2
1. HW Sec. 1.3: 13.
dy
p
= 3 y ; y (0) = 1:
dx
The rate function is given by
f (x; y ) =
p
3
y:
p
It is known from calculus that f (x; y ) = 3 y is a continuous function for
all x and y , in particular in a neighborhood of the point x0 = 0 and y0 = 1.
So, by the existence theorem, solution exists close to the point (0; 1).
@p
1
3
y=
2
@y
3y 3
is continuous in a neighborhood of the point (0; 1). Indeed, 1 is a constant,
2
thereby a continuous function for all x and y ; 3y 3 is a continuous function
2
for all x and y . Notice that 3y 3 does not vanish in a neighborhood of
2
the point of (1; 0). Hence, the ratio of continuous function 1 and 3y 3 is
continuous because the denominator is not zero around (0; 1). So, by the
uniqueness theorem, we see that there is a unique solution of the foregoing
initial value problem.
Answer: solution exists and is unique.
(students do not have to include the sketch).
1
2. Sec. 1.3: 14.
dy
p
= 3 y ; y (0) = 0:
dx
The rate function is given by
f (x; y ) =
p
3
y:
p
3
It is known from calculus that f (x; y ) = y is a continuous function for
all x and y , in particular in a neighborhood of the point x0 = 0 and y0 = 0.
So, by the existence theorem, solution exists close to the point (0; 0). If
y 6= 0, then
1
@p
3
y=
2:
@y
3y 3
It is readily seen that
lim
y !0
@p
1
3
y = lim
2 = 1;
y !0 3y 3
@y
hence, the partial derivative fy cannot be continuous at (0; 0). The condition for uniqueness fails. So, the uniqueness theorem does not guarantee
the uniqueness of solution.
3. Sec. 1.4:13.
dy
= y 4 + 1 cos x:
dx
This is a separable equation:
y3
y3
dy = cos xdx:
y4 + 1
By integrating,
Z
y3
dy =
4+1
y
Z
cos xdx = sin x + C:
Now we calculate the integral in the left-hand side:
Z
Z
Z
d y4 + 1
y3
1
y3
dy =
; whence
dy
y4 + 1
4
y4 + 1
y4 + 1
1
1
= ln y 4 + 1 + C = ln y 4 + 1 + C:
4
4
Answer (implicit solution):
ln y 4 + 1 = 4 sin x + C:
2
4. Sec. 1.4:17.
y 0 = 1 + x + y + xy:
First we factor the right-hand side:
1 + x + y + xy = (1 + x) + y (1 + x) = (1 + x) (1 + y ) :
Hence we have
dy
dx
dy
= (1 + x) (1 + y ) )
=
)
dx
1+y
1+x
Z
Z
x2
dy
= (1 + x) dx ) ln j1 + y j = x +
+ C.
1+y
2
Then
ln j1 + y j = x +
x2
+ C.
2
Answer: (implicit solution)
ln j1 + y j = x +
x2
+ C:
2
5. Sec. 1.4:49. According to Newton law of cooling, the time rate of change
s
of the temperature T (t) of a cake is proportional to the dierence A T ,
where A is constant temperature:
dt
= k (A
dt
T ); k > 0:
We have
A = 70; T (0) = 210; T (30) = 140:
Hence,
Z
dt
70
T
=
Z
dt
= k (70
dt
kdt;
`nj70
T );
T j = kt + c; 70
Now T (0) = 210 implies 70 210 = Be0 ; B =
140 yields
70
140 =
140e
30k
;
1
=e
2
30k
; k=
T = Be
100 =
t=
1
1
ln
30 2
0:0231:
140e 0:0231t ) 3=14 = e 0:0231t )
1
3
ln
= 66:67 min:
0:0231 14
Thus the cake will be 100 F after 66 min 40 sec.
3
:
140. Substitution T (30) =
To nd when the cake will be 100 F we solve the equation
70
kt
6. Sec. 1.4:51.
A0 = kA ) A (t) = A (0) ekt ; where A (0) = 15:
(a)
A (t) = 15ekt ) 10 = 15e5k ) k =
A (t) = 15ekt ; k =
8: 109 3
8: 109 3
10
2
10
2
:
:
(b)
2
8: 109 3 10
A (8) = 15e
8
= 7: 840 5:
(c)
2
8: 109 3 10
1 = 15e
t
) t = 33: 394month:
7. Sec. 1.4:53. The initial number of languages is 1; each 6000 years the
number of languages is multiplied by 1.5. Hence, the number of languages
L (t) is given by
L (t) = 1:5t :
By solving the equation
1:5t = 150;
we nd:
t = 12: 358 ) 6000 12: 358 = 74148 )
the ancestors arrived about 74000 years ago:
8. Sec. 1.4:57. By Toricelli law,
s
A
dy
=
dt
p
k ) y y0 =
dy
p
a y)p =
y
adt;
k
where a = A , A is the area of the cross-section of the cylinder (notice that
A = const).Then, by integrating,
p
2 y = at + c:
p
Let y (0) = h, where h is the height of the cylinder. Then 2 h = c. Hence
p
p
2 y = at + 2 h:
Now we nd a. We have
Finally
p
2 y (T ) = 0 =
p
2 y=
p
p
2h
aT + 2 h ) a =
.
T
p
p
p
2h
p
t+2 h) y = h 1
T
) y (t) = h 1
4
t
T
2
:
t
T
Now
t
T
V (t) = Ay (t) = Ah 1
2
:
By recalling that V0 = Ah; we get
t
T
V (t) = V0 1
2
:
9. Sec. 1.4:59.The tank is shown on the sketch:
y
Rotate the curve
x 2 = by
0
x
The horizontal cross-sectional area of the tank is
A (y ) = x2 = by:
By Toricelli Law for a draining tank,
s
A (y )
dy
=
dt
p
a 2gy;
where a is the area of the circular hole at the bottom of the tank. Hence,
p
p
p
dy
a 2g
a 2g
p dy
by
= a 2gy ) y
=
: Denote k =
:
dt
dt
b
b
Hence, we have the following equation:
p dy
y
=
dt
5
k:
By integrating both sides of the equation, we nd
p dy
y
=
dt
k)
23
y2 =
3
k t + C:
Using y (0) = 4;
2
23
16
4 2 = +C ) C = 8 =
:
3
3
3
Using y (1) = 1;
2
=
3
k+
16
16
)k=
3
3
2
14
=
.
3
3
Now we can answer a): Solution is
2
16
14
t+
) y3 = 8
3
3
22
y3 =
3
7t ) y = (8
2
7t) 3 .
b). The tank will be empty at
0 = (8
2
7t) 3 ) t =
8
hour ) t = 1 : 08 : 34pm:
7
c). If the initial radius is x = 2 and the initial depth is y = 4; then
2
22 = 4b ) b = 1:g = 32 (3600) f t=hr2
Next
2
g = 32 (3600) f t=hr2 :
Since
q
p
2
a 2 32 (3600)
14
a 2g
14
k=
=
=
)a=
= r2 ;
3
b
3 3600 8
whence
1
r=
60
r
7
12
0:0127293 7693 f t
0: 15275 25232 in
Answer:
r
10. Sec. 1.5:7.
0: 15275 25232 in:
p
2xy 0 + y = 10 x:
1
y
= 5x 2 :
2x
R
1
p (x) =
) (x) = e
2x
y0 +
6
dx
2x
=
p
x:
p
xy 0 +
yp
x = 5x
2x
1
2
p
x:
Notice that
p
xy 0 +
p
yp
y
dp
x = xy 0 + p =
xy :
2x
dx
2x
p
xy =
Z
5dx = 5x + C )
p
p
y = 5 x + C= x:
11. Sec. 1.5:17.
(1 + x)y 0 + y = cos x; y (0) = 1: ) y 0 +
P (x) =
1
cos x
; Q(x) =
;
1+x
1+x
R
(x) = e
Multiplying by
y
cos x
=
:
1+x
1+x
dx
1+x
= eln j1+xj = 1 + x:
(x) ;
0
(1 + x)y 0 + y = cos x ) [(1 + x)y ] = cos x )
(1 + x)y =
Z
cos x dx )
y = sin x + C ) y (x) =
sin x + C
1+x
From the initial condition:
1 = y (0) =
0+C
=) C = 1:
0+1
Answer:
y (x) =
sin x + 1
1+x
12. Sec. 1.5:37. The volume of brine in the tank after t sec is V (t) = 100 + 2t
gal, so the initial value problem is
x0 (t) = 5 1
3
x
; x(0) = 50;
100 + 2t
where x(t) is the amount of salt in the tank after t sec. Hence
dx
3
+
x = 5:
dt
100 + 2t
7
The integrating factor is
R
(t) = e
3dt
100+2t
3
= e 2 `nj100+2tj = (100 + 2t)3=2 :
Hence
Dt (x(t) (100 + 2t)3=2 ) = 5(100 + 2t)3=2 ;
x(t)(100 + 2t)3=2 = 5
x(t) =
21
(100 + 2t)5=2 + C ;
52
(2t + 100)5=2 + C
:
(100 + 2t)3=2
Substitution at x(0) = 50 gives
x(0) = 50 =
100; 000 + C
; c=
1000
50; 000:
Tank is full after 150 sec since
V (t) = 100 + 2t = 400 () t = 150:
Hence
x(150) = 100 + 2:150 50; 000(100 + 2:150) 3=2
50; 000
50
= 400
= 400
= 400 6:25 = 393:75 lb
8; 000
8
13. (a)
8
(b) It is evident that y = 5 is an equilibrium solution (5
(c)
dy
Separable equation
=5 y
)
dt
dy
= dt )
5y
Z
Z
dy
= dt )
5y
ln j5 y j = t + C )
jy
y
Set A =
5j = e
5=
t+C
= eC et )
eC e t :
eC . Then we get
y (t) = 5 + Ae t :
If t = 0, then y0 = 5 + A ) A = y0
5)
y (t) = 5 + (y0
9
5) e t :
y = 0 , y = 5).
(d) The sketch from part (a) indicates that solutions should approach to
the equilibrium as t ! +1. Indeed,
y (t) = 5 + Ae t :
Because e
t
! 0 as t ! +1, we see that
lim y (t) = lim
t!+1
t!+1
10
5 + Ae
t
= 5: