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# 681 Review Sheet for MATH M0050 at Purdue

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This 5 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 94 views.

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Date Created: 02/06/15

MA 266 Fall 00 REVIEW 6 LAPLACE TRANSFORMS You should be able to express functions that are de ned by different formulas 07 t lt c 1 t 2 c After the vector t has been de ned7 the MATLAB statement 6 gt 0 gives value zero for t g c and value one for t gt c The statement 6 gt c can then be used for the step function uct to graph functions that involve step functions EXAMPLE To obtain the graph of y uwt sint 7T U37rt sint 377 on different intervals in terms of the step functions uct 7 gvaph u grown 7 pl 7 Mammmap The jump caused by a step function Will appear as a nearly vertical line on your plot EXAMPLE To obtain the graph of y 1 u1t 2u2t U3t Gvaph uly1 M710 2U72 U73 2 You should be able to use the de nition of the Laplace transform as an im proper integral to evaluate Laplace transforms You should be able to use Table 621 in the text to evaluate Laplace transforms and inverse Laplace transforms 0 Formula 13 tells us that L uctft 0 e CSFs7 where Fs Lft To use Formula 13 to evaluate L uctgt we rst note that the factor uct corresponds to the factor 6 05 in the Laplace transform We then note that we need to evaluate the Laplace transform of f 7 where ft c gt We can substitute t c for t in the formula ft c gt to obtain the formula for ft and then evaluate Fs L EXAMPLE Evaluate L u1tt2 SOLUTION We rst note that Formula 13 applies with c 1 Also7 the function f in Formula 13 must satisfy ft 1 7 Substituting t 1 for t in this formula7 we obtain ft t 12 t2 216 1 Table 621 then gives 2 2 1 2 is 2 2 1 Lft8 3S 2ESO U1LLLL 0 Formula 13 also says L 1e CSFs uctft c where ft L 1 To use Formula 13 to evaluate L 1 e CSFs7 we rst note that the factor 6 05 corresponds to the factor uct in the inverse Laplace transform We then use Table 621 to evalaute ft L 1 We then substitute t c for t in the formula for ft to obtain ft c EXAMPLE Evaluate L 1 e25 1 8 12 SOLUTION We rst note that Formula 13 applies with c 2 Table 621 1 gives ft L 1 D2 tet Substitution of t 2 for t in the formula 8 for ft gives ft 2 t 2et 27 so L 1e258112 u2tt 2et 2 PARTIAL FRACTIONS The partial fraction expansion of a rational function P Q consists of the sum of a polynomial and terms that have a power of a factor of Q in the denominators To obtain the partial fraction expansion of P Q 0 Check the degrees of the numerator and denominator If deg P 2 deg Q divide P by Q to obtain polynomials S and R with degR lt deg Q and Pa Sx C25 QW If degP lt deg Q then Sx 0Rx Px and it is not necessary to divide 0 Factor the denominator into powers of distinct linear terms and powers of irreducible quadratic terms This factorization determines the form of the expansion of R Q A theorem of algebra guarantees that any polynomial with real number coef cients can be factored into powers of linear terms and powers of irreducible quadratic terms The linear factors can be written in the form x r7 where r is a zero of the denominator The quadratic ax2 bx c is irreducible if it has no real zeros That is7 if b2 4ac lt 0 o If contains exactly m identical linear factors x r the expansion of RQ contains a sum of the form A1 A2 Am x r x r x rm where A1 A2 Am are unknown constants o If contains exactly n identical irreducible quadratic factors ax2 bx c the expansion of R Q contains a sum of the form B1 UCl B2 UCg 2 T 2 2Jrquot39Jr2 7 ax bxc ax bxc ax bxc where B1 C1Bg Cg Bn7 Cm are unknown constants Each distinct factor of Q contributes a term or terms of the type indicated to the expansion of R Q In each case7 the denominators are powers of the factors7 where the powers run from the rst power up to the power of the factor in Q Terms that have powers of linear factors in the denominator have numerators that are numbers Terms that have powers of irreducible quadratic factors in the denominator have numerators of the form numberx number 0 Complete the partial fraction expansion by solving for the values of the un known constants A theorem of algebra guarantees a unique solution for the values 4 MA 266 SPR 00 REVIEW 6 PRACTICE QUESTIONS 1 sweet 11 12 13 14 15 16 17 18 19 0 evaluate the Laplace transform of ft fwy 2 1gtltz y 3y2yQz D1y yW y5 wm1w m 0 L Otu1t T6Hcos27d7 10 L L2 L sin2t C0s2t L e t1 C0s3t L e2tt3 sint Lu14 2 a 0 tltL f l 0 tgz 7 17 y J 27 ylO 17 yH0 17 y Use the de nition of the Laplace transform as an improper integral to e t t2 2 7 0gtltz MA 266 SPR 00 REVIEW 6 PRACTICE QUESTION ANSWERS 2 2 1 s 83 2 s 2 824 824 1 81 37 s1 s129 6 1 4 8 24s 221 s 725 2 4 4 5 6 i3i2i 8 8 8 8 1 26 5 725 6 i2 8 7 82 3s 2 3 2 2 8 s 4 8 8 8 1 67571 s 9 s 1 82 4 637 6t 10 2 L3 11 6 1 12 GS 275 5 sin2t C 13 e2tcos2t sin2t t 267 15 736 t 2 t 22gt 16 u1t1 C0st 1 17 6t 1 6t 2 1 1 t 18 t T2T267d7 0 2 2 19 L ft f5 fte 5tdt ff e te stdt 6s1b 62s1 62s1 1 s1gtt 51120er dt 191310 s1 51 14

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