# hw10 - Homework 10 Solutions 10.3.1 10.3.13 10.3.15 y 2 y =...

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Homework 10 Solutions 10.3.1:

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10.3.13:
10.3.15: y 00 + ω 2 y = f ( t ) , y (0) = 0 , y 0 (0) = 0 f ( t ) = 1 , 0 < t < π 0 , t = 0 , π, 2 π - 1 , π < t < 2 π f ( t ) is odd so a 0 = a n = 0 and b n = 2 L Z L 0 f ( t ) sin nπt L dt = 2 π Z π 0 sin ntdt = 2 ( - cos nx | π 0 ) = 2 (1 - ( - 1) n ) = 4 π (2 n - 1) n = 1 , 2 , 3 , ... f ( t ) = 4 π X n =1 sin (2 n - 1) t 2 n - 1 y 00 + ω 2 y = 4 π X n =1 sin (2 n - 1) t 2 n - 1 y = c 1 cos ωt + c 2 sin ωt + Y Y = X n =1 A 2 n - 1 sin (2 n - 1) t Y 00 = X n =1 - (2 n - 1) 2 A 2 n - 1 sin (2 n - 1) t Y 00 - ω 2 Y = X n =1 - (2 n - 1) 2 A 2 n - 1 sin (2 n - 1) t + ω 2 X n =1 A 2 n - 1 sin (2 n - 1) t = X n =1 A 2 n - 1 ( ω 2 - (2 n - 1) 2 ) sin (2 n - 1) t = 4 π X n =1 sin (2 n - 1) t 2 n - 1 A 2 n - 1 = 4 π 1 ( ω 2 - (2 n - 1) 2 )(2 n - 1) y = c 1 cos ωt + c 2 sin ωt + 4 π

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Unformatted text preview: 2 )(2 n-1) y (0) = 0 = c 1 y = ωc 2 cos ωt + 4 π ∞ X n =1 cos (2 n-1) t ω 2-(2 n-1) 2 y (0) = 0 = ωc 2 + 4 π ∞ X n =1 1 ω 2-(2 n-1) 2 c 2 =-4 π ∞ X n =1 1 ( ω 2-(2 n-1) 2 ) ω y = 4 π ∞ X n =1 1 ω 2-(2 n-1) 2 ± 1 2 n-1 sin (2 n-1) t-1 ω sin ωt ²...
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