188_pdfsam_math 54 differential equation solutions odd

# 188_pdfsam_math 54 differential equation solutions odd - y...

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Chapter 4 Solving the initial value problem yields y (0) = 1 = c 1 + c 2 , y 0 (0) = 0 = c 1 r 1 + c 2 r 2 c 1 = r 2 / ( r 2 r 1 ) , c 2 = r 1 / ( r 1 r 2 ) , and so the desired solution is y ( t )= 50 + 2499 2 2499 e ( 50 2499) t + 50 + 2499 2 2499 e ( 50+ 2499) t . Since both powers in exponential functions tend to −∞ as t →∞ , y ( t ) 0. (c) The corresponding mass-spring model has negative damping b = 6 and positive stiFness k = 8. Thus the magnitude | y ( t ) | of the displacement y ( t ) will increase without bound, as t →∞ . Moreover, because of the positive initial displacement and initial zero velocity, the mass will move in the negative direction. Thus, our guess is that y ( t ) →−∞ as t →∞ . Now we ±nd the actual solution. Since the roots of the auxiliary equation are r =2and r = 4, a general solution to the given equation is y ( t )= c 1 e 2 t + c 2 e 4 t . Next, we ±nd c 1 and c 2 satisfying the initial conditions. y (0) = 1 = c 1 + c 2 , y 0 (0) = 0 = 2 c 1 +4 c 2 c 1 =2 , c 2 = 1 . Thus, the desired solution is
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Unformatted text preview: y ( t ) = 2 e 2 t − e 4 t , and it approaches −∞ as t → ∞ . (d) In this problem, the stiFness k = − 3 is negative. In the mass-spring model, this means that the spring forces the mass to move in the same direction as the sign of the displace-ment is. Initially, the displacement y (0) = − 2 is negative, and the mass has no initial velocity. Thus the mass, when released, will move in the negative direction, and the spring will enforce this movement. So, we expect that y ( t ) → −∞ as t → ∞ . To ±nd the actual solution, we solve the auxiliary equation r 2 + 2 r − 3 = 0 and obtain r = − 3, 1. Therefore, a general solution is given by y ( t ) = c 1 e − 3 t + c 2 e t . We ±nd c 1 and 184...
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