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# 40 Class Note for MA 26600 with Professor Yu at Purdue

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COURSE
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KARMA
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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 13 views.

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Date Created: 02/06/15
A Brief Review Chapter 6 The Laplace Transform The Laplace transform is very useful for solving linear differential equations with discontinuous or impulsive forcing terms A table of elementary Laplace transforms will be provided in the final you needn7t memorize the formulae in that table however you should be familar with them and know how to use them Actually remember them are more helpful 1 Know how to rewrite a piecewise continuous function in terms of unit step function uct For a two step function 7 f1t 0lttlt1 m f Mt tgt1 where f1t f2t can by any given function We can rewrite it as 1 1u1f1tu1f2t f1tu1f2t f1t For a three step function f1t 0lttlt1 ft f2t1lttlt2 f3t7 tgt2 where f1t f2t f3t can be any given function We can rewrite function ft as fa 1 u1f1t ui 7 W f2t u2 f3t f1tgtu1f2tgt f1t u2fst 1029 Practice Problems Section 63 1 2 11 2 Be familar with partial fractions You need do partial fraction expansion to find the inverse trans form of a fraction function Practice Problems Section 62 2 4 8 3 Solve an initial value problem with piecewise continuous forcing term Section 64 Example 2 y4y9tv 1 ND 07 MO 07 where 0 0lttlt5 gt t755 5lttlt10 1 t 10 Step 1 calculate the Laplace transform of nonhomogeneous term gt first rewrite gt in terms of unit step function gt 171150u57u10t755u101 710 7 u tisi t 7 5 5 10 5 then the Laplace transform of gt is 7 fl l fl 755il710si7 i Gltsgt7 glttgti5 u5ltt 5 Samoa 10gt75e 82 5e 827 5 32 Step 2 Take the Laplace transform on the both side of the first equation of 1 using the formula y t 8238 i 8 90 MO we get 82 4 Y8 7 8 MW 7 y 0 CS plugging in the intial value7 we get 32 4Ys Cs solve it for Ys7 Cs 76755767105 7 l 5247fHQ9 whereHs7m Step 3 The solution is given by the inverse transform of Ysi Suppose ht is the inverse trans form of Hs7 then the solution is W 7 1YS 7 glus 7 5 7 u10tht7 10 2 Y8 7 To find ht7 we need we take the partial fraction expansion of Hs7 which is l l l l H3gt7m71lt7mgt ht iti sinet SO Other Practice Problems Section 64 6 7 Solve an initial value problem with impulsive forcing term Section 65 example 1 2y y 2y75t75 yltogto7 yltogtoi 3 Practice Problems Section 65 1 i The convolution integral Know the definition of convolution t t f9 ft7TyTdT7 fTyt7Tdn and the Laplace transform of f 6 g f y 7F8 CSL 1FSGS7 f 97 where F8 7 ft7 08 7 9t Practice Problems Section 66 4 9

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