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# 374 Class Note for MATH 251 at PSU

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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 Pennsylvania State University taught by a professor in Fall. Since its upload, it has received 24 views.

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Date Created: 02/06/15

Differential Equations LECTURE 28 IVPS with Laplace Transforms Now that we have a good grasp of how to take Laplace transforms and inverse transforms let s return to differential equations First we should recall the following formula with H denoting the nth derivative of f E W s Fltsgt 7 s 1flt0gt7 st m 7 77 sf 20 7 f 10 We ll be dealing exclusively in this lecture with second order differential equations so in particular we ll need E 2 8Y8 7 240 and E 2 8238 7 SMO 7 MO You should be familiar with the general formula however REMARK Notice that we must have our initial conditions at t 0 to use Laplace transforms EXAMPLE 281 Solve the following VP using Laplace transforms y 7 52 7 6y 5t 240 1 20 2 We begin by transforming both sides of the equation E 34 7 SE 3 7 6E 5E 25 3213 7 sy0 7 we 7 53145 5340 7 6Ys 3275376Yss7275 As we ve already begun doing now we solve for Ys 5 77s YS3232753768275376 7 5 77s 3237 6s1 s76s1 5717732733 323 7 6s 1 We now have an expression for Ys which is the Laplace transform of the solution yt to the initial value problem We ve simpli ed it as much as we can now it s time to take the inverse transform The partial fraction decomposition is A B C D Y 7 7 7 S 332s76s1 Setting numerators equal gives us 6 732 7 33 Ass 7 6s 1 B3 7 6s 1 1 0323 1 1 D32s 7 6 Differential Equations Lecture 28 IVPs with Laplace Transforms We can nd constants by choosing key values of s 30 676B 371 36 2m0 0 571 1477D D7 W 1 31 1277mAgi A715 So 11 1 1 1 11 Y i77i 771 2 ms 326s76 2s1 1 la 1 t yt 7t6e 755 D EXERCISE Solve the initial value problem in the previous example using Undetermined Coef cients Do you get the same thing Which method took less work EXAMPLE 282 Solve the following initial value problem y 1 2371 5y cost710t y0 0 yO 1 We begin by transforming the entire equation and solving for Ys E 34 2E 3 SE cost710 t 10 8 7 8210 7 we we 7 210 ms 7 7 g 2 i s 10 s 2s5Ys717 8217 So we have s 10 1 Y 7 S 321822s5 32322s5 32717237175 Y1S Y2ltSgt Y3S Now we ll have to take inverse transforms This will require doing partial fractions on the rst two pieces Let s start with the rst one s A8 B Cs 1 D Y 18 321822s5 327171 322s5 After putting everything over a common denominator7 we set numerators equal 3 As32 1 23 5 B82 1 23 5 1 0332 1 Ds2 1 4AQEQABDFHEMJBCFGBD This gives us the following system of equations7 which we solve AC0 2ABD0 1 1 1 1 5Aw01 AE BE 073 D 5BD0 Differential Equations Lecture 28 IVPs with Laplace Transforms Thus our rst term becomes 1 3 1 1 1 3 1 1 Y 7 i if if 18 532110321 532235 232235 We ll hold off on taking the inverse transform for the time being Now7 let s deal with Y23 10 A B 08D Y 7 7 i 28 3232235 3 32 32235 We put everything over a common denominator and set numerators equal 710 14332 1 23 5 B32 1 23 5 C33 1 D32 A 133 2A B Ds2 5A 2B3 513 This gives the following system of equations AC0 2ABD0 4 4 2 5A2BO A E 3 72 0 73 D E 5B710 Thuswehave 41 2 4 3 2 1 Y 777777 7 28 53 32 532235532235 Let s return to our original function Y8 Y18 358 Y38 ili lig 717 2 7121 75321 10321 53 32 5 5 32235 2 5 32235 1 3 1 1 41 2 3 9 1 5321E32153 32 3124E3124 Now we have to adjust the last two terms to make them suitable for the inverse transform Namely7 we need to have 3 1 in the numerator of the second to last7 and 2 in the numerator of the last 1 3 1 1 41 2 3171 9 1 gminmigg s1w4in Hm 1 3 1 1 41 2 31 19 1 3mmm3 s124m mwl 1 3 1 1 41 2 31 19 2 5321E32153 32 31243124 So our solution is 7 1 1 i 4 it 19 t i yt 7 g cost E s1nt g 7 2t 7 e cos2t 1 Es s1n2t D We could have done both of the preceding examples using Undetermined Coefficients ln fact7 it would have been a lot less work Let s do some involving step functions7 which is where Laplace transforms really shine Differential Equations Lecture 28 IVPs with Laplace Transforms EXAMPLE 283 Solve the following initial value problem 24 7 52 6y 2 7 U2t62t 4 240 0 20 0 As before we begin by transforming everything Before we do that however we need to write the coefficient function of u2t as a function evaluated at t 7 2 gm 7 52 77 6y 2 7 772t62t72 Now we can transform a 7maiwuw7Muritht 2 32Ys 7 sy0 7 yO 7 53Ys 5y0 7 6Ys g 7 eizs 52 32 7 53 6Ys g 7 57298 i 2 So we end up with 7 2 7 625 1 Y 8373s72 373x3722 Y1s eist s Since one of these terms has an exponential we ll need to deal with them separately I ll leave it to you to check all of the partial fractions 11 2 1 1 Y 33573 372 1 1 1 Y2Sis73 372 s722 Thus we have 7 11 2 1 1 729 1 1 1 YltSgti2g 33737372e 7373s72s722gt and Mt g 7 53 7 52 7 1205 751072 7 52t72 7 t7 252t72gt 1 2 5 7 363 7 62 1205 7634 7 5274 2552194 once we observe that E 1 teat D EXAMPLE 284 Solve the following initial value problem y 4y8tu4t y0 0 yO 0 We need to rst write the coefficient function of u4t in the form ht 7 4 for some function ht So we write ht 7 4 t t7 4 4 and conclude ht t 4 So our equation is y 4y 8 t 7 4 4U4t Now we want to Laplace transform everything 319quot M y 7 SE 1 E t 7 4 4U4t 8 32Ys 7 sy0 7 yO 4Ys g 5 4SE t 4 8 1 4 2 4 Y 7 74s 7 7 ltsgtlta Se 9 8 Differential Equations Lecture 28 IVPs with Laplace Transforms So we have 8 1 4 Y 74s 8 3324 6 32327174 3324gt 7 8 749 1 43 7 3324 6 32324 where we ve consolidated the two fractions being multiplied by the exponential to reduce the number of partial fraction decompositions we need to compute After doing partial fractions leaving the details for you to check7 we have 7 ms emu23 2 23 Y1587824 and 1 11 s 1 1 Y 7 777 77 28 3432 327174 43271747 so 72 23 749 1 s 1 1 YltSgtis 324e 3 327174 4327174 2 s 1 s 1 2 772 745 7 iii if s 824e 3 3 327174 8327174 and the solution is yt 7 2 7 2 cos2t u4t lt1 7 g 7 4 7 cos2t 7 4 7 sin2t 7 4 2 7 2cos2t u4t it 7 cos2t 7 8 7 sin t 7 8

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