Laplace transforms give s 2 Y s 2 Y s 3 Y s 2 s 3 2 e 3 s s 3 6 e 3 s s 2 9 e 3

Laplace transforms give s 2 y s 2 y s 3 y s 2 s 3 2 e

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Laplace transforms give: s 2 Y ( s ) + 2 Y ( s ) - 3 Y ( s ) = 2 s 3 - 2 e - 3 s s 3 - 6 e - 3 s s 2 - 9 e - 3 s s . (Alternately, one could evaluate F ( s ) = Z 3 0 t 2 e - st dt,
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which gives the same answer.) It follows that Y ( s ) = 2 - 2 e - 3 s - 6 se - 3 s - 9 s 2 e - 3 s s 3 ( s + 3)( s - 1) The partial fractions is tedious but straight forward, giving 2 s 3 ( s + 3)( s - 1) = 1 54( s + 3) + 1 2( s - 1) - 2 3 s 3 - 4 9 s 2 - 14 27 s - 2 - 6 s - 9 s 2 s 3 ( s + 3)( s - 1) = - 65 108( s + 3) - 17 4( s - 1) + 2 3 s 3 + 22 9 s 2 + 131 27 s The inverse Laplace transform gives the solution y ( t ) = 1 54 e - 3 t + 1 2 e t - 1 3 t 2 - 4 9 t - 14 27 - u 3 ( t ) 65 108 e - 3( t - 3) + 17 4 e ( t - 3) - 1 3 ( t - 3) 2 - 22 9 ( t - 3) - 131 27 5. a. If L{ f } = F ( s ), then F ( s ) = Z 0 f ( t ) e - st dt. Leibnitz’s rule of differentiation of the integral states that d dx Z b a f ( x, y ) dy = Z b a ∂f ( x, y ) ∂x dy. In other words, the differentiation operator can commute with the integral. Thus, dF ( s ) ds = Z 0 ∂f ( t ) e - st ∂s dt = - Z 0 t f ( t ) e - st dt. It follows that L{ t f ( t ) } = - dF ( s ) ds . b. Since L sin(2 t ) = 2 s 2 +4 , then - d ds 2 s 2 + 4 = 4 s ( s 2 + 4) 2 , which is L{ t sin(2 t ) } . c. For the initial value problem y 00 + 4 y = 2 cos(2 t ) , y (0) = - 1 , y 0 (0) = 4 , we take Laplace transforms and obtain: s 2 Y ( s ) + s - 4 = 4 Y ( s ) = 2 s s 2 + 4 or Y ( s ) = 4 - s s 2 + 4 + 2 s ( s 2 + 4) 2
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The inverse Laplace transform gives y ( t ) = 2 sin(2 t ) - cos(2 t ) + 1 2 t sin(2 t ) . 6. a. The differential equation can be written y 00 + y = δ ( t ) + δ ( t - π ) + δ ( t - 2 π ) + δ ( t - 3 π ) + ... The Laplace transform becomes L{ y 00 } + L{ y 0 } = s 2 L{ y } - sy (0) - y 0 (0) + L{ y 0 } = L{ δ ( t ) + δ ( t - π ) + δ ( t - 2 π ) + δ ( t - 3 π ) + ... } or ( s 2 + 1) L{ y } = 1 + e - πs + e - 2 πs + e - 3 πs + ... Thus, L{ y } = 1 s 2 + 1 + e - πs s 2 + 1 + e - 2 πs s 2 + 1 + e - 3 πs s 2 + 1 + ... or y ( t ) = sin( t ) + u π ( t ) sin( t - π ) + u 2 π ( t ) sin( t - 2 π ) + u 3 π ( t ) sin( t - 3 π ) + ... = sin( t )(1 - u π ( t ) + u 2 π ( t ) - u 3 π ( t ) + ... ) = sin( t ) X n =0 ( - 1) n u ( t ) . b. The differential equation can be written y 00 + y = δ ( t ) + δ ( t - 2 π ) + δ ( t - 4 π ) + δ ( t - 6 π ) + ... The Laplace transform becomes L{ y 00 } + L{ y 0 } = s 2 L{ y } - sy (0) - y 0 (0) + L{ y 0 } = L{ δ ( t ) + δ ( t - 2 π ) + δ ( t - 4 π ) + δ ( t - 6 π ) + ... } or ( s 2 + 1) L{ y } = 1 + e - 2 πs + e - 4 πs + e - 6 πs + ... Thus, L{ y } = 1 s 2 + 1 + e - 2 πs s 2 + 1 + e - 4 πs s 2 + 1 + e - 6 πs s 2 + 1 + ... or y ( t ) = sin( t ) + u 2 π ( t ) sin( t - 2 π ) + u 4 π ( t ) sin( t - 4 π ) + u 6 π ( t ) sin( t - 6 π ) + ... = sin( t )(1 + u 2 π ( t ) + u 4 π ( t ) + u 6 π ( t ) + ... ) = sin( t ) X n =0 u 2 ( t ) . c. In the first case, sine functions alternately cancel each other out. Thus, on alternate intervals of π , the function is positive part of the sine curve followed by an interval of the zero function. The second case is the resonance case. In this case, each interval of 2 π has the solution increasing the amplitude of the sine function by 1. This solution becomes unbounded, so the bridge would collapse.
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