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Unformatted text preview: i 30. The homogeneous equation, y’ = —3y has solution yh(t) = e‘3’. We look for a particular solution in t, the form yp(t) = v(t)yh (t), where v is an unknown»
function. Since
y; = v’yh + vyi
= 1’th — 30y}:
= v'yh — By,” and y; = 3y,, + 4, we have 11’ = 4/yh : 4e3r,
Integrating we see that v(t) = 4e3t /3, and 4 4
M!) = v(t)yh(t) = 363t .e—ar = _3_' The general solution is 4
3'0) : Ypa) + £301“): 3‘ + Ce‘3’. 6~ dF ~_— (1/x + 2xy3)dx + (1/Y + 3x2y2)dy 10. With P = 1—— ysinx and Q = cosx, we see that 36. so the equation is exact. We solve by setting FOR”: [P(x,y)dx=/(1—ysinx)dx
=r+yeosx+¢(y). To ﬁnd qt, we differentiate Q06, y) = EJ55— = cosx +¢’(y). y Thus 45’ = 0, so we can take 4) = 0. Hence the
solution is F(x, y) = x + ycosx = C. x2 2
F(x,y) = —(1/2)1n<" “2) + arctan(y/x) —1nx = C 40. The homogeneous equation x’ = (2/ t2)x has solu
tion xh (t) = 6—2” . We look for a particular solution
of the form xp(t) = v(t)xh(t), where v is an un—
known function. Since x; = v’x;l + inc;t
= v'xh + 212xh/t2
= v'xh +2xp/t2,
and x; : 2xP/t2 +1/t2, we have 1/ : l/(tth) =
e2/’/t2. Integrating we ﬁnd that v(t) : —62/‘/2,
and xp (t) = —1/2. The general solution is x(t) = xp(t) + th(t) = «1/2 + Ce“2/‘. Since 36(1) 2 0,
we ﬁnd that C = (22/2, and the solution is 1 _
xv) = §(1+em W). 12. With P = x and Q _ y
x2 + _ “7* , WC 00m
pute y W
a P —2xy a Q 3y_m72”a? so the equation is exact. To ﬁnd the solution we
integrate F(x7y) = / P(x,y)dx x
‘/
=‘m+¢(y). To ﬁnd (I), we differentiate Q(x,y)=3l: y a *f— = ¢’()')~ y w/x +y2 Thus (15’ = 0, so we can take (15 = 0. Hence the
solution is F(x, y) = t/xz + 2 = C. 40. y(x) : x ln(C + 21m x) meﬁﬁ‘lu 4 ea waNimywunwinwn “WNW w u «mmpmymmmmm‘ «mm :mm cmwumu it . ., i _ t ., 2,9. 18. (i) In this case, f(y) = 6 + y —— )22 factors as f (Y) = (2 + y)(3 — y), whose graph is shown in the
next ﬁgure. f (y)=6+yy2 (ii) The phase line is easily captured from the previ— ous ﬁgure, and is shown in the next ﬁgure. (iii) The phase line in the second ﬁgure indicates that solutions decrease if y < —2, increase for
——2 < y < 3, and decrease if y > 3. This allows us
m MS; 1y angtrnrt the phase portrait shown in the ty plane in the next ﬁgure. Note the unstable equilib—
rium solution; ytt) = ——2e and the stable equilibrium solution, y(t) = 3. 24. Writing the equation as y’ = 5 — 2y the right hand side is f (y) = 5 —— 2)/.“'/I?h::ee 2h:
$61213; 1n the next ﬁgure. We have also merit;
h arm of the‘solutions on the yaxis, which
Is bows that. y = 5/21s an asymptotically stable equi
1 rium pornt. Thus any solution curve will approach y = 5/2 as 1 increases. f()’)=5—2y (5/2, 0) The exact solutio
iS separable
that it is = in :12er In AhwuﬁilWﬁWmﬁmmﬂ ( dri‘can be found since the eauatio s
an inear . ' l “W
) With some work we ﬁnd J“ ~ e: Una, Clearly the solution has the indicated limiting behavior. 26. Separating variables, versus y. dy a = (3 H)“ “W f(y)=(3+y)(1—y) ,t
d (3 + y)(1 — y) E
A partial fraction decomposition allows us to con— tinue. 1 1 1 ~3 0) (1 0)
_ _____.._ ____ d —_—_ ( w 1
4[3+y+1—y] y y
lnl3+yl—1nI1—yl:4t+C
3
m +W=m+c
1—y
3
+ 3’ ___ eCe4r
1y
3 + y __ A 4!
1 , y ‘“ e Note the equilibrium points at y = —3 and y = 1.
Moreover, note that between —3 and 2, solutions in—
with y(0) = 2, crease to the stable point at y = 1. Thus,
lim t = 1.
§i3=AﬂmeA=~5 Hm“)
I — 2
and
2:): _ _5e4t
1y t yw+5
y:5_e—4r O 5
lim =—+~:1
tfoo 5—0 Using qualitative analysis, plot the graph of the right—
hand side of ...
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 '08
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 Differential Equations, Linear Algebra, Algebra, Equations, Equilibrium point, homogeneous equation, Stability theory

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