Differential Equation and omega

[iggybaseball]

I am having trouble with the following problem:

Find the value(s) of \omega [\tex] for which y = \cos(\omega * t) [\tex] satisfies<br /> <br /> \frac{d^2*y}{d*t^2} + 9y = 0[\tex]<br /> <br /> 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.<br /> <br /> Find the value(s) of omega for which y = cos(omega*t) satisfies:<br /> <br /> (d^2t)/(dt^2) + 9y = 0<br /> <br /> -----------------------------------------------------------------------<br /> <br /> I am not entirely sure what I am supposed to do here. My ideas have been <br /> 1.) switch 9y over to the right side<br /> 2.)Take the integral of both sides<br /> 3.)Take the integral of both sides again to solve for y(t)<br /> <br /> 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