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286practicefinalsolns - 286-E1 Spring 2010 Practice Final...

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286-E1, Spring 2010 Practice Final Exam - Solutions 1 1 (a) Find all a R so that the solution of the initial value problem x 0 = x 4 - 1 , x (0) = a satisfies lim t →∞ x ( t ) = - 1 . Solution. Since x 4 - 1 = ( x - 1)( x + 1)( x 2 + 1), the autonomous equation x 0 = x 4 - 1 has equilib- rium solutions x ( t ) = - 1 and x ( t ) = 1. x 4 - 1 is negative only on ( - 1 , 1), so we conclude that x ( t ) = - 1 is a stable equilibrium, while x ( t ) = 1 is unstable. (Sketch the phase diagram!) Therefore we have that lim t →∞ x ( t ) = - 1 precisely when x (0) = a ( -∞ , 1). (b) Classify the type of origin as a critical point of the system: d dt ~x = 0 - k k 0 ~x where k 6 = 0 is an arbitrary nonzero real number. Describe geometrically the trajectories of this system. Solution. The characteristic polynomial is: p ( λ ) = det - λ - k k - λ = λ 2 + k 2 = ( λ + ik )( λ - ik ) Since the egeinvalues λ = ± ik are purely imaginary, (0 , 0) is a center. Solution trajectories are circles centered at the origin, since for ( x 1 ( t ) , x 2 ( t )) satisfying the system we get: d dt x 1 ( t ) x 2 ( t ) = - kx 2 ( t ) kx 1 ( t ) d dt ( x 2 1 + x 2 2 ) = 2 x 1 x 0 1 + 2 x 2 x 0 2 = - 2 kx 1 x 2 + 2 kx 1 x 2 = 0 2. Find the general solution of the equation: ( x 2 + 1) dy dx + xy = x Solution. The equation is linear, so we use the integrating factor method. dy dx + x x 2 + 1 y = x x 2 + 1 r ( x ) = exp Z x x 2 + 1 dx = exp 1 2 Z du u = exp 1 2 ln( x 2 + 1) = ( x 2 + 1) 1 2 d dx ( x 2 + 1) 1 2 y = x ( x 2 + 1) 1 2 ( x 2 + 1) 1 2 y = Z x ( x 2 + 1) 1 2 dx ( x 2 + 1) 1 2 y = Z dv 2 v 1 2 = v 1 2 + C ( x 2 + 1) 1 2 y = ( x 2 + 1) 1 2 + C y = 1 + C ( x 2 + 1) - 1 2
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286-E1, Spring 2010 Practice Final Exam - Solutions 2 3. Find the general solution of the equation: (2 x sin y cos y ) y 0 = 4 x 2 + sin 2 y Solution. After substitution v = sin y , the equation becomes homogeneous: 2 xv dv dx = 4 x 2 + v 2 dv dx = 4 x 2 + v 2 2 xv = 2 x v + 1 2 v x u + x du dx = 2 u + u 2 where u = v x = sin y x x du dx = 2 u - u 2 2 udu 4 - u 2 = dx x - ln | 4 - u 2 | = ln | x | + C 4 - u 2 = K x 4 - sin 2 y x 2 = K x 4. Compute the Fourier series of the 4 π -periodic extension of the function: f ( t ) = ( 0 , 0 t 2 π 2 , - 2 π t 0 Solution. Fourier series with half-period L = 2 π are of the form: f ( t ) = a 0 2 + X n =1 a n cos nt 2 + b n sin nt 2 a 0 = 1 2 π Z 2 π - 2 π f ( t ) dt = 1 2 π Z 0 - 2 π 2 dt = 2 a n = 1 2 π Z 2 π - 2 π f ( t ) cos nt 2 dt = 1 2 π Z 0 - 2 π 2 cos nt 2 dt = 2 sin nt 2 0 - 2 π = 0 b n = 1 2 π Z 2 π - 2 π f ( t ) sin nt 2 dt = 1 2 π Z 0 - 2 π 2 sin nt 2 dt = - 2 cos nt 2 0 - 2 π = - 2 (1 - ( - 1) n ) f ( t ) = 1 + X n =1 - 2 (1 - ( - 1) n ) cos nt 2
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286-E1, Spring 2010 Practice Final Exam - Solutions 3 5. Suppose that a mass m = 2 kg is attached to a spring with spring constant k = 32 N/m whose other end is fixed to a wall. (a) Find the value of damping coefficient c for which this system is critically damped. What is the general solution for the position function x ( t ) of the mass at time t in this case?
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