189_pdfsam_math 54 differential equation solutions odd

189_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.3 c 2 from the initial conditions. y (0) = 2= c 1 + c 2 , y 0 (0) = 0 = 3 c 1 + c 2 c 1 = 1 / 2 , c 2 = 3 / 2 . Thus, the solution to the initial value problem is y ( t )= e 3 t 2 3 e t 2 , and, as t →∞ , it approaches −∞ . (e) As in the previous problem, we have negative stiFness k = 6. But this time the initial displacement, y (0) = 1, as well as the initial velocity, y 0 (0) = 1, is positive. So, the mass will start moving in the positive direction, and will continue doing this (due to the negative stiFness) with increasing velocity. Thus our prediction is that y ( t ) when t . Indeed, the roots of the characteristic equation in this problem are r = 2 and 3, and so a general solution has the form y ( t c 1 e 2 t + c 2 e 3 t . To satisfy the initial conditions, we solve the system y (0) = 1 =
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