Rekha Santhanam MA 108 Ordinary Differential Equations Recap Lecture 17 Ex Find

# Rekha santhanam ma 108 ordinary differential

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Rekha Santhanam MA-108 Ordinary Differential Equations
Recap Lecture 17 Ex. Find inverse Laplace transform of F ( s ) = e - s 1 2 s - e - 2 s s + 1 ( s + 1) 2 + 1 . L - 1 1 2 s = 1 2 , L - 1 s + 1 ( s + 1) 2 + 1 = e - t sin t Hence L - 1 ( F ( s )) = 1 2 u ( t - 1) - u ( t - 2) e - ( t - 2) sin( t - 2) = 0 , 0 t < 1 1 2 , 1 t < 2 - e - ( t - 2) sin( t - 2) + 1 2 , t 2 Rekha Santhanam MA-108 Ordinary Differential Equations
Recap Lecture 17 1 14D170018 SARTHAK MITTAL 2 140110005 HIMANI PRANAV MEHTA 3 140110011 DESHMUKH TANMAY HEMANT 4 140010012 ANSHUMAN ACHARYA 5 140110079 SARAGDAM VINAY KUMAR 6 14D170021 ABHINEET SINGH KALER 7 140110071 HIMMAT SINGH RAJPUT 8 140010031 AAKRITI VARSHNEY 9 140110047 DESHPANDE SHACHI SHAILESH 10 140110063 ANUPAM SIDHANT 11 140010018 ANMOL YADAV 12 14D110013 AKHILESH VYAS 13 140110053 SANAT SAMEER MALHOTRA 14 140110033 AJINKYA GORAD 15 14D170027 KETHAVATH MONISH CHANDRA NAIK 16 140110028 SHARMA ANKUR 17 140110008 ARAK PRASAD RAMSWARUP Rekha Santhanam MA-108 Ordinary Differential Equations
Recap Lecture 17 IVP with peicewise continuous forcing functions Ex. Consider the differential equation of the form y 00 + 3 y 0 + 2 y = e t , 0 < t 2 e - t , 2 < t , y (0) = 1 , y 0 (0) = - 1 . From what we know, this IVP has a unique solution in the interval (0 , 2) and if the IVP was defined on t 0 (2 , ) then we would have a unique solution on (2 , ) . But its still possible to get a solution which is continuous on [0 , ) . Let y 1 be the unique solution to the given IVP on [0 , 2) . Then evaluate y 1 (2) and y 0 1 (2) . Define a new IVP as y 00 + 3 y 0 + 2 y = e - t , y (2) = y 1 (2) , y 0 (2) = y 0 1 (2) . This has a unique solution y 2 on [2 , ) . Rekha Santhanam MA-108 Ordinary Differential Equations
Recap Lecture 17 This gives us a solution y ( t ) of original IVP on (0 , ) as y ( t ) = y 1 ( t ) , 0 t < 2 y 2 ( t ) , 2 t < such that y, y 0 are continuous on [0 , ) . Note that, since r ( t ) is discontinuous at t = 2 , ( r (2+) = e - 2 and r (2 - ) = e 2 ;) though, y and y 0 are continuous on (0 , ) , y 00 is not defined at t = 2 .

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