex]
[tex]k^2 = \frac{100}{9}[/tex]
[tex]k = \pm \sqrt{\frac{100}{9}}[/tex]
[tex]k = \pm \frac{10}{3}[/tex]
Those are the values.
A similar problem is given at https://brainly.com/question/24348029
