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by: Adam Crona


Adam Crona
GPA 3.73


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This 6 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 3D at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 48 views. For similar materials see /class/201875/math-3d-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

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Date Created: 09/12/15
Math 3D Additional Judicious Guessing Examples Paul Macklin February 4 2005 The method ofjudicious guessing can be tricky to get the hang of but it is rewarding It can save you a lot of work when compared to variation of parameters even in the longer examplesl Here I hope to help clarify some additional cases of judicious guessing h l7m going to work several of the odd exercises in your text page 164 as well as problem 10 the stumper problem77i Many thanks to those who contributed ideas on how to solve that problem Exercises Find a particular solution to each of the following equations 1 y 3ytgill Solution Because the righthand side is a polynomial we expect 1 to be of the form of a polynomial Next notice that the solutions to the homogeneous problem are cost and sint and so neither is nor 1 is a solution to the homogeneous DE So we only expect to need a 3rd degree polynomial 1W abtct2dt3 t b 3a 3d2 t 36 6dt and so 1b Sc 6dt 311 3a Sbt 3M 3 t31t20t0171 t33d t230 t3b 6d135 2a By equating coef cients of these polynomials we get 1 2 l d c0b7 a7 andso 1 2 1 7777 73 wt 3 3t3tl 3 y 7 y tZEL Solution Because the righthand side is a polynomial times an exponential we expect a particular solution of the form 1W W 1bt v e vet 1b t UHEE 216 vet where v is a polynomial of degree at least 2 Letls plug this in to get a better feeling for what degree 1 needs to e 1 611 21 v 11 511 e ltt21t010gt 61 27 1 Notice that if the degree of v is m then the degree of v 21 is m 7 11 Because we need this to be a seconddegree polynomial we need m 7 1 2 ie m 31 Therefore 1 vte e a bt ct2 dt3i If we plug in this re ned guess into 1 we get ett2 e 25 6dt 212 4ct 642 e ltt26d t6d 4c12c 2b By equating coef cients we have and a is undeterminedi Therefore we get a whole family of solutions 1 1 1 7t 7 7t 713 1 6 a 4 4 6 We might as well pick the simplest solution and set a 01 1 1 1 1 1 1 it 7 72 73 7 t 7 7 72 w 6 lt41 4t 6 16 lt4 4t 6 y 2y yequot Solution Because the righthand side is an exponential we might expect a particular solution of the form ce h However we see that both 6quot and t6quot are solutions to the homogeneous equationi So instead we expect a particular solution of the form 1111 111 MO 16quot 7 v6quot t ve7t 7 2167 v67 where v is a polynomial Let7s plug this in to get a better feeling for what degree 1 needs to be 1 6711 7 21 0 211 equot2v 7 21 j 6711 6711 6711 Notice that if the degree of v is m then the degree of v is m 7 21 Because we need this to be a constant polynomial we need m 7 2 0 ie m 21 Therefore 1 vtequot equota bt ct2i Notice that since 6quot and t6quot are solutions to the homogeneous equation we have aLequot bLtequot cLt2equot 0 0 cLt2equoti Thus our choices for a and b are irrelevant as they cannot contribute towards a nonzero righthand side Thus we can set a b 0 right away If this is the case and if we plug in this re ned guess we get equot1 e lt20 By equating coef cients we have and so 1 7t2eiti T 2 Alternatively notice that in our work above 67 e7tvn Therefore 1 v 1 vtcl 1 t2clt02 where cl and Cg are arbitrary We set cl Cg 0 to obtain 1 1 t2 7ti y 4y tsin2t Solution For the moment let us work on the related problem y 4y temp Then the righthand side is an exponential times a polynomial so we expect 11 of the form 11 vte2quot IJ 1621 2ive2n 1 vHeZzt 42306221 7 4062117 where v is a polynomial Now if we plug this guess into the modi ed DE we have IJH 62211 4230 7 41 411 2it4v 2itt 2itv4Zv Now if v is a polynomial of degree m then 1 421 is of degree m 7 11 Therefore as the polynomial of the RHS is of degree 0 m 7 1 1 ie m 21 Therefore 11 is of the form 11 a bt Ct2 2it and if we plug this re ned guess into our DE we have emu e2i ltt85i 120 4b and so 7 1 b 7 1 c 7 82 7 161 Since there were no conditions on a it is a free parameter So we set a 01 Thus 7 1 2it 11 716t1722te Next we nd that 7 1 2it 11 7 Et1722te 1 t coset i sin2t 7 Zit cos2t i sin2tgt 1 Et coset z s1n2t 7227 cos2t 2t s1n2tgt tltcos2t 2t sin2t i72t cos2t sin2t Now if Ly y 4y is the linear operator for the LHS then LG 762 tcos2t it sin2ti Therefore by equating imaginary and real parts and by using rules of linear operators 1 2 L lttcos 2t 2t s1n2tgtgt 7 tcos2t and L ltlttsin2t 7 2t2cos2tgtgt tsin2t Because our original RHS was tsin2t we choose 1 2 t 7 E t s1n2t 7 2t cos2tgti iy 72y5y2cos2t Solution First recall the halfangle identity 1 cos 2t 2 i cos2t Therefore the DE is equivalent to Ly y 7 2y By 1 cos2ti Let us consider the related DE I Ly 1 6213 Notice rst that if y then y 72y5y 511 So if we can nd a 11 such that I L 11 e2quot then 1 L ltgiigt 1e2quot1 Now we assume that 11 is of the form 11 vte2quot IJ 1621 2ive2n 1 vHeZzt 42306221 7 4062117 where v is a polynomial Now if we plug this guess into the modi ed DE we have IJH 62211 4230 7 41 72 11 e2it72v 7 421 511 e2i 5v 2i 1 e2itv 14239 7 2 v17 Now we see that if v is a m degree polynomial then 1 14239 7 2 v1 7 4239 is also m h degreei We need this to be a zeroth degree polyonmial ie m 0 and so 1 is of the form va whereby 1 1 4i 1 4i 00a174271 a7 174i 7116 7 717 Thus 14i 2 11 z 17 e 1 cos2t 7 4s1n2tgt zlts1n2t 4cos2tgti Thus L 11 1 62 1 cos2t isin2ti Therefore by equating real parts and using linear operator theory 1 L re 1 cos2t and so 1W g 117 cos2t 7 4sin2tgt does the trick iiei 1 cos2t 2 cos22ti y 7 2y By 2cos2 te Solution First notice that by the halfangle identity 1 2t 2 cos2 te 2 e e e cos2ti So let us study two related problems My y 7 2y 5y 6 7 2 and 3 Lltygt y 7 2y 5y 616211 e12it 6047 where a 1 2239 Suppose we have 1 that satis es 2 and 2 that satis es Then L11 2 L121 L12 e em etlt1 21quot etlt1 cos2t isin2tgti If so then the real part of 11112 satis es the real part of the righthand side namely e lt1cos2tgt 2 cos2 te i Let us nd such a 11 Assume that 1 is of the form 1 611005 1 e p e p 1 6 10 e 2p 6 10 where p is a polynomial of degree mi Then 1 6 10 210 0 721M 6 7210 7 20 5w1 e 5p 6 0 6 40 Now for the equality to hold we must have 1 p 411 The left side of this equation is a polynomial of degree 0 and the right is a polynomial of degree mi Therefore m 0 and p is a constant Hence 1 17410710777 4 and so 1 z 111 15 A Now let s look for a 2 which we assume is of the form 2 emq 1 emq ea aq w eatq eatgaq 60411247 where q is a polynomial of degree n Then 3 6W4 2a4 4 72 72q 7 2amp4 5112 e 5q em1 emltqHZqa71qa2 72a5gti Before we proceed let s see if we can simplify this equation Notice that the coef cient for q is the character tic A 39 or the equation evaluated at 1 Hmmmmw 1 This suggests that we take a peek at the characteristic equation and nd its roots T272r507r1i221 Notice then that a 1 2239 is a root of the characteristic equation and so a2 7 2a 5 0 Thus the equation above simpli es to 60411 em gm 42 since a 71 22quot Thus 1 q 42111 The left side is a polynomial of degree 0 and the right side is a polynomial of degree n 7 1 Therefore n 1 so qt a bt Hence 1 1 1 a ht 4ia bt 4239 a b I 712 2 Since a is undertermined we set a 0 Thus 71042711 2 1 12 2 7 712m 7 lte lt72cos2t 7 z s1n2tgt 7 lte s1n2t 7 lite cos2ti Therefore the real part of 1 2 is 7 1 z 1 z 7 1 z 1 7 16 lte s1n2t 7 16 lt1ts1n2tgti You should indeed check to see that this statis es the original differential equation as a particular solution 1 did in Maple It works 11 y y 7 6y sint tam Hint Consider the two problems I My equot and Ly temi Find a 1 that satis es the rst and a 2 that satis es the second Take the imaginary part of 1 and add it to 21


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