Differential Equation and omega

[iggybaseball]

I am having trouble with the following problem:

Find the value(s) of [tex] \omega [\tex] for which [tex] y = \cos(\omega * t) [\tex] satisfies

[tex]\frac{d^2*y}{d*t^2} + 9y = 0[\tex]

I am trying to use latex but it doesn't seem to be working when I do "preview post" so I will rewrite what I am saying to make it more understandable in case Latex doesn't work upon posting.

Find the value(s) of omega for which y = cos(omega*t) satisfies:

(d^2t)/(dt^2) + 9y = 0

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I am not entirely sure what I am supposed to do here. My ideas have been
1.) switch 9y over to the right side
2.)Take the integral of both sides
3.)Take the integral of both sides again to solve for y(t)

This approach however left me lost in the dark and I feel is incorrect. I also tried substituting y = cos(omega*t) in for y but I can't solve the following equation. Could someone give me some ideas what I should do?

 

[TD]

You need to use a normal slash in your closing tag, so [ /tex ], instead of the backslash \.

Here you go:

$\omega$ for which $$y = \cos(\omega t)$$ satisfies

$$\frac{d^2 y}{d t^2} + 9y = 0$$

The equation is homogenous, the char. polynomial gives $$\lambda ^2 + 9 = 0$$ so the solutions are $$\pm 3i$$

That gives the following lineair combination of cos & sin as solution: $$y = c_1 \cos 3t + c_2 \sin 3t$$

Does that help?

 

[iggybaseball]

Thank you for pointing out my mistake. I am a little confused by your answer because we haven't talked about "homogenous equations", or had any solutions with imaginary numbers (3i). However would this be right...

$$y = \cos(\omega t)$$

$$\frac{dy}{dt} = -\omega\sin(\omega t)$$

$$\frac{d^2y}{dt^2} = \omega^2cos(\omega t)$$

$$\omega^2\cos(\omega t) + 9\cos(\omega t) = 0$$

$$\cos(\omega t)(\omega^2 + 9) = 0$$

$$\omega = \pm3$$

$$\cos(\omega t) = 0$$

$$\omega t = 1$$

$$\omega = \frac{1}{t}$$

Therefore would the solutions
$$\omega = \pm3 , \frac{1}{t}$$
be my solutions? Thanks again for the help with Latex. Life is much more easier now.

 

[TD]

Watch out, if $\frac{dy}{dt} = -\omega\sin(\omega t)$ then $\frac{d^2y}{dt^2} = -\omega^2cos(\omega t)$ since the derivative of sinx is just cosx, the sign doesn't change there.

Other then that, your solutions would indeed be 3 and -3, as found before

After that, be careful. $\cos 0 = 1$ but not $\cos 1 = 0$

So $\cos \omega t = 0$ would give the following solutions for $\omega$ as well, as a function of $t$:

$$\cos \omega t = 0 \Leftrightarrow \omega t = \frac{\pi }{2} + 2k\pi \,\,\vee \,\, \omega t = \frac{{3\pi }}{2} + 2k\pi \Leftrightarrow \omega = \frac{{\pi \left( {4k + 1} \right)}}{{2t}} \,\,\vee \,\, \omega = \frac{{\pi \left( {4k + 1} \right)}}{{2t}}$$

 

[iggybaseball]

Oh you are right that is a silly mistake on my part. I am a little confused and don't understand what you are writing (I'm not as advanced hehe) but after revision I get :

$$-\omega^2\cos(\omega t) + 9\cos(\omega t) = 0$$

$$\cos(\omega t)(-\omega^2 + 9) = 0$$

So my solutions are still valid (luckily) but are you saying if I add
$$2 \pi k$$
to
$$\frac{1}{t}$$

Those are also solutions? Sorry I just don't understand the last part. Thank you again.

 

[TD]

You had $$\cos(\omega t)(-\omega^2 + 9) = 0$$, which seems correct.

$$(-\omega^2 + 9) = 0$$ gives as solutions 3 and -3.

$$\cos(\omega t) = 0$$ gives the solutions I described above, since the argument of which a cosine is 0 has to be either $\frac{\pi }{2}$ or $\frac{3\pi }{2}$, and of course you can add $2k\pi$ to those solutions.

 

[iggybaseball]

Now I understand everything you have written. Thank you so much for the help. Have a nice day

 

[TD]

Glad I could help, a nice day to you too