Problem with differential equations

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In Example 3 and Example 4 what is the meaning of omega? Why is it 1 and 3 respectively? I don't know how these results are obtained. I differentiated the expression above Example 3 and filled in with the given information in Example 3 but it didn't make anything clear.

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In mathematical terms, Omega is the same to-be-determined variable as in Example 2. In terms of Physics, it is the https://en.wikipedia.org/wiki/Angular_frequency.

Example solution for Example 3:

If I take x = x0 cos (wt) + v0/w sin (wt) and plug in x(t=0) = 0, then I get the condition

x0 cos (0) + v0/w sin(0) = 0.

Hence, x0 = 0.

The first derivative of the function is dx/dt = w x0 (-sin(wt)) + v0 cos (wt).

Plugging in the constraint dx/dt (x=0) = 1 yields

w x0 (-sin(0)) + v0 cos (0) = 1.

Hence, v0 = 1.

The 2nd derivative of the function is d^2 x/dt^2 = -w^2 x0 cos(wt) - w v0 sin(wt).

That one is a bit tricky, but by comparison you will see that this is -w^2 * (original function). So this satisfies the condition d^2 x/dt^2 = -x if w^2 = 1. I don't know why w=-1 is not considered here; maybe I missed something.

Hope that helps. If so, it would be nice to hear what the actual step was that you got stuck on.

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52 minutes ago, timo said:

Hope that helps. If so, it would be nice to hear what the actual step was that you got stuck on.

Thanks for your help. I just wanted to stick to the book procedure but sometimes the book makes it harder than it really is, this has happened before.

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• 2 weeks later...

Actually, the book specifically tells you what omega is in problem 2!  There the equation is d^2x/dt^2= omega x.  In problem 3 the equation is d^2x/dt^2= -x.  Comparing that to problem 2 you should immediately see that problem 3 is the same as problem 2 with omega= 1.

Edited by HallsofIvy
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1 hour ago, HallsofIvy said:

Actually, the book specifically tells you what omega is in problem 2!  There the equation is d^2x/dt^2= omega x.  In problem 3 the equation is d^2x/dt^2= -x.  Comparing that to problem 2 you should immediately see that problem 3 is the same as problem 2 with omega= 1.

" the book specifically tells you what omega is in problem 2!  There the equation is d^2x/dt^2= omega x."

Seems to me that example 2 starts with

$\frac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$

Did you know that you can expand the attachment so that you don't need a magnifying glass to read it. Edited by studiot
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With my eyes, that probably wouldn't help!  But thanks for the correction.

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