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Unformatted text preview: Our final answer is therefore L { f } = 1 − e s 1 s +1 + e s − e 2 s s 1 e − 2 s . 7 Find the first six terms in the power series expansion at x = 0 for the general solution to the differential equation ( x 2 + 1) y ′′ + y = 0. Solution: We need to find the first six terms, so we make the following substitutions y = a + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ... , y ′ = a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + 5 a 5 x 4 + ... , y ′′ = 2 a 2 + 6 a 3 x + 12 a 4 x 2 + 20 a 5 x 3 + ... , MAP 2302, Fall 2011 — Final Exam Review Problems 7 Now we make the substitution: ( x 2 +1) y ′′ + y = ( x 2 +1)(2 a 2 +6 a 3 x +12 a 4 x 2 +20 a 5 x 3 + ... )+( a + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ... ) . We now expand to get (2 a 2 + a ) + (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 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 , 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 / 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 / 2 = 0, so a 4 = a / 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 + a 1 x a 2 x 2 a 1 6 x 3 + a 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. 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.clearly indicate the amplitude, period, and phase angle of this motion....
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 Fall '08
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 Laplace, lim, Constant of integration, Exam Review Problems

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