nagle_differential_equations_ISM_Part22

# nagle_differential_equations_ISM_Part22 - CHAPTER 4 Linear...

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CHAPTER 4: Linear Second Order Equations EXERCISES 4.1: Introduction: The Mass-Spring Oscillator 2. (a) Substituting cy ( t ) into the equation yields m ( cy ) 00 + b ( cy ) 0 + k ( cy ) = c ( my 00 + by 0 + ky ) = 0 . (b) Substituting y 1 ( t ) + y 2 ( t ) into the given equation, we obtain m ( y 1 + y 2 ) 00 + b ( y 1 + y 2 ) 0 + k ( y 1 + y 2 ) = ( my 00 1 + by 0 1 + ky 1 ) + ( my 00 2 + by 0 2 + ky 2 ) = 0 . 4. With F ext = 0, m = 1, k = 9, and b = 6 equation (3) becomes y 00 + 6 y 0 + 9 y = 0 . Substitution y 1 = e - 3 t and y 2 = te - 3 t yields ( e - 3 t ) 00 + 6 ( e - 3 t ) 0 + 9 ( e - 3 t ) = 9 e - 3 t - 18 e - 3 t + 9 e - 3 t = 0 , ( te - 3 t ) 00 + 6 ( te - 3 t ) 0 + 9 ( te - 5 t ) = (9 t - 6) e - 3 t + 6(1 - 3 t ) e - 3 t + 9 te - 3 t = 0 . Thus, y 1 = e - 3 t and y 2 = te - 3 t are solutions to the given equation. Both solutions approach zero as t → ∞ . 6. With F ext = 2 cos 2 t , m = 1, k = 4, and b = 0, the equation (3) has the form y 00 + 4 y 0 = 2 cos 2 t. For y ( t ) = (1 / 2) t sin 2 t , one has y 0 ( t ) = 1 2 sin 2 t + t cos 2 t, y 00 ( t ) = 2 cos 2 t - 2 t sin 2 t ; y 00 + 4 y 0 = (2 cos 2 t - 2 t sin 2 t ) + 4 ± 1 2 t sin 2 t ² = 2 cos 2 t. 101

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Chapter 4 Hence, y ( t ) = (1 / 2) t sin 2 t is a solution. Clearly, this function satisﬁes the initial condi- tions. Indeed, y (0) = 1 2 t sin 2 t ± ± ± ± t =0 = 0 y 0 (0) = 1 2 sin 2 t + t cos 2 t ± ± ± ± t =0 = 0 . As t increases, the spring will eventually break down since the solution oscillates with the magnitude increasing without bound. 8.
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nagle_differential_equations_ISM_Part22 - CHAPTER 4 Linear...

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