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# Ordinary and Partial Differential Equations MATH 251

Penn State

GPA 3.88

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This 0 page Class Notes was uploaded by Elaina Osinski on Sunday November 1, 2015. The Class Notes belongs to MATH 251 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/233005/math-251-pennsylvania-state-university in Mathematics (M) at Pennsylvania State University.

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Date Created: 11/01/15

Differential Equations LECTURE 25 Inverse Laplace Transforms At the end of last class we saw that upon using Laplace transforms on an initial value problem we end up with an expression for Ys the Laplace transform of the desired solution Thus we need to know how to go backwards given a transformed function Ys how do we nd the original function yt7 This is a slightly more complicated process than taking transforms which was quite straightforward We refer to ft as the inverse Laplace transform of Fs and use the notation 1 05 3198 Our starting point is that the inverse Laplace transform is linear just like the original transform was Theorem 251 Given two Laplace transforms Fs and 13 c1aFs 1293 ac1Fs b 1Gs for any constants a I So we ll decompose our original transformed function into pieces inverse transform and then put everything back together This is where familiarity with the basic Laplace transforms and the table becomes handy The strategy is to try to identify the desired inverse transform by looking at the denominator In most cases this will tell us what the original function will have to be but occasionally we will have to look at the numerator to distinguish between two potential inverses eg the denominators for the transforms of sinat and cosat are the same but the numerators differ Then we know precisely how we have to write our function Fs so that it is the inverse transform of the function we ve identi ed as the inverse Sometimes this requires a little bit of algebra or arithmetic Let s look at some examples Example 251 Find the inverse transforms of the following 3 1 F3 g i a 7 This one is quite straightforward The denominator of the rst term indicates that this will be the Laplace transform of 1 Since E 1 we ll factor out the 3 before taking the inverse transform For the second term this is just the Laplace transform of eZt and there s nothing else to do with it The third term is also an exponential can and we ll need to factor out the 4 in the numerator before we inverse transform So we have 1Fs3 117 1 124L11373 E 87 ft 31 7 521 4 53 3 7 ezt 4531 Differential Equations Lecture 25 Inverse Laplace Transforms H 5 1 7 H GltSgtis3 2s 43339 The rst term is just the transform of e inverse transforming The second term looks like it ought to be an exponential but it s got a 25 instead of an s in the denominator and transforms of exponentials should just have s We can x this by factoring a 2 out of the denominator and then taking the inverse transform The third term has 53 as its denominator This indicates that it will be related to the transform of t2 The numerator isn t quite correct though since E t2 92211 So we would need the numerator to be 2 and right now it s 7 How do we x this We ll multiply by absorb the top 2 into the transform and keep the out front Let s start by rewriting the transform with these xes incorporated 3 multiplied by 5 which we ll factor out before 1 1 27 G 5777 if 8 3773 2372233 51 1 33 7 3773 2372 283 Now we can take the inverse transform 1 7 t5 7377 2t 752 9 5 26 2 2s 3 H 1 S 32 25 32 16 The denominator of the rst term 52 25 indicates that this should be the transform of either sin5t or cos5t The numerator is 23 though which tells us that once we factor out the 2 it will be the transform of cos5t The second term s denominator is 52 16 so it will be the transform of either sin4t or cos4t The numerator is a constant 3 so it will be the transform of sin4t The only problem is that the numerator of E sin4t should be 4 while here it is 3 We ll x this as in the previous example by multiplying by 3 We rewrite the transform 3 4 3 H 27 77 S 32 52 4 32 42 1 2 s 1 a 4 7 32 52 4 32 42 Then we take the inverse ht 2cos5t 2mm Let s do some now that require a little more work Example 252 Find the inverse Laplace transforms for each of the following 3s 7 7 1 F m Looking at the denominator we recognize that this will involve a sine or a cosine as it has the form 52 a2 It s not quite either though since it has both an s and a constant in the numerator while a cosine just wants the s and the sine just wants the constant Differential Equations Lecture 25 Inverse Laplace Transforms This is easy to account for however If we split up this fraction into the difference of two fractions then x them up as we did in the previous example we ll be able to take our inverse tranform FS3328176 7 3s 7 321673216 73 s 7 3216713216 3Z 7 8216 43216 Now each term is in the correct form and we can take the inverse transform ft 3cos4t 7 sin4t 1 7 33 G S 32 2s 10 If we look at our table of Laplace transforms we might see that there are no denomina tors that look like a full quadratic polynomial Also this polynomial doesn t factor nicely However there are terms in the table that have denominators of the form 3 7 12 1 2 those for 5 cosbt and eat sinbt We can put this denominator in that form if we complete the square Then we can x the numerators to gure out the inverse transform 522s10 322s17110 32 2s 1 9 s 12 9 Thus our transformed function can be written as 17 33 G S s 12 9 We won t split this up into two pieces yet First we ll x the s in the numerator to be 3 1 which we ll need for the numerator of 5 cos 325 We do this by adding and subtracting 1 to the s This will produce some other constant term which we ll combine with the already present constant and then we can worry about xing that numerator to correspond to the numerator of the transform of 5 sin 3t 7 1 7 38 1 7 1 T s 12 9 1 7 38 1 3 s 12 9 738 1 4 s 12 9 Now we can break our transform up into two pieces one of which will correspond to the cosine and the other to the sine At that point xing the numerators is the same as in the Cs Differential Equations Lecture 25 Inverse Laplace Transforms last couple of examples 3 s l 3 s129 s129 4 gt 7357 cos 3t 3572 sin 3t Cs7 3 s 2 111 H s 7 m This should appear similar at rst glance to the previous example but there s a difference this time we can factor the denominator This requires us to deal with the inverse transform differently Factoring we see 3 2 18 7 s 3s 7 4 We know that if we have a linear denominator that will correspond to an exponential In this case we have the product of two linear factors This by itself isn t the denominator of any particular Laplace transform but we know a method for turning certain rational functions with factored denominators into a sum of more simple rational functions with those factors in each denominator partial fractions We start by writing 1118 A i s3si4 33 374 We have to put our partial fraction decomposition over a single denominator 32 7Asi4Bs3 s3si4 7 s3si4 39 This needs to be true for any value of s in particular the numerators must match for every value of s 32Asi4Bs3 As a result we can choose values of s to plug in that will isolate an individual constant Let s do this for each 373 71 77A a 14 6 s 4 6 7B B Thus our transform can be written as 1 6 H L L S s 3 s 7 4 and taking the inverse transforms we get 1 6 ht 573 54 D Remark We could have done the last part of the previous example as we had the previous parts by completing the square However this would have left us with expressions involving the hyperbolic sine sinh and the hyperbolic cosine cosh These are interesting functions which can be written in terms of exponentials as we got in the form of our answer to the previous example but it will Differential Equations Lecture 25 Inverse Laplace Transforms Factor in Denominator Partial Fractions Term A b 7 0m A A am 1 b A b k 71 72 t t 39 7k Gal ambazb2 ambk 2 Ax B am 1 bm c 2 am lm 1 0 A196 Bi A296 2 AM Bk am2 1 bm ck am2bmc az2bmc2 nax2bmck TABLE 251 Translation from factored denominator to partial fractions be much easier for us to work with the exponentials so we re better offjust doing partial fractions even though it s slightly more work Partial fractions and completing the square are a part of life when it comes to Laplace trans forms Being comfortable with these techniques is especially important when we re working with initial value problems since most of our answers will involve some combinations of exponentials sines and cosines Let s quickly review partial fractions The rst step is to factor the denominator as much as you can Then using Table 251 we can nd each of the terms for our partial fractions decomposition This table isn t exhaustive but we ll only worry about having linear or quadratic factors Let s do some more examples that require partial fractions Example 253 Find the inverse transform of each of the following i 23 1 1 F s 87283871 The form of the decomposition will be A B C G 7 7 7 S 372s3371 since all of the factors in our denominator are linear Putting the right hand side over a common denominator and setting numerators equal we have 281AS3871B872871C87283 We can once again use the method from the previous example where we choose key values of s that will isolate the coefficients 32 5A51 1 1 s 73 75 B7574 71 3 31 30714 07Z Thus the partial fraction decomposition for this transform is 1 l i FS 7 7 i 7 i Differential Equations Lecture 25 Inverse Laplace Transforms ii iii 2 7 33 G S s 7 2 32 3 Now we have a quadratic in the denominator Looking at Table 2517 we see that the form of the partial fractions decomposition will be Gltsgt A s 7 2 If we put everything over a common denominator and setting the numerators equal7 we will get B3C 32339 2 73s 1432 3 7 35 Cs7 2 Notice tht we can t use the method from the previous examples there are only two key values77 for s but there are 3 constants7 so we would be stuck at some point Thus we need to use the more formal method7 which requires us to multiply out the right hand side and compare coefficients 2733A323Bs0s72 A323AB3272Bs03720 A Bs2 723 13 3A 7 20 We have the following system of equations to solve 32 AB0 1 8 8 5 8 721370773 7 A77E BE 077 so 3A72C2 Thus the partial fraction decomposition is g 8 g G 7 7 7 7 7 S s72323 32 5 i 7 7872 782 7 SZ3 and the inverse transform is 8 2t 8 5 i t 77 7 3t 7 7 3t g 75 7cosltfgt 7 sinltfgt 2 HS 83 The partial fraction decomposition in this case is A B C D H 7 7 7 S s 32 33 1 Setting numerators equal and multiplying out gives 2 A32s 71 333 71 Cs 71 D33 A337A32B327B3C3987C39D33 Differential Equations Lecture 25 Inverse Laplace Transforms and we have to solve the system of equations 33 A D 0 32 7A B 0 31 7B C 0 so 70 2 Thus our partial fractions decomposition becomes 2 2 2 2 HSgt 2 s 2 2 2 2 2 7 s 32 2I33 371 a 1472 B72 072 D2 and the inverse transform is ht 72 7 2t 7 t2 25

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