Teacher has an old test up for review for an exam tomorrow. One problem:
•2. (a) Solve DE x’ ’ + 4x = 0 with the initial conditions x(0) = 2 and x’(0) = -2
This is simple harmonic motion and the general solution is x(t) = c1 cos(2t) + c2 sin(2t), with
x’(t) = -2c1 sin(2t) + 2c2 cos(2t) and the initial conditions yield
x(0) = 2 = c1
x’(0) = -2 = 2 c2 or c2 = -1
So the solution to the IVP is x(t) = 2 cos(2t) - sin(2t)
I understand all of that, part b though…
• b) Express the solution in phase amplitude form (it’s an image, so my writing it will look a bit awkward) :
A = [(2)^3 + (-1)^3]^ (1/2)
Where exactly did the ^3 come from? I was under the impression that A = the square root of x(t)^2 + (x’(t)/w)^2. I apoligize for the incredibly boring question.