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dwilmer
Nov18-09, 03:57 PM
In Boyce and Diprima textbook it says that:

mu'' + ku = 0 .

Then it says the general solution of this is:

u = Acos (w^2)t + B sin (w^2)t , where w^2 = k/m


It provides no explanation how it arrives at this. Where does the cos and sin come from?????

PS: it is supposed to be the greek letter w, whatever that is called and it also has a sub-0 on it, but i didnt include it because it looks confusing without the right fonts.

please help, thanks

tiny-tim
Nov18-09, 04:53 PM
Hi dwilmer! Welcome to PF! :smile:

(have a little omega: ω and try using the X2 tag just above the Reply box :wink:)
mu'' + ku = 0 .

Then it says the general solution of this is:

u = Acos (w^2)t + B sin (w^2)t , where w^2 = k/m


It provides no explanation how it arrives at this. Where does the cos and sin come from?????

Because [cos(ωt)]'' is obviously -ω2cos(ωt), and [sin(ωt)]'' is obviously -ω2sin(ωt) :smile:

n1person
Nov18-09, 06:39 PM
Perhaps a more helpful explanation can be found here:

http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf

page 4 in particular offers a little bit of an explanation, and if you are still confused I would advise reading the entire thing, it isn't too long and is relatively well done.