=========================preview======================
(MATH150)2001_spring_final_exam_solution.pdf
Back to MATH150 Login to download
======================================================
Math150,FinalExam,Solutions,Spring2001

1.SolvethefollowingEuler'sequation(8pts)
200
ty+5ty0+13y.0.t.0:
Solution.Let
z.lnt:

Then . dydydz1dyd2y1d2ydy
....;: dtdzdttdzdt2 t2dz2dz
Theequationbecomes
d2ydydy d2ydy;+5+13y.0.)+4+13y.0: dz2dzdz dz2 dz
Thecharcteristicequationis r 2+4r+13.0.)(r+2)2.;9.)r.;2.3i. Thegeneralsolutionis y.e;2z(c1cos3z+c2sin2z): Since
1
;2z;2lnt
e.e..
t2
wehavethe.nalsolution c1cos3lnt+c2sin2lnt
y. : t2
2.Findthesolutionofthefollowinginitialvalueproblem(8pts)
000
y+2y+5y..(t;2)+5u2(t).y(0).1.y0(0).;1: Solution.Lety(t)bethesolution.Set Y(s).Lfy(t)g: ApplyLaplacetransformtotheoriginalequation:
e;2s
s 2Y(s);s+1+2(sY(s);1)+5Y(s).e;2s+5: s
5e;2s 2;2s
.)(s+2s+5)Y(s).s+1+e+: s
e;2s 5e;2
s+1 s
Y(s).++

s2+2s+5s2+2s+5s(s2+2s+5)
.
e;2s
s+1 1s+2
;2s
.++e;
2222 2
(s+1)+2(s+1)+2ss+2s+5
.
e;2s
s+1 1s+11
;2s
.++e;;:
2222 2222
(s+1)+2(s+1)+2s(s+1)+2(s+1)+2
;t2;t
y(t).ecos2t+1 u2(t)esin2(t;2)+u2(t)(1;e 2;t(cos2t+(1.2)sin2t)):
2 3.Solvethefollowinginitialvalueproblem(Hint:youmayusetheconvolutioninte-gral.8pts)
(4);t4000
y;81y.e.y(0).y00(0).y(0).0.y0(0).6:
Solution.Lety(t)bethesolution.Set
;t4
Y(s).Lfy(t)g.G(s).Lfeg:
ApplyLaplacetransformtotheoriginalequation:
4Y(s);6S242
s;81Y(s).G(s).)(s;81)Y(s).6s+G(s)
6s2 G(s)
Y(s).+
s4;81s4;81
6s2 G(s)
.+
(s2;9)(s2+9)(s2;9)(s2+9)
.
33G(s)11
.++;:
s2;9s2+918s2;9s2+9
Therefore,the.nalsolutionis
1 Z t ;4
y(t).sinh3t+sin3t+e(sinh3(t;.);sin3(t;.))d.:
54
0
4.Findthegeneralsolutionofthesystem(8pts)
.
3;4
0
~x.~x:
1;1
Solution.Let.
3;4

A.:
1;1
Thenthecharacteristicpolynomialis
.

3;.;4
jA;.Ij..(3;.)(;1;.)+4..2;2.+1.(.;1)2:

1;1;.
Sowehavearepeatedeigenvalue1.
Since
.
2;4
A;I..
1;2
~
thesystem(A;I).~.0isequivalentto
.1;2.2.0:
Taking.2.1,wehave.1.2.Sowegetaneigenvector
.
2
~
..:
1
Hence . 2t
e
1
isasolution. To.ndsecondlinearlyindependentsolution,wetry
~t
~x..te+.~e t:
Thenwehave
.
2;4.1 2
~
(A;I)~.......1;2.2.1:
1;2.2 1
Taking.2.0,weget.1.1.Sothesecondsolutionis
.
~tt2t+1t
.te+.~e.e:
t
Thusthegeneralsolutionis
.
2t2t+1t2(c2t+c1)+c2 t~x.c1 e+c2 e.e: 1tc2t+c1
5.Findthesolutionofthefollowinginitialvalueproblem(8pts)
.
01;13
~x.~x.~x(0). :
5;32
Solution.Let.
1;1

A.:
5;3
Thenthecharacteristicpolynomialis
.

1;.;1 2
. .2
jA;.Ij..(1;.)(;3;.)+5.+2.+2.(.+1)+1:

5;3;.
Sowehavecomplexeigenvalue;1.i.
Since
.
2;i;1
A;(;1+i)I..
5;2;i
~
thesystem(A;(;1+i)I).~.0isequivalentto
(2;i).1;.2.0:
Taking.1.1,wehave.1.2;i.Sowegetaneigenvector
.
110
.+i:
2;i2;1
Thusthegeneralsolutionis
.
1010
;t
~x.c1 cost+sint+c2 sint;coste:
2121
Theinitialcondition
.
3103c1 3
~x(0). .)c1 ;c2 ...
22122c1;c2 2
.)c1.3.2c1;c2.2.)c1.3.c2.4:
Therefore,thesolutionis
. .
1010
;t
~x.3cost+sint+4sint;coste:
2121
6.Findthegeneralsolutionofthesystem(10pts)
.
11e;2t 0~x.~