We now expand to get 2 a 2 a 6 a 3 a 1 x 12 a 4 2 a 2

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We now expand to get (2 a 2 + a 0 ) + (6 a 3 + a 1 ) x + (12 a 4 + 2 a 2 + a 2 ) x 2 + (20 a 5 + 6 a 3 + a 3 ) x 3 + · · · = 0 Note that we can stop here because we have enough equations to solve for a 2 , a 3 , a 4 , and a 5 in terms of our two arbitrary constants, which will be a 0 and a 1 (these are always the arbitrary constants in power series solutions of second order equations!). Our system of equations is then 2 a 2 + a 0 = 0 , 6 a 3 + a 1 = 0 , 12 a 4 + 3 a 2 = 0 , 20 a 5 + 7 a 3 = 0 . We proceed from top to bottom. The first equation shows that a 2 = - a 0 / 2 . The second shows that a 3 = - a 1 / 6 . The third then becomes 12 a 4 + 3 a 2 = 12 a 4 - 3 a 0 / 2 = 0, so a 4 = a 0 / 8 . The fourth becomes 20 a 5 + 7 a 3 = 20 a 5 - 7 a 1 / 6 = 0, so a 5 = a 1 / 120 . Our final answer is therefore y = a 0 + a 1 x - a 0 2 x 2 - a 1 6 x 3 + a 0 8 x 4 + a 1 120 x 5 + · · · . 8 Find the singular points of the equation ( x 2 - 1) y ′′ - ( x - 1) y + 17 y = 0 and classify them as regular or irregular. Solution: To find the singular points, we first write the differential equation in standard form: y ′′ - x - 1 x 2 - 1 y + 17 x 2 - 1 y = 0 . From this we see that the singular points are x = ± 1. We need to test each of them separately. The point x = 1 will be a regular singular point if both of the limits lim x 1 ( x - 1) x - 1 x 2 - 1 and lim x 1 ( x - 1) 2 17 x 2 - 1 exist. It is easy to check that both of these limits exist (in fact, they are both equal to 0), so x = 1 is a regular singular point.
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MAP 2302, Fall 2011 — Final Exam Review Problems 8 For x = - 1, our two limits are lim x →− 1 ( x + 1) x - 1 x 2 - 1 and lim x →− 1 ( x + 1) 2 17 x 2 - 1 . Again, both limits exist (the first equals 1, the second equals 0), so x = - 1 is also a regular singular point. 9 A 2 kg mass is attached to a spring with stiffness k = 10 N/m. The mass is displaced 1 m to the right of the equilibrium point and given an initial leftward velocity of 1 m/s. Let y ( t ) denote the position of the spring at time t seconds, where positive y indicates distance to the right of the equilibrium point. Solve for y , express it in the form y = A sin( ωt + φ ), and then clearly indicate the amplitude, period, and phase angle of this motion. Solution: First we need to write this in the standard spring equation form of my ′′ + by + ky = f ( t ). In this case, m = 2, b = 0, k = 10, and f ( t ) = 0, so we have 2 y ′′ + 10 y = 0 . This has the characteristic polynomial 2 r 2 + 10 = 0, with roots r = ± 5 i . Therefore the solution is y = C cos 5 t + D sin 5 t. The initial conditions can be interrupted as y (0) = 1 and y (0) = - 1. From these we find that C = 1 and D = - 1 / 5 , giving the solution of y = cos 5 t - sin 5 t 5 . Now to answer the question, we want to express y in the form A sin( ωt + φ ). We do this using the angle addition formula; we want to find A , φ , and ω so that y = 1 cos 5 t + - 1 5 sin 5 = A sin φ cos ωt + A cos φ sin ωt, so we need to solve the system of equations A 2 = 1 2 + parenleftbigg - 1 5 parenrightbigg 2 = 6 5 , tan φ = 1 - 1 / 5 = - 5 .
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MAP 2302, Fall 2011 — Final Exam Review Problems 9 This shows that the solution is y ( t ) = radicalbig 6 / 5 sin parenleftBig 5 t + arctan( - 5) parenrightBig .
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  • Fall '08
  • TUNCER
  • Laplace, lim, Constant of integration, Exam Review Problems

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