Prof. Hunter 22B Homework 2 Solutions
Steffen Docken and Paige Buck-Moyer
October 14, 2014
2.1 20. Solve the initial value problem. ty 0 + (1 + t)y = t where y(ln 2) = 1
Ans: The first thing to do is get the equation into one of the standard forms. Dividing through
by t gives
1+t
0
y +
y=1
t
We can now see that our equation is of the form y 0 + p(t)y = q(t) where in our case p(t) = 1+t
t
and q(t) = 1. So we need to use an integrating factor. The first thing we will do is to integrate
p(t). This gives
Z Z 1+t
1
dt =
+ 1 dt = ln(t) + t
t
t
Therefore µ(t) = eln(t)+t = tet Multiplying the equation by µ(t) gives
1+t
t 0
(tet ) = tet
te y + y
t
tet y 0 + y(t + 1)(et )
Integrating both sides, and then using integration by parts on the right side gives
Z
tet y = (tet )dt
tet y = tet −
Z
et dt
tet y = tet − et + C
Dividing this equation by tet gives
1
y(t) = 1 − +
t
C
t
e−t
Now we use the given initial conditions to solve for C
1
C
+
e− ln 2
ln 2
ln 2
1
C
1
+
1=1−
ln 2
ln 2
2
1 = y(ln 2) = 1 −
2 ln 2 − 2 + C
2 ln 2
2 ln 2 = 2 ln 2 − 2 + C
1=
C=2
Plugging this back in to our expression for y(t) gives
y(t) = 1 −
1
1 2 −t
+ e
t
t
33. Show that if a and λ are positive constants, and b is any real number, then every solution of the
equation
y 0 + ay = be−λt
has the property that y → 0 as t → ∞.
Ans: We will split this into two cases, a = λ and a 6= λ. First, for a = λ we have
y 0 + ay = be−at
R
Let µ(t) be an integrating factor. Then µ(t) = exp adt = eat . Multiplying the equation by
µ(t) gives
eat y 0 + aeat y = be−at eat
or
eat y
0
=b
Therefore, integrating both sides gives eat y = bt + C, where C is a constant. So,
y(t) = e−at (bt + C)
Also, since a > 0, the exponential is a decaying exponential, and y(t) → 0 as t → ∞, because the
exponential decays faster than the linear term grows.
Now, for a 6= λ, we again have µ(t) = eat as an integrating factor. But when we multiply through
by µ(t), we get
eat y 0 + aeat y = be−λt eat = be(−λ+a)t
or
eat y
0
= be(a−λ)t
Integrating both sides gives
eat y =
be(a−λ)t
+C
a−λ
where C is a constant. So,
y(t) =
be(a−λ−a)t
be−λt
+ Ce−at =
+ Ce−at
a−λ
a−λ
Therefore, y(t) → 0 as t → ∞, since a, λ > 0, meaning y(t) is the sum of two decaying exponentials.
2.2
4. Solve the differential equation y 0 = (3x2 − 1)/(3 + 2y)
Ans: Separating the variables gives (3 + 2y)dy = (3x2 − 1)dx and integrating both sides gives
Z
Z
(3 + 2y)dy =
(3x2 − 1)dx
3y + y 2
9
y 2 + 3y +
4
2
3
y+
2
3
y+
2
y
= x3 − x + C
= x3 − x + C +
9
= x3 − x + C
4
= x3 − x + C
=
x3 − x + C
21
=
x3 − x + C
21
2
−
3
2
the
9
4
is absorbed into C
21. Solve the initial value problem
y 0 = (1 + 3x2 )/(3y 2 − 6y),
y(0) = 1
and determine the interval in which the solution is valid.
Ans: Separating the variables gives (3y 2 − 6y)dy = (1 + 3x2 )dx and integrating both sides gives
Z
Z
2
3y − 6ydy =
1 + 3x2 dx
y 3 − 3y 2
y − 3y − x3 − x
3
2
= x + x3 + C
= C
Now, plugging in the initial condition gives 13 −3(1)2 −03 −0 = C, so C = −2 and y 3 −3y 2 −x3 −x =
−2 is the solution. A graph of the solution is given below. The integral curve becomes vertical at
x = 1, so the interval in which the solution is valid is [0, 1).
2.3
2. A tank initially contains 120L of pure water. A solution of γg/L of salt water enters the tank at
2L/min and the well mixed solution flows out of the tank at the same rate. Find an expression
in terms of γ for the amount of salt in the tank. Also, find the limiting amout of salt in the tank
as t → ∞.
Ans: Let y(t) denote the amount of salt present in the tank at time t. To describe y 0 we need
to determine the change in the amount of salt in the tank. The change in salt in the tank, y 0 , is
given by y 0 = (grams of salt coming into the tank) - (grams of salt leaving the tank)
For salt flowing into the tank we have
2γg
2L γg =
min
L
min
For salt flowing out of the tank we have
2L
y(t)g
y(t)g
=
min
120L
60min
3
This is because to get the total amount of salt flowing out we multiply the amount of solution
flowing out of the tank my the concentration of salt in that solution. This gives us the differential
equation
y
y 0 = 2γ −
60
To solve this we can separate variables.
120γ − y
60
60
dy = 1dt
120γ − y
y0 =
Integrating both sides gives
Z 60
1
120γ − y
Z
dy =
1dt
−60 ln(120γ − y) = t + C
ln(120γ − y) =
−t
+C
60
Exponentiating gives
−t
120γ − y = Ce 60
−t
y(t) = Ce 60 + 120γ
Now we can use out initial data to solve for C. We know that at time t = 0 there is no salt in the
tank. This gives us the condition y(0) = 0 Substituting this in gives
0
y(0) = 0 = Ce 60 + 120γ
0 = C + 120γ
C = −120γ
−t
Replacing C with this value in our expression for y(t) gives us y(t) = 120γ(−e 60 + 1)
Now we need to find the limiting amount of salt in the tank. This is found by finding the value
of y(t) where there is no change in the amount of salt. In other words, the value of y(t) where
y 0 = 0. Using our differential equation
y0 =
Setting y 0 = 0
120γ − y
60
120γ − y
60
0 = 120γ − y
0=
y = 120γ
This is our limiting value for y(t)
20. A ball with mass 0.15kg is thrown upward with initial velocity 20m/s from the roof of a building
30m high. Neglect air resistance.
(a) Find the maximum height above the ground that the ball reaches.
Ans: Let y be the height of the ball above the ground. Then our differential equation comes
from Newton’s law of motion F = ma, and F = −mg is just the force due to gravity. So, the
differential equation is my 00 = −mg, or
y 00 = −g
4
and the initial conditions y(0) = 30m and y 0 (0) = 20m/s. Integrating both sides twice gives
Z Z
Z Z
d dy
dtdt = −
gdtdt
dtZdt
Z
dy
dt = − gt + adt
dt
1
y = − gt2 + at + b
2
Now plugging in the first initial condition gives 30m = y(0) = − 12 g(0)2 + a(0) + b, which
means that b = 30m. Taking the derivative gives y 0 = −gt + a, so plugging in the second
initial condition gives 20m/s = y 0 (0) = −g(0) + a, which gives a = 20m/s. So, the height of
the ball at time t is given by
1
y(t) = − gt2 + (20m/s)t + 30m
2
Now, the ball reaches its maximum height when its velocity changes sign, or when y 0 = 0.
Taking the derivative gives y 0 = −gt + 20m/s. So, y 0 = 0 at t = 20m/s
, and
g
ymax
20m/s
)
g
2
20m/s
20m/s
1
+ 20m/s
+ 30m
− g
2
g
g
400m2 /s2
400m2 /s2
−
+
+ 30m
2g
g
400m2 /s2
+ 30m
2g
50.4m
= y(
=
=
=
≈
using g = 9.8m/s2 .
(b) Assuming that the ball misses the building on the way down, find the time that it hits the
ground. Ans: For our coordinate system, the ground is at y = 0, so the ball hits the ground
at a time t, such that 0 = y(t) = − 21 gt2 + (20m/s)t + 30m. Solving for t gives
t =
≈
−20m/s ±
p
−1.17s, 5.25s
Since t must be larger than 0, t ≈ 5.25s.
5
400m2 /s2 + g60m
−g
(c) Plot the graphs of velocity and position versus time.
The graph of the position versus time is
And the graph of the velocity versus time is
6