Solutions for Chapter 2
Section 2.1
(4) Second-order, linear, nonhomogeneous, variable coe¢ cients
(14) L(y) = y0 + ty
L (y1 + y2 ) = (y1 + y2 )0 + t (y1 + y2 )
= y10 + y20 + ty1 + ty2
= (y10 + ty1 ) + (y20 + ty2 )
= L (y1 ) + L (y2 )
L (cy) = (cy)0 + t (cy)
= cy 0 + cty
= cL (y)
(22)
(28) 2y 00 + y 0
y = 0. Sol:
1
(46)
Section 2.2
(2) y 0 + 2y = 3et : Sol:
sin t
3
(12) y 0 + y = 3 :Sol: We multiply each side of the equation by the
t
t
integrating factor
Z
3
= exp
dt = exp (3 ln jtj) = jtj3 = t3 for t > 0;
t
giving
t3 y
Integrating, we …nd t3 y =
0
= sin t:
cos t + c .
(14) (t2 + 9)y 0 + ty = 0:Sol: We rewrite the equation as
y0 +
t
= 0;
(t2 + 9)
and then multiply each side of the equation by the integrating factor
Z
Z
Z
t
t 1
11
dt
=
exp
du = exp
du
= exp
2
(t + 9)
u 2t
u2
u = t2 + 9
= exp
giving
1
ln juj
2
du = 2tdt
p
p
= juj = t2 + 9
Integrating, we …nd
p
h p
t2 + 9 y
i0
= 0:
t2 + 9 y = c;or y (t) = p
2
c
:
t2 + 9
(18) Sol: We …nd the integrating factor to be
Z
3
dt = t 3
= exp
t
(for t > 0):
0
Multiplying the DE by this, we get (t 3 y) = 1:hence, t 3 y = t + c; or
y = t4 + ct3 : Substituting y (1) = 4 gives c +1= 4 or c = 3. Hence, the
solution of the IVP is y = t4 + 3t3
(20) We solved this DE in Problem 13 and found
y=
c
:
1 + et
c
Substituting y (0) =1 gives = 1 or c = 2 . Hence, the solution of the
2
2
IVP is
:
1 + et
= et : Multiplying
(26) Sol: Here p(t ) =1 therefore the integrating factor is
each side of the equation by yields
0
yet
=
et
:
1 + et
Integrating (using substitution u = et ) gives yet = ln (1 + et ) + c:
(36) y 0 y = et y 2 ;so that y 2 y 0 y 1 = et ;or y 2 y 0 = y
v 0 = y 2 y 0 = y 1 et = v et :Or
v0 + v =
1
+et : Let v = y 1 ;so
et :
To solve this one for v;we see that the integrating factor is
et v
0
So
1 t
e + ce
2
v=
and
y=v
1
1 2t
e + c:
2
e2t ; or et v =
=
=
1
ce
t
3
et =2
=
t
2
2ce
t
et
:
= et :So
Section 2.4
(6) Sol: x (t) = amount of salt at time t: Then V (t) = 300 + 2t
x0 = 3
x
; x (0) = 0
300 + 2t
Rewrite DE as
x0 +
x
= 3:
300 + 2t
The integrating factor is
Z
1
= exp
dt = exp
300 + 2t
Thus
h
1
ln (300 + 2t)
2
=
p
(300 + 2t):
i0
(300 + 2t)1=2 x = 3 (300 + 2t)1=2 :
Integrating both sides leads to
Z
1
2
1=2
(300 + 2t) x = 3 (300 + 2t)1=2 = 3
(300 + 2t)3=2 + c
2
3
3=2
= (300 + 2t) + c
or
x = (300 + 2t) + c (300 + 2t)
1=2
:
p
Inserting p
the initial condition,
we see 0 = x (0) = 300 + c= 300; or
p
c = 300 300 = 3000 3:; and the solution is
p
x = (300 + 2t) 3000 3 (300 + 2t) 1=2 :
The tank will be full when V (t) = 300 + 2t = 600; or t = 150 min. At
that time,
p
x (150) = (300 + 300) 3000 3 (300 + 300) 1=2 387: 87 (lb)
(8) The model is
dx
=0
dt
0:03x; x (0) = 20:
The solution is
x = 20e
4
0:03t
:
Because she wants to reduce salt to 10 lbs, we set
10 = 20e
So
t=
100
ln 2
3
0:03t
:
23 (min)
(10) (a) x (t) = salt in tank A at time t: Then
x0 = 0
x
; x (0) = 50
50
x
2; or x0 =
100
(b) Solution of IVP is x = 50e
t=50
:
(c) Let y (t) = salt in tank B at time t: The input amount of salt into tank
B=
1
x
2=
50e t=50 = e t=50
100
50
The output is
y
y
2= :
100
50
So the IVP is
y
y 0 = e t=50
; y (0) = 0:
50
(d) The solution of the above IVP is
y (t) = te
t=50
Section 2.5
(14) The logistic DE is
y 0 = ry 1
y
; L=5
L
109 ; y0 = 0:2
109
If initially the population doubles every hour (assuming y = 0:2 109 ert
), we have
2y0 = y0 er
So
r = ln 2:
5
The solution is
L
y=
L
y0
1+
1 e
rt
and
y (4) = 4:9
109
(22) y = 0 is the only equilibrium solution. it is a node (semistable).
(24) y = 0 is stable. y = M is stable if M > L and is unstable if M < L:
y = L is stable if M < L and is unstable if M > L:
6
7