2nd order differential equation homework help please!
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2nd order differential equation homework help please!

[From: ] [author: ] [Date: 11-09-27] [Hit: ]
take the derivative...i figured the amplitude would be 1? but its not........
x"+4x=0, x(0)=1, x'(0)=-2

please show me how to find the amplitude, phaseangle, and period!

i got as far as utilizing the characteristic equation...
x^2+4 =0
x = +/- 2i
x(t) = Acos(2t)+Bisin(2t)
then using the initial conditions..
x(0)=1 => cos(2t)+Bisin(2t)
take the derivative...
x'(t)=-2sin(2t)+2Bicos(2t)
-2 = x'(0) => 0 +2Bi
divide both sides by 2i
B= -1/i
x(t) = cos(2t)-sin(2t)

i figured the amplitude would be 1? but it's not... and i think the period 2pi? no clue as to what a phaseangle is =T help please! thanks ^_^

-
x(t)
= cos(2t) - sin(2t)
= sqrt(2) [ cos(2t) cos(pi/4) - sin(2t) sin(pi/4) ]
= sqrt(2) cos(2t+pi/4)

The rest should be easy for you.
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