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(MATH150)2000_and_2001_spring_midterm_exam_solution.pdf
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Math150,MidtermExam,Solutions,Spring2000

1.SolvethefollowingBernoulliequation(20pts)
1

y0;y.2y 2tant2.t.0:
t
Solution.Rewritetheequationas
y
0
1
y2;ty .2tant2:
Changevariable 1
u.:
y
Then
0
u.;y0 :
y2
Sotheequationbecomes 11
;u0;u.2tant2.)u0+u.;2tant2:
tt
Moreover,
R
1
choose
dt
et .t:
Hence R ;2ttant2dt+Clnjcost2j+C
u..:
tt
Therefore,the.nalsolutionis 1t
y..or0:
ulnjcost2j+C
2.Sketchthesolutionsofthefollowingequationandindicatestableandunstable equilibriumsolutions(qualitativeanalysis)(20pts):
dy.;(y;1)(y;2)2(y;3):
dt
Solutions.Let
f(y).;(y;1)(y;2)2(y;3):

Thenthezerosoff(y)are y.1.2.3: (1pt)
Thus y.1.y.2andy.3areequilibriumsolutions:(1pt) Moreover,
f0(y).;(y;2)[(y;2)(y;3)+2(y;1)(y;3)+(y;1)(y;2)]
.;(y;2)[y 2;5y+6+2y 2;8y+6+y 2;3y+2]
.;2(y;2)(2y 2;8y+7)
!.!
.
pp4;24+2
.;4(y;2)y;y;:
22
Thezerosoff0(y)are
pp4;24+2
y..2.:
22
Partition[0.1)by

pp4;24+2
y.1..2..3:
22 Note
00
y.f(y)f0(y):
Wehavethat
000
on[0.1).y.0.y.0.
.!
p4;2 000
on1..y.0.y.0.
2
.!
p4;2 000
on.2.y.0.y.0.
2
.!
p
4+2
000
on2..y.0.y.0.
2
.!
p
4+2
000
on.3.y.0.y.0.
2
000
on(3.1).y.0.y.0:
Thuswehavethefollowingsolutioncurves.Thesolutiony.1isanunstableequi-libriumsolution.Thesolutiony.2isansemistable(orunstable)equilibriumsolution. Thesolutiony.3isanstableequilibriumsolution.
3.Solvethefollowingproblembythemethodofexactequationsandintegrating factors(20pts):
ydx+(2xy 2+2y 3+4y+x)dy.0.y(0).2:
Solution.Note
M.y.N.2xy 2+2y 3+4y+x:
My.1.Nx.2y 2+1:
2
Thus
Nx;My
.2yispurelyafunctionofy:
M
Sowecanlookforanintegeratingfactoroftheform ...(y): Solving
.0Nx;My
..2y.
.M
weget
y2
..e: Hencetheequation
y2y2
yedx+(2xy 2+2y 3+4y+x)edy.0 isexact. Let Z .(x.y).yey2dx+f(y).xyey2+f(y):
Wehave
y2y2
.y.x(1+2y 2)e+f0(y).(2xy 2+2y 3+4y+x)e
y2
.)f0(y).2y(2+y 2)e: Thus ZZ
y2y2choosey2
f(y).2y(2+y 2)edy.(2+y 2)edy2.(y 2+1)e: So
y2
..(xy+y 2+1)e: Thegeneralsolutionis
y2
(xy+y 2+1)e.C: Theconditiony(0).2implies
4
5e.C: The.nalsolutionis
y2
(xy+y 2+1)e.5e 4: 4.Findthesolutionofthefollowinginitialvalueproblem(20pts): dyx2+3y2
..y(1).;2:
dx2xy
Solution.Note

..
x2+3y2 13yy
f(x.y)..+.F.
2y
2xy2xx
x
where
13u
F(u).+:
2u2 3
Changevariable
y
u.:
x
Then
00
y.(xu)0.u+xu: Thustheequationbecomes 13u
0
u+xu.F(u).+:
2u2 Hence
2
1u1+u
0
xu.+.:
2u22u So ZZ
2ududx
.:
2
1+ux .)1+u 2.Cx: Substituteu.y.x:
2 p
y
3
1+.Cx.)y 2.Cx 3;x 2.)y..Cx;x2:
2
xTheconditiony(1).;2impliesthatweshouldchoosenegativesignand p;2.;C;1.)C.5:
The.nalsolutionis
py.;5x3;x2:
5.Findthesolutionofthefollowinginitialvalueproblem(20pts):
000
y;4y;5y.0.y(0).1.y0(0).1: Solution.Thecharacteristicequationis r 2;4r;5.0.)(r;5)(r+1).0: Therootsare y.;1.5: Thegeneralsolutionoftheequationis
;t5t
y.c1e+c2e:
Note

0;t5t
y.;c1e+5c2e:
Theconditiony(0).1implie c1+c2.1:
Theconditiony0(0).1implie
;c1+ 5 c2.