231_pdfsam_math 54 differential equation solutions odd

# 231_pdfsam_math 54 differential equation solutions odd - to...

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Exercises 4.7 d 2 Y dt 2 = d [ y 0 ( t )] dt = y 0 ( t ) d ( t ) dt = y 0 ( t ) . Therefore, denoting t = s ,weobta in Y 0 ( t )+ tY ( t )= y 0 ( t )+ ty ( t )= y 0 ( s ) sy ( s )=0 . 2. Comparing the given equation with (13) on page 202 in the text, we conclude that inertia m =1 , damping b =0 , stiFness “ k ”= 6 y. ±or y> 0, the stiFness “ k ” is negative, and it tends to reinforce the displacement. So, we should expect that the solutions y ( t ) grow without bound. 3. As in Problem 2, this equation describes the motion of the mass-spring system with unit mass, no damping, and stiFness “ k ”= 6 y . The initial displacement y (0) = 1 is negative as well as the initial velocity y 0 (0) = 1. So, starting from t =0 , y ( t ) will decrease for a while. This will result increasing positive stiFness, 6 y , i.e., “the spring will become stiFer and stiFer”. Eventually, the spring will become so strong that the mass will stop and then go in the positive direction. While y (
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Unformatted text preview: to approach zero displacement point, y = 0. Thereafter, with y ( t ) > 0, the stiFness becomes negative, which means that the spring itself will push the mass further away from y = 0 in the positive direction with force, which increases with y . Thus, the curve y ( t ) will increase unboundedly. ±igure 4.23 con²rms our prediction. 5. (a) Comparing the equation y = 2 y 3 with equation (7) in Lemma 3, we conclude that f ( y ) = 2 y 3 , and so F ( y ) = Z 2 y 3 dy = 1 2 y 4 + C, where C is a constant. We can choose any particular value for C , say, C = 0. Thus F ( y ) = (1 / 2) y 4 . Next, with constant K = 0 and sign “ − ” in front of the integral, equation (11) on page 201, becomes t = − Z dy p 2(1 / 2) y 4 = − Z y − 2 dy = y − 1 + c, 227...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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