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(MATH150)Final_Exam_Review_Problems_solutions(Part1).pdf
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Lecture26
SolvethefollowingEulerequation:
2000
xy;3xy+5y.0.x.0: (26:13) Solution.Letz.lnx.Then
..
dy1dyd2y1d2ydy
...;: (26:14)dxxdzdx2 x2dz2 dz
Substitutingthemintotheequation,wehave
d2ydydyd2ydy
;;3+5y.0.;4+5y.0.(26:15)
dz2 dzdzdz2 dz
whosecharacteristicequationis:
r 2;4r+5.0: (26:16)
Therootsare:
r.2.i: (26:17)
Thegeneralsolutionis: y.e 2lnx(c1coslnx+c2sinlnx).x 2(c1coslnx+c2sinlnx):2(26:18)
Solvethefolloingproblemsbythemethodofundeterminedcoe.cients:
2.
000
y+y;2y.2t: (26:19) Solution.Thecorrespondinghomogeneousequationis
000
y+y;2y.0. (26:20) whosecharacteristicequationis r 2+r;2.0.)(r+2)(r;1).0.)r.1.;2:(26:21) Thusthegeneralsolutionoftheequation(26.2)is
x;2x
y.c1e+c2e: (26:22)
0
Lety0.At+Bbeaparticularsolutionoftheequation(26.19).Moreover,y0.Aand
00
y0.0.So 1
A;2(At+B).2t.);2A.2.A;2B.0.)A.;1.B.;:(26:23)
2
Hence
1
y0.;t;: (26:24)
2
Thusthe.nalsolutionis 1
x;2x
y.c1e+c2e;t;:2 (26:25)
2
3.
00
y+y.2cost.y(0).1.y0(0).3: (26:26) 3
00
Solution.Thegeneralsolutionofthecorrespondinghomogeneousequationy+y.0
is:
y.c1cost+c2sint: (26:27) Let
y0.Atcost+Btsint (26:28) beaparticularsolution.
0
y0.(A+Bt)cost+(B;At)sint. (26:29)
00
y0.(2B;At)cost;(2A+Bt)sint: (26:30) Substitutingthemintotheequation,weget 2Bcost;2Asint.2cost: (26:31)
A.0.B.1.y0.tsint: (26:32) Thusthegeneralsolutionis: y.c1cost+(c2+t)sint: (26:33)
0
y.(c2+t)cost+(1;c1)sint: (26:34) y(0).1.)c1.1. (26:35) y0(0).3.)c2.3: (26:36)
The.nalsolutionis:
y.cost+(3+t)sint:2 (26:37)
4.
00022t):
y;4y+4y.4(t+e (26:38) Solution.Thecorrespondinghomogeneousequationis
000
y;4y+4y.0. (26:39) Whosecharacteristicequationis: r 2;4r+4.0.)r.2isarepeatedroot: (26:40) Thusthegeneralsolutionis
y.(c1+c2t)e 2t: (26:41) Firstwewantto.ndaparticularsolutionoftheequation:
0002
y;4y+4y.4t: (26:42) Let
y0.At2+Bt+C (26:43) beaparticularsolution.Then
0 00
y0.2At+B.y0.2A: (26:44) Substitutethemintotheequation, 2A;4(2At+B)+4(At2+Bt+C).4t2.4At2+(4B;8A)t+2A;4B+4C.4t2:(26:45) 3
4A.4.4B;8A.0.2A;4B+4C.0.)A.1.B.2.C.
:(26:46)
So y0.t2+2t+3 : (26:47)
2
Nextwewantto.ndaparticularsolutionoftheequation:

0002t
y;4y+4y.4e: (26:48) Let y0.At2 e 2t (26:49) beaparticularsolution.Then
0 2t00 2t
y0.2A(t+t2)e.y0.2A(1+4t+2t2)e: (26:50) Substitutethemintotheequation,
2t2t2t2t2t
2A(1+4t+2t2)e;8A(t+t2)e+4At2 e.4e.)2Ae2t.4e:(26:51) SoA.2and y0.2t2 e 2t: (26:52) The.nalsolutionis y.(c1+c2t+2t2)e 2t+t2+2t+3 : (26:53)
2
5.
000;x
y+2y+5y.4ecos2x.y(0).1.y0(0).0:(26:54) Solution.Thecorrespondinghomogeneousequationis
000
y+2y+5y.0. (26:55) whosecharacteristicequationis r 2+2r+5.0.)(r+1)2+4.0.)r.;1.2i:(26:56) Sothegeneralsolutionoftheequation(26.55)is y.c1e;xcos2x+c2e;xsin2x: (26:57)



Let
y0.Axe;xcos2x+Bxe;xsin2x (26:58)
beaparticularsolutionoftheequation(25.54).Notethat
0;x;x
y0.(A;Ax+2Bx)ecos2x+(B;Bx;2Ax)esin2x.(26:59)
00 ;x;x
y0.(4B;2A;3Ax;4Bx)ecos2x+(;2B;4A;3Bx+4Ax)esin2x:(26:60) Substitutingtheabovethreeexpr