Well I really don't know that much about Kepler's laws and so I cannot say I am 100 percent sure about this but this is what I got out of it so far.(adsbygoogle = window.adsbygoogle || []).push({});

All the question is asking you to do is to verify that [tex]x(t)[/tex] and [tex]y(t)[/tex] are in fact solutions to the diffeqs. You do not need to solve the diffeqs to check their solutions. All you need to do is check'em. So lets do that and see what happens.

Like I said before [tex]x''(t)=-Rw^2cos(wt) \mbox{ and } y''(t)=-Rw^2cos(wt)[/tex].

And we are given those exact same functions above just in a different form. So we just plug in x and y for the given diffeqs and then check to see if they match our diffeqs.

[tex]\frac{d^2X}{dt^2}=\frac{-GMx}{(x^2+y^2)^{3/2}}[/tex]

is our first given diffeq, so where ever we see an x we put it there and where ever we see a y we put that there and in unsimplified form it looks like,

[tex]-GM*Rcos(wt)/([Rcos(wt)]^2+[Rsin(wt)]^2)^(3/2)[/tex]

Which simplifies to [tex]-GM*cos(wt)/R^2[/tex]

Well we can set that equal to our solution and we have

[tex]x''(t)=-Rw^2cos(wt)=-GM*cos(wt)/R^2[/tex]

Which simplifies to[tex] w^2=GM/r^3[/tex]

The exact same procedure for[tex] y''(t)[/tex] will give the same answer [tex]w^2=GM/R^3[/tex]

Now if someone knows Keplers third law then we could be sure of our answer.

I could be wrong though so I would be careful how you are using this solution.

Regards

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# Helping out a friendor at least trying to

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