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(MATH150)Final_Exam_Review_Problems_solutions(Part2).pdf
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Lecture27
SolutionsofReviewProblemII
Solvethefolloingproblemsbythemethodofundeterminedcoe.cients:
ThedatainProblem6arenoteasytocalculate.Wechangetheproblemas:
6.
. ;t
4eif0.t.1.
y;2y;3y. y(0).0.y0(0).3:(27:1)
0003t2+4t;2ift.1.
Solution.Thecorrespondinghomogeneousequationis

000
y;2y;3y.0. (27:2) whosecharacteristicequationis r 2;2r;3.0.(r;3)(r+1).0.)r.;1.3:(27:3) Thegeneralsolutionoftheequation(27.2)is
;t3t
y.c1e+c2e: (27:4) Ontheinterval[0.1],theproblembecomes
000;t
y;2y;3y.4e.y(0).0.y0(0).3:(27:5) Let y0.Ate;t (27:6) beaparticularsolutionoftheaboveequation.Wehave
0;t00;t
y.A(1;t)e.y.A(t;2)e:(27:7)
00
Substitutingthemintotheequation(27.5),wehave
;t;t;t
A(t;2)e;2A(1;t)e;3Ate;t.4e(27:8)
.);4Ae;t.4e;t.)A.;1.)y0.;te;t:(27:9) Thegeneralsolutionoftheequationin(27.5)is
y.(c1;t)e;t+c2e 3t:(27:10)
Moreover,
0;t3t
y.(t;c1;1)e+3c2e:(27:11)
Theinitialconditionsin(27.5)imply
. .
c1+c2.0c1.;1
.) (27:12)
;c1;1+3c2.3c2.1:
Sothesolutionis y.;(1+t)e;t+e 3t: (27:13)
Furthermore,
03t
y.te;t+3e: (27:14) ;13;13
y(1).;2e+e.y0(1).e+3e:(27:15) Ontheinterval[1.1),theequationin(27.1)becomes
y
00;2y0;3y.3t2+4t;2:(27:16)
Let y0.At2+Bt+C (27:17)
beaparticularsolution.Wehave
000
y0.2At+B.y0.2A:(27:18)
By(27.16), 2A;2(2At+B);3(At2+Bt+C).3t2+4t;2(27:19) .)A.;1.B.0.C.0.)y0.;t2:(27:20) Thegeneralsolutionof(27.16)is
;t3t2
y.c1e+c2e;t: (27:21)
Moreover,
0;t3t
y.;c1e+3c2e;2t:(27:22) By(27.15),wehave
;13;13;13;13
c1e+c2e;1.;2e+e.;c1e+3c2e;2.e+3e(27:23) 11
.)c1.(e;7).c2.(4+3e;3;e;4):(27:24)
44The.nalsolutionis
.
;(1+t)e;t+e3t ift.1.
y.122(27:25)
1 ;t;33t
(e;7)e+(4+3e;e;4)e;tift.1:
44
Solvethefollowingproblemsbythemethodofvariationofparameters
7.
9.
00
y+9y..0.t.:(27:26)
cos3t6
00
Solution.Thegeneralsolutionofthecorrespondinghomogeneousequationy+9y.0
is
y.c1sin3t+c2cos3t:(27:27) Let
y.u1(t)sin3t+u2(t)cos3t(27:28) beasolutionoftheequationin(27.26).Note
y
0
.

sin3t+
2
0u
cos3t+3u1cos3t;3u2sin3t(27:29)
1
0u
Weimposethecondition
Thenwehave
1
0u
sin3t+
2
0u
cos3t.0:(27:30)
.3u1cos3t;3u2sin3t:(27:31)
0
y
.;(3u
2
0
9u1)sin3t+(3
1
0u
;9u2)cos3t:(27:32)
+

y
00
Substitutingthemintotheequation(27.26),weget
9
;(3u
2
0
9u1)sin3t+(3
1
0u
;9u2)cos3t+9(u1sin3t+u2cos3t).(27:33)
cos3t
+

9

.)3;0u2
sin3t+3cos3t.:(27:34)
0u
1
cos3t
So . sin3t+cos3t.0.
01020u
2
u01
u
9 (27:35)cos3t;sin3t.
:

u
cos3t
.)

0u
1
.3

.

.;3tan3t:(27:36)
0u
2
.)u1.3t+c1. (27:37)
Z
3sin3t
u2.;dt.lncos3t+c2:(27:38)
cos3t
Thusthegeneralsolutionis y.(3t+c1)sin3t+(lncos3t+c2)cos3t:2 (27:39)
8.
t
e
y
00
;2y
0
+y.: (27:40)
1+t2
Solution.Thegeneralsolutionofthecorrespondinghomogeneousequation
y
00
is
;2y
0
+y.0 (27:41)
y.(c1+c2t)e t: (27:42)
We let
y.(u1(t) + u2(t)t)e t (27:43)
b e asolutionofthe originalequation(27.40). Then

tt
y
0
.(

1
0u
+

2t)e
0u
(u1+(1+t)u2)e
:

(27:44)

+

Weimpose
The