Math 308 Week 1 Solutions
Here are solutions to the even-numbered suggested problems. The answers to the oddnumbered problems are in the back of your textbook, and the solutions are in the Solution
Manual, which you can purchase from the campus bookstore.
NSS Section 1.1
14. Write a differential equation that fits the physical description: the velocity at time
t of a particle moving along a straight line is proportional to the fourth power of its
position x.
dx
Answer:
= kx4
dt
16. Write a differential equation that fits the physical description: the rate of change of
the mass A of salt at time t is proportional to the square of the mass of salt present
at time t.
dA
Answer:
= kA2
dt
NSS Section 1.2
18. Let c > 0. Show that the function φ(x) = (c2 − x2 )−1 is a solution to the initial value
problem dy/dx = 2xy 2 , y(0) = 1/c2 , on the interval −c < x < c. Note that this
solution becomes unbounded as x approaches ±c. Thus, the solution exists on the
interval (−δ, δ) with δ = c, but not for larger δ. This illustrates that in Theorem 1
the existence interval can be quite small (if c is small) or quite large (if c is large).
Notice also that there is no clue from the equation dy/dx = 2xy 2 itself, or from the
initial value, that the solution will “blow up” at x = ±c.
Answer: We can check that φ(x) = (c2 − x2 )−1 satisfies the initial value problem
dy/dx = 2xy 2 , y(0) = 1/c2 :
2x
φ0 (x) = 2
(c − x2 )
And:
2xφ(x)2 = 2x(c2 − x2 )−2 =
(c2
2x
− x2 )
Thus, φ0 (x) = 2xφ(x)2 , so φ(x) is a solution to the differential equation. Additionally,
φ(0) =
1
c2
Thus, φ(x) is a solution to the initial value problem.
The rest of the problem does not contain any questions to be answered. It’s just
explaining why this initial value problem and solution are interesting. The differential
equation satisfies the conditions of Theorem 1: f (x, y) = xy 2 and ∂f /∂y = 2xy are
continuous at every point (x0 , y0 ). Thus, by Theorem 1, every point (x0 , y0 ) has a
solution to the differential equation that goes through it. Theorem 1 tells us that the
solution is defined on some interval surrounding x0 , x0 − δ < x < x + δ, but it does
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not say anything about how large that interval is. For the given differential equation,
we see that the solution through the point (0, 1/c2 ) has domain −c < x < c, which
could be really large or really small, depending on the value of c. So, we see that
for a differential equation where f (x, y) and ∂f /∂y are continuous everywhere, the
solution to a given initial value problem may be defined on an arbitrarily small or
arbitrarily large interval.
22. Verify that the function φ(x) = c1 ex + c2 e−2x is a solution to the linear equation
d2 y dy
+
− 2y = 0
dx2 dx
for any choice of the constants c1 and c2 . Determine c1 and c2 so that the following
initial condition is satisfied.
(a) y(0) = 2, y 0 (0) = 1
Answer: First, we need to verify that φ(x) is a solution to the differential equation.
φ0 (x) = c1 ex − 2c2 e−2x
φ00 (x) = c1 ex + 4c2 e−2x
Thus:
¡
¢ ¡
¢
¡
¢
φ00 (x) + φ0 (x) − 2φ(x) = c1 ex + 4c2 e−2x + c1 ex − 2c2 e−2x − 2 c1 ex + c2 e−2x
= (c1 + c1 − 2c1 ) ex + (4c2 − 2c2 − 2c2 ) e−2x
= 0
Thus, φ(x) satisfies the differential equation for any choice of the constants c1 and c2 .
(a) The condition y(0) = 2 tells us that 2 = c1 e0 + c2 e0 , which is the same as
2 = c1 + c2
The condition y 0 (0) = 1 tells us that 1 = c1 e0 − 2c2 e0 , which is the same as
1 = c1 − 2c2
Subtracting these two equations, we get that 1 = 3c2 , so c2 = 1/3. Plugging
this into the first equation, we get that 2 = c1 + 1/3, so c1 = 5/3. Thus, the
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1
answer is c1 = , c2 =
3
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26. Determine whether Theorem 1 implies that the following initial value problem has a
unique solution.
p
dy
= 3x − 3 y − 1,
y(2) = 1
dx
√
Answer: The function f (x, y) = 3x − 3 y − 1 is continuous, so we look at ∂f /∂y:
1
∂f
=
∂y
3(y − 1)2/3
2
This is not continuous when y = 1, so it is not continuous at the point (2, 1). Thus,
the intial value problem does not satisfy the conditions of Theorem 1. So, Theorem
1 does not imply that the given initial value problem has a unique solution. The
answer is No
NSS Section 1.3
4. If the viscous force in Problem 3 is nonlinear, a possible model would be provided by
the differential equation
dv
v3
=1−
dt
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Redraw the direction field in Figure 1.14 to incorporate this v 3 dependence. Sketch
the solutions with initial conditions v(0) = 0, 1, 2, 3. What is the terminal velocity in
this case.
Answer:
3
2
1
v(t)
0
–1
–2
–3
–3
–2
–1
0
1
2
3
t
From the direction field, we see that the terminal velocity is v = 2
6. Consider the differential equation
dy
= x + sin y
dx
(a) A solution curve passes through the point (1, π/2). What is its slope at this
point?
(b) Argue that every solution curve is increasing for x > 1.
(c) Show that the second derivative of every solution satisfies
d2 y
1
= 1 + x cos y + sin 2y
2
dx
2
3
(d) A solution curve passes through (0, 0). Prove that this curve has a relative
minimum at (0, 0).
Answer:
(a) At (1, π/2), the slope is 1 + sin(π/2) = 1 + 1 = 2.
(b) Since sin y is always between −1 and 1, x+sin y for x > 1 will always be positive.
Thus, every solution curve has positive derivative for x > 1, so every solution
curve is increasing for x > 1.
(c) First, we take the derivative of the differential equation (using implicit differentiation). We get
d2 y
dy
= 1 + cos y
2
dx
dx
dy
Now, we plug x + sin y in for
:
dx
d2 y
= 1 + cos y (x + sin y)
dx2
Thus,
d2 y
= 1 + x cos y + cos y sin y
dx2
Now, we use the trig identity sin(2θ) = 2 sin θ cos θ to change cos y sin y into
1
sin 2y:
2
d2 y
1
= 1 + x cos y + sin 2y
2
dx
2
(d) From the original differential equation, we can compute the first derivative of
the solution through the point (0, 0). It is 0 + sin 0 = 0. Thus, the point (0, 0)
is a critical point.
From part (c), we can compute the second derivative of the solution through the
point (0, 0). It is 1 + 0 cos 0 + 12 sin 0 = 1. Thus, the second derivative at the
point (0, 0) is positive, so the critical point is a local minimum.
NSS Section 1D, p. 34-37
1. (d) Sketch the phase line for y 0 = (y − 1)(y − 2)(y − 3) and state the nature of its
equilibria.
Answer: To find the equilibria, we look at (y − 1)(y − 2)(y − 3) = 0. This
equation is satisfied for y = 1, 2, 3. These are the equilibria. To see what kind
of equilibria these are, we look at what happens to solutions in between these
values of y:
For y < 1, (y − 1)(y − 2)(y − 3) is negative (so solutions are decreasing).
For 1 < y < 2, (y − 1)(y − 2)(y − 3) is positive (so solutions are increasing).
For 2 < y < 3, (y − 1)(y − 2)(y − 3) is negative (so solutions are decreasing).
For y > 3, (y − 1)(y − 2)(y − 3) is positive (so solutions are increasing).
Thus, we get the following phase line:
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C œ $ source
C œ # sink
C œ " source
Thus, the equilibria are y = 1, 2, 3. The equilibrium y = 1 is a source (because
it repels the solutions above and below it). The equilibrium y = 2 is a sink
(because it attracts the solutions above and below it). The equilibrium y = 3 is
a source (because it repels the solutions above and below it).
(f) Sketch the phase line for y 0 = y sin y and state the nature of its equilibria.
Answer: To find the equilibria, we look at y sin y = 0. This equation is satisfied
for y = 0, π, −π, 2π, −2π, . . . . Thus, the equilibria are all multiples of π.
For y > 0, y sin y is positive between an even multiple of π and and odd multiple
of π — for (2k)π < y < (2k + 1)π. It is negative for (2k + 1)π < y < (2k + 2)π.
Thus, the positive even multiples of π are sources, and the positive odd multiples
of π are sinks.
For y < 0, y sin y is positive for −(2k + 1)π < y < −(2k)π. It is negative for
−(2k + 2)π < y < −(2k + 1)π. Thus, the negative even multiples of π are sinks,
and the negative odd multiples of π are sources.
The equilibrium y = 0 attracts solutions below it and repels solutions above it.
Thus, it is a node (an equilibrium that is neither a sink nor a source).
Thus, we get the following phase line:
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the pattern continues
C œ $1
sink
C œ #1
source
Cœ1
sink
Cœ!
node
C œ −1
source
C œ −#1 sink
C œ −$1 source
the pattern continues
(g) Sketch the phase line for y 0 = y sin y + 0.1 and y 0 = y sin y − 0.1. Discuss the
effect of the small perturbation ±0.1 on the equilibria.
Answer: For y 0 = y sin y + 0.1, there is not an equilibrium near y = 0, but
all the other multiples of π have a nearby equilibrium, with sources and sinks
remaining the same as in (f). The phase line looks like:
the pattern continues
sink (near C œ $1)
source (near C œ #1)
sink (near C œ 1)
source (near C œ −1)
sink (near C œ −#1)
source (near C œ −$1)
the pattern continues
For y 0 = y sin y − 0.1, there are two equilibria near y = 0 (one is approximately
y = −.3189288248 and one is approximately y = .3189288248). In between these
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two equilibria, y sin y − 0.1 is negative, so the solution curves are decreasing.
Thus, the positive equilibrium is a source, and the negative equilibrium is a
sink.
All the other multiples of π have one nearby equilibrium, with sources and sinks
remaining the same as in (f). The phase line looks like:
the pattern continues
sink
source
sink
source
2 equilibria
near C œ !
sink
source
sink
source
the pattern continues
We saw in this problem, that when a differential equation with a node equilibrium is changed slightly, the node equilibrium could disappear, or it could split
into two equilibria, one a source and one a sink.
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