Solutions
1. Use the substitution y = um to solve the initial-value problem
y � + y = y 2/3 , y(0) = 8.
Solution:
y = um , y � = mum−1 u�
mum−1 u� + um = u2m/3
u� +
1
1 2m/3−m+1
u=
u
m
m
2m/3 − m + 1 = 0 if m = 3
m = 3 =⇒ u� +
1
1
u=
=⇒ u = C e−t/3 + 1 =⇒ y = (C e−t/3 + 1)3
3
3
y(0) = 8 =⇒ C + 1 = 2 =⇒ C = 1 =⇒ y = (e−t/3 + 1)3
2. Use the substitution y = tz to find the solution of
�y� y
π
y � = sec
+ , y(1) = .
t
t
6
Solution:
y = tz =⇒ y � = z + tz � =⇒ z + tz � = sec z + z
=⇒ cos z dz =
1
dt =⇒ sin z = ln t + C =⇒ z = sin−1 ( ln t + C)
t
=⇒ y = t sin−1 ( ln t + C)
y(1) = π/6 =⇒ C = sin(π/6) = 1/2 =⇒ y = t sin−1 ( ln t + 1/2)
3. Use the substitution y = xz to find an implicit general solution of
Solution:
y = xz =⇒ y � = z + xz � =⇒ z + xz � =
=⇒ xz � =
=⇒
z
z2
−z =−
1+z
1+z
1+z
1
dz = − dx
2
z
x
1
=⇒ − + ln |z| = − ln |x| + C
z
1
− + ln |xz| = C
z
x
− + ln |y| = C
y
dy
y
=
.
dx
x+y
z
1+z
4. A water tank has the shape of the surface of revolution obtained by revolving the graph of y = x2 ,
0 ≤ x ≤ 2, about the y-axis. It is full of water at time t = 0 and thereafter drains through a hole at the
bottom. Assume that the constant ρ in Toricelli’s law is 2π
3 .
a) Solve for the depth of the water at time t.
b) Find the draining time.
Solution:
2π √
a) a(y) = πx2 = πy =⇒ πy y � = −
y, y(0) = 4
3
2
2 3/2
2
√
y dy = − dt =⇒
y
= − t + C =⇒ y = (C − t)2/3
3
3
3
y(0) = 4 =⇒ C 2/3 = 4 =⇒ C = 43/2 = 8 =⇒ y = (8 − t)2/3
b) y = 0 when t = 8
5. Set up the differential equation for the depth of water in
a) a draining spherical tank with radius 1,
b) a draining conical tank with height 3 and radius 1.
Solutions:
a) a(y) = π x2 = π (1 − (y − 1)2 ) = π (2y − y 2 ), so π (2y − y 2 )
b) a(y) = π r2 = π (y/3)2 =
π 2
π dy
√
y , so y 2
= −ρ y
9
9
dt
dy
√
= −ρ y
dt
6. Compute two steps of Euler’s method, with stepsize h = 0.1, for the initial-value problem
y� = t −
Solution:
√
y, y(0) = 1.
y0 = 1
√
y1 = 1 + (.1)(0 − 1) = 0.9
√
y2 = 0.9 + (.1)(.1 − .9) ≈ 0.8151
7. Compute two steps of the improved Euler method, with stepsize h = 0.1, for the initial-value problem
y� = t −
Solution:
√
y, y(0) = 1.
√

k1 = f (0, 1) = 0 − 1 = −1






k2 = f (.1, 1 + (.1)(−1))
Step 1:
√


= .1 − .9 = −0.8486




y1 = 1 + (.05)(−1 − 0.8486) = 0.9076

k1 = f (.1, 0.9076) = −0.8527






k2 = f (.2, 0.9076 + (.1)(−0.8527))
Step 2:
√


= .2 − .8223 = −0.7068




y2 = 0.9076 + (.05)(−0.8527 − 0.7068) = 0.8296
√
8. Make a sketch showing several isoclines and the direction field for the equation y � = t− y in the square
0 ≤ t ≤ 2, 0 ≤ y ≤ 2. Then sketch the graph of the solution satisfying y(0) = 1.
Solution:
√
√
The isoclines are the curves t − y = m, or t = y + m. These are horizontal shifts of the right half
of the parabola y = t2 . The picture below shows the ones corresponding to m = −1, −1/2, 0, 1/2, 1.
2.0
1.5
1.0
0.5
0.5
1.0
1.5
9. Suppose that a population satisfies the logistic equation P � =
Assume that time is measured in years.
2.0
1
2
�
P 1−
P
400
�
in the absence of harvesting.
1
(a) If P (0) = 400, and 10
of the population is harvested each year, what stable equilibrium value will
P (t) approach as t → ∞?
�
�
1
P
1
�
Solution: Find the positive equilibrium solution of P = P 1 −
−
P.
2
400
10
1
P
2
�
1−
P
400
�
−
1
P
1
4
P = 0, P �= 0 =⇒ 1 −
− = 0 =⇒ P = (400) = 320
10
400 5
5
(b) What fraction of the population harvested each year will result in stable equilibrium of 250?
�
�
P
Solution: We want to find β so that 12 P 1 − 400
− β P = 0 when P = 250.
�
�
�
�
1
250
1
250
3
250 1 −
− 250 β = 0 =⇒ β =
1−
=
2
400
2
400
16
�
�
P
10. Suppose that a population satisfies the logistic equation P � = k P 1 − 700
in the absence of harvesting.
Find k if harvesting 10% of the population per year reduces the stable equilibrium from 700 to 600.
Solution:
�
�
P
1
0 = kP 1−
−
P when P = 600
700
10
�
�
600
1
=⇒ 0 = k · 600 1 −
−
600
700
10
�
�
6
1
k
1
=⇒ 0 = k 1 −
−
= −
7
10
7 10
7
=⇒ k =
10