pset_3sol

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 Cauchys Integral Formula to compute the first inverse transform as L bracketleftbigg 1 ( s + 1) 2 bracketrightbigg = 1 2 i integraldisplay c + i c i exp( st ) 1 ( s + 1) 2 ds = 1 2 i 2 i 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...
View Full Document

This note was uploaded on 12/12/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

Page1 / 11

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online