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pset_3sol

# pset_3sol - ACM 95/100b Problem Set 3 Solutions Faisal...

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Unformatted text preview: ACM 95/100b Problem Set 3 Solutions Faisal Amlani February 2, 2009 Problem 1 (10 points) Solve the initial value problem y ′′ + 2 y ′ + y = u ( t ) y (0) = 0 y ′ (0) = 1 where u ( t ) is the unit step function discussed in class: u ( t ) = braceleftBigg t < 1 1 t ≥ 1 using the Laplace transform. SOLUTION : Since the initial value problem is solving for t ≥ , we have for the Laplace transform of the right hand side, U ( s ) = L [ u ( t )] = integraldisplay ∞ exp( − st ) u ( t ) dt = integraldisplay ∞ 1 exp( − st ) u ( t > 0) + integraldisplay 1 exp( − st ) u ( t < 1) dt = integraldisplay ∞ 1 exp( − st ) dt = − 1 s exp( − st ) vextendsingle vextendsingle vextendsingle ∞ 1 = bracketleftbigg lim t →∞ − 1 s exp( − st ) bracketrightbigg + 1 s exp( − s ) = exp( − s ) s . Define Y ( s ) , Y ′ ( s ) , Y ′′ ( s ) the respective Laplace transforms of y ( t ) , y ′ ( t ) , y ′′ ( t ) . We know from the initial conditions that Y ′ ( s ) = sY ( s ) − y ′ (0) = sY ( s ) , Y ′′ ( s ) = s 2 Y ( s ) − sy (0) − y ′ (0) = s 2 Y ( s ) − 1 . 1 Thus, the Laplace transform of the ODE is s 2 Y − 1 + 2 sY + Y = U = ⇒ ( s + 1) 2 Y = 1 + U = ⇒ Y ( s ) = 1 ( s + 1) 2 + exp( − s ) s ( s + 1) 2 . We want to compute the inverse to get back the original solution, i.e. compute y ( t ) = L − 1 bracketleftbigg 1 ( s + 1) 2 bracketrightbigg + L − 1 bracketleftbigg exp( − s ) s ( s + 1) 2 bracketrightbigg . Taking some c to lie to the right of all singularities and closing the contour, we can invoke generalized Cauchy’s Integral Formula to compute the first inverse transform as L bracketleftbigg 1 ( s + 1) 2 bracketrightbigg = 1 2 πi integraldisplay c + i ∞ c − i ∞ exp( st ) 1 ( s + 1) 2 ds = 1 2 πi 2 πi d ds [exp( st )] s = − 1 = t exp( − t ) This then gives y ( t ) = t exp( − t ) + L − 1 bracketleftbigg exp( − s ) s ( s + 1) 2 bracketrightbigg . To compute the remaining inverse transform, note by partial fraction decomposition 1 s ( s +1) 2 = 1 s − 1 s +1 − 1 ( s +1) 2 = ⇒ L − 1 bracketleftbigg exp( − s ) s ( s + 1) 2 bracketrightbigg = L − 1 bracketleftbigg exp( − s ) s bracketrightbigg − L − 1 bracketleftbigg exp( − s ) s + 1 bracketrightbigg − L − 1 bracketleftbigg exp( − s ) ( s + 1) 2 bracketrightbigg = u ( t ) − L − 1 bracketleftbigg exp( − s ) s + 1 bracketrightbigg − L − 1 bracketleftbigg exp( − s ) ( s + 1) 2 bracketrightbigg noting the Laplace transform of the step-function u ( t ) computed before. Now, noting 2 that L [exp( − t )] = 1 s +1 ] and from above L [ t exp( − t )] = 1 ( s +1) 2 , we have that exp( − s ) s + 1 = exp( − s ) integraldisplay ∞ exp( − st ) exp( − t ) dt = integraldisplay ∞ exp( − ( t + 1) s ) exp( − t ) dt = integraldisplay ∞ 1 exp( − vs ) exp(1 − v ) dv change of variables v = t + 1 = integraldisplay ∞ 1 (1) exp( − vs ) exp(1 − v ) dv + integraldisplay 1 (0) exp( − vs ) exp(1...
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pset_3sol - ACM 95/100b Problem Set 3 Solutions Faisal...

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