```520
CHAPTER TEN /SOLUTIONS
Solutions for Section 10.4
dP = 0:02P implies that dP = 0:02 dt:
dt
P
R dP
R
P = 0:02 dt implies that ln jP j = 0:02t + C:
jP j = e0:02t+C implies that P = Ae0:02t ; where (A0:02=)0eC :
0:02t
We are given P (0) = 20. Therefore, P (0) = Ae
= A = 20. So the solution is P = 20e
:
dQ
Q
dQ
dt
2. dt = 5 implies that Q = 5 :
R dQ
R dt
1
Q = 5 implies that ln jQj = 5 t + C:
1
1
1
So jQj = e 5 t+C = e 5 teC implies that Q = Ae 5 t ; where A = eC : From the initial conditions we know that
( 15 )0
Q(0) = 50, so Q(0) = Ae = A = 50. Thus Q = 50e 51 t :
3. dm
dt = 3m: As in problems 1 and 2, we get
m = Ae3t :
Since m = 5 when t = 1; we have 5 = Ae3 ; so A = e53 : Thus m = e53 e3t = 5e3t,3 :
1.
4.
5.
6.
7.
R dI R
= 0:2 dx implies that
0:2I implies that dI
I
I = 0:2 dx implies that ln I = 0:2x + C .
So I = Ae0:2x , where A = eC . According to the given boundary condition, I ( 1) = 6. Therefore, I ( 1) =
Ae0:2(,1) = Ae,0:2 = 6 implies that A = 6e0:2 : Thus I = 6e0:2 e0:2x = 6e0:2(x+1) :
R dy
R 1
dy y
dy
y
dx + 3 = 0 implies dx = 3 implies y =
3 dx:
dI
dx
=
,
,
,
,
1
1
Integrating and moving terms, we have y = Ae, 3 x : Since y(0) = A = 10; we have y = 10e, 3 x :
1 dz
z dt
5 implies dz
z = 5 dt:
Integrating and moving terms, we have z = Ae5t . Using the fact that z (1) = 5; we have z (1) =
Therefore, z = e55 e5t = 5e5t,5 :
dP
dt
=
=
R dP
Ae5 = 5; so A = e5 :
5
P + 4 implies that PdP+4 = dt:
P +4
=
R
dt implies that ln jP + 4j = t + C:
+ 4 = Aet implies that
Therefore P = 104et 4:
P
8.
jj
,
dy = 2y , 4 = 2(y , 2).
dx
P
=
Aet , 4: P
=
100 when
t
=
0; so
P (0 )
=
Ae0 , 4
=
100; and
A
=
104:
R dy
R
Factoring out a 2 makes the integration easier: ydy
,2 = 2 dx implies that y,2 = 2 dx implies that ln y 2 = 2x + C:
y 2 = e2x+C implies that y 2 = Ae2x where A = eC : The curve passes through (2,5), which means 3 = Ae4;
so A = e34 : Thus, y = 2 + e34 e2x = 2 + 3e2x,4 :
j , j
9.
10.
,
Factoring out the 0.1 gives dm
:1m + 200
dt R= 0dm
R = 0:1(m + 2000).
0:1t
dm
m+2000 = 0:1 dt implies that m+2000 = 0:1 dt, so ln m + 2000 = 0:1t + C: So m = Ae
initial condition, m(0) = Ae(0:1)0 2000 = 1000; gives A = 3000: Thus m = 3000e0:1t 2000:
dB + 2B = 50 implies dB = 2B + 50 = 2(B 25) implies R dB = R 2 dt:
dt
dt
B,25
,
,
j
,
j
,
,
After integrating and doing some algebra, we have B 25 =
so A = 75e2 : Thus B = 25 + 75e2 e,2t = 25 + 75e2,2t :
11.
j ,j
,
Ae,2t : Using the initial condition, we have 75 = Ae,2;
We know that the general solution to a differential equation of the form
dy = k(y , A)
dt
is
,
, 2000: Using the
y = Cekt + A:
10.4 SOLUTIONS
Thus, in our case, we get
y = Cet=2 + 200:
We know that at t = 0 we have y = 50, so solving for C we get
y = Cet=2 + 200
50 = Ce0=2 + 200
,150 = Ce0
C = ,150:
Thus we get
12.
y = 200 , 150et=2:
We know that the general solution to a differential equation of the form
dQ = k(Q , A)
dt
is
H = Cekt + A:
To get our equation in this form, we factor out a 0:3 to get
dQ = 0:3 Q 120 = 0:3(Q
dt
0:3
,
, 400):
Thus, in our case, we get
Q = Ce0:3t + 400:
We know that at t = 0 we have Q = 50 so solving for C we get
Q = Ce0:3t + 400
50 = Ce0:3(0) + 400
,350 = Ce0
C = ,350:
13.
Thus we get
Q = 400 , 350e0:3t :
Rearrange and write
Z
or
Z
dr
, ln j1 , Rj = r + C
which can be written as
1
or
dR =
1,R
1
, R = e,C ,r = Ae,r
R(r) = 1 , Ae,r :
The initial condition R(1) = 0:1 gives 0:1 = 1 , Ae,1 and so
A = 0:9e:
Therefore
R(r) = 1 , 0:9e1,r :
521
522
14.
CHAPTER TEN /SOLUTIONS
Z
Write
and so
ln
y dy =
1
1
dt
3+t
jyj = ln j3 + tj + C
or
ln
where ln D = C . Therefore
The initial condition y(0) = 1 gives D =
Z
jyj = ln Dj3 + tj
y = D(3 + t):
1
3
and so
y(t) = 13 (3 + t):
15.
R
=
2
16.
17.
R
tez implies e,z dz = tdt implies e,z dz = t dt implies ,e,z = t2 + C:
Since the solution passes through the origin, z = 0 when t = 0, we must have ,e,0 = 02 + C; so C = ,1: Thus
,e,z = t2 , 1; or z = , ln(1 , t2 ):
R
R
dy=dx = 5y=x implies dy=y = 5dx=x. So ln jyj = 5 ln jxj + C = 5 ln x + C implies jyj = e5 ln x eC , and thus
y = Ax5 where A = eC . Since y = 3 when x = 1, so A = 3. Thus y = 3x5:
dy = y2 (1 + t) implies that R dy = R (1 + t) dt implies that , 1 = t + t + C implies that y = , 1 .
dt
y
2
y
t+ t +C
2
= , t +2t,4 :
Since y = 2 when t = 1, then 2 = , 1+ 1+C : So 2C + 3 = ,1; and C = ,2: Thus y = , t 1
+t,2
dz
dt
2
2
2
2
2
2
1
2
18.
19.
R
2
2
R
2
z + zt2 = z (1 + t2) implies that dzz = (1 + t2 )dt implies that ln jz j = t + t3 + C implies that z = Aet+ t :
z = 5 when t = 0; so A = 5 and z = 5et+ t :
dw = w2 sin 2 implies that R dw = R sin 2 d implies that , 1 = , 1 cos 2 + C . According to the initial
d
w
2
w
conditions, w(0) = 1; so ,1 = , 12 + C and C = , 12 : Thus , w1 = , 12 cos 2 , 12 implies that w1 = cos 2 +1 implies
that w = cos 2 +1 :
du = u2 implies R du = R dx = R ( 1 , 1 )dx implies , 1 = ln jxj , ln jx + 1j + C:
x(x + 1) dx
x(x+1)
x 1+x
u
u
u(1) = 1; so , 11 = ln j1j, ln j1 + 1j + C: So C = ln 2 , 1: Solving for u yields , u1 = ln jxj, ln jx + 1j + ln 2 , 1 =
ln jx2j+xj1j , 1, so u = ln j ,x1 j,1 .
x
Separating variables and integrating with respect to r gives
dz
dt
=
3
3
3
3
3
2
2
2
20.
2
2
+1
21.
Z
z dz =
1
so that
ln
Z
(1 +
r2 )dr
jz j = r + 13 r3 + C:
The initial condition z (0) = 1 gives C = 0 so that
z (r) = er+(1=3)r :
3
22.
Separating variables and integrating with respect to
Z
1
gives
Z
w2 dw =
Now set
2
=
t, then this becomes
Z
1
w2 dw = 2
1
cos
Z
2
d :
cos tdt
10.4 SOLUTIONS
and so
, w1 = 12 sin t + C
2
w = sin(,
t) + D
or
Using the initial conditions give D =
23.
24.
25.
26.
27.
28.
29.
30.
32.
w = sin ,2 2+ D :
,2, so the solution is
w = sin ,22, 2 :
dR = kR implies that dR = k dt which implies that R dR = R k dt. Integrating gives ln jRj = kt + C; so jRj =
dt
R
R
ekt+C = ekt eC . R = Aekt ; where A = eC :
dQ , Q = 0 so dQ = Q : This is now the same problem as Problem 25, except the constant factor on the right is 1
dt
k
dt
k
k
1
instead of k: Thus the solution is Q = Ae k t for any constant A:
dP = P , a, implying that dP = dt so R dP = R dt. Integrating yields ln jP , aj = t + C; so jP , aj = et+C =
dt
P ,a
P ,a
et eC . P = a + Aet; where A = eC or A = 0.
dQ = b , Q implies that dQ = dt which, in turn, implies R dQ = R dt. Integrating yields , ln jb , Qj = t + C; so
dt
b,Q
b,Q
jb , Qj = e,(t+C ) = e,te,C . Q = b , Ae,tR; where A = e,C or A = 0.
dP = k(P , a),so dP = k dt, so R dP = k dt. Integrating yields ln jP , aj = kt + C so P = a + Aekt where
dt
P ,a
P ,a
A = eC or A = 0.
dR = aR + b: If a = 0; then this is just dR = b; where b is a constant. Thus in this case R = bt + C is a solution for
dt
dt
any constant C:
b
If a 6= 0; then dR
dt = a(R + a ):
Now this is just the same as Problem 27, except here we have a in place of k and , ab in place of a, so the solutions are
R = , ab + Aeat where A can be any constant.
dy = y(2,y) which implies that dy = , dt, implying that R dy = , R dt, so , 1 R ( 1 , 1 )dy = , R dt.
dt
y(y,2)
(y ,2)(y )
2
y y,2
2j
Integrating yields 12 (ln jy , 2j , ln jyj) = ,t + C , so ln jyj,
=
,
2
t
+
2
C
.
yj
2
Exponentiating both sides yields j1 , y2 j = e,2t+2C ) y2 = 1 , Ae,2t ; where A = e2C . Hence y = 1,Ae
,2t . But
2
2
y(0) = 1,A = 1; so A = ,1, and y = 1+e,2t :
R cos x
R 1 2 ln t
dx
1+2 ln t
t dx
dt = (1 + 2 ln t) tan x 2implies that tan x = ( t ) dt which implies that sin x dx = ( t + t ) dt.
j
j
ln sin x = ln t + (ln t) + C:
2
sin x = eln t+(ln t) +C = t(eln t )ln t eC = t(tln t )eC . So sin x = At(ln t+1) , where A = eC . Therefore x =
ln t+1
arcsin(At
).
dx = x ln x , so R dx = R dt and thus ln ln x = ln t + C , so ln x = eC eln jtj = eC t . Therefore ln x = At, where
dt
t
x ln x
t
A = eC , so x = eAt :
dy = y ln( y ) implies that dyy = dt implies that R dyy = R ( dt):
dt
2
y ln( )
y ln( )
j
31.
523
j
j
,
j
jj
j
,
j
jj
,
2
2
Substituting w = ln( y2 ); dw = y1 dy gives:
R
R dw
that ln jwj = ln j ln( y2 )j = ,t + C . Since y(0) = 1; we have C
w = (, dt) implies
y
ln(ln 2): Thus ln j ln( )j = ,t + ln(ln 2), or
2
j ln( y2 )j = e,t
+ln(ln 2)
Again, since y(0) = 1, we see that
not satisfy y(0) = 1.)
=
j
j
ln ln 12 = ln
j , ln 2j =
e,t
= (ln 2)
, ln(y=2) = (ln 2)e,t and thus y = 2(2,e,t ). (Note that ln(y=2) = (ln 2)e,t does
524
33.
CHAPTER TEN /SOLUTIONS
(a), (b)
y
x
(c)
dy
dx
=
(b).
34.
R
x
y , so
R
y dy = x dx and thus y2
2
= x2 +
2
C , or y2 , x2 = 2C: This is the equation of the hyperbolas in
(a), (b)
y
x
(c)
35.
, xy , which implies that
A
y = x ; where A = eC :
=
R dy
y
=
,
R dx
, ln jxj+C
x ; so ln jyj = , ln jxj+C implies that jyj = e
jxj),1eC :
=(
By looking at the slope fields, we see that any solution curve of y0 = yx intersects any solution curve to y0 = xy :
Now if the two curves intersect at (x; y); then the two slopes at (x; y) are negative reciprocals of each other, because
y
1
x=y = x . Hence, the two curves intersect at right angles.
,
36.
dy
dx
,
,
If y(x) = xv(x), then y0 = v + xv0 and the differential equation becomes
Solving for v0 gives
+ xv
1+v
v + xv0 = xx ,
xv = 1 , v :
dv = 1 + v , v = 1 + v2
x dx
1,v
1,v
10.4 SOLUTIONS
525
which is a separable equation. Rearranging and integrating with respect to x gives
Z
Z
1 v
1
dv
=
dx
1 + v2
x
Z
Z
Z
1
dv 12 1 +2vv2 dv = x1 dx
1 + v2
1
ln(1 + v2 ) = ln x + C:
arctan(v)
2
,
,
,
Now put v =
y
x to find
arctan
37.
jj
Starting off with the homogeneous equation
we replace y(x) with xv(x) and get
y 1 2 2
x , 2 ln(x + y ) = C:
f xy = dy
dx ;
y
dy d(xv(x))
dv
f (v) = f xv
=f
x
x = dx = dx = v + x dx :
Thus we have
dv
f (v) , v = x dx
which is clearly separable.
38.
Making the substitution s = x + a, t = y + b. the differential equation becomes
dy = x + y + a + b + 5 :
dx x , y + a , b , 1
Now select a and b so that a + b + 5 = 0 and a , b , 1 = 0, that is, a = ,2, b = ,3. The differential equation becomes
dy = x + y :
dx x , y
Now set y(x) = xv(x), then y0 = v + xv0 and the differential equation becomes
+ xv
1+v
v + xv0 = xx ,
xv = 1 , v :
Solving for v0 gives
dv = 1 + v , v = 1 + v2
x dx
1,v
1,v
which is a separable equation. Rearranging and integrating with respect to x gives
Z
Z
1,v
1
dv
=
1 + v2
x dx
Z
Z
Z
1
1
2v
1
dv
,
dv
=
1 + v2
2
1 + v2
x dx
1
arctan(v) , ln(1 + v2 ) = ln jxj + C:
2
Now put v = xy to find
y 1 2 2
arctan
x , 2 ln(x + y ) = C:
Transforming back to the original variables x = s + 2 and y = t + 3 gives
t + 3 , 1 ln((s + 2)2 + (t + 3)2) = C:
arctan
s+2 2
```