Math 211 Fall 2007 Exam 1 Solutions
Instructor: S. Cautis
Thursday, October 4, 2007
Instructions: This is a closed book, closed notes exam. Use of calculators is
not permitted. You have one hour and fifteen minutes. Do all 7 problems.
Please do all your work on the paper provided. You must show your work
to receive full credit on a problem. An answer with no supporting work or
explanation will receive little to no credit.
Please print you name clearly here.
Print name:
Upon finishing please sign the pledge below:
On my honor I have neither given nor received any aid on this exam.
Grader’s use only:
1.
/15
2.
/15
3.
/15
4.
/10
5.
/22
6.
/10
7.
/13
1. Consider the ordinary differential equation f (t) · f ′ (t) = [3f (t)2 + 1]t2 .
(a) [10 points] Find the general solution.
This is a separable ODE. Separating we have:
f df
= t2 dt
3f 2 + 1
Integrating we get
t3
1
ln(3f 2 + 1) =
+ C1
6
3
which simplifies to
ln(3f 2 + 1) = 2t3 C2
or better
3
3f 2 + 1 = C3 e2t .
One can solve for f explicitly but there’s no need to.
(b) [5 points] Does the existence theorem imply the existence of a solution
through (t, f ) = (1, 0)? Explain.
In standard form the ODE is:
f′ =
t2 (3f 2 + 1)
= h(t, f )
f
so at (t, f ) = (1, 0) the right side h(t, f ) is not continuous. This means
that the existence theorem does not guarantee a local solution there.
y
2π
π
π/2
−π/2
t
−π
Figure 1: Sketch solution for 2c.
2. Consider the ordinary differential equation y ′ = 5 sin(y).
(a) [5 points] Find all the equilibrium (i.e. constant) solutions.
We want all solutions of the form y(t) = c where c is a constant. If this
were the case then y ′ (t) = 0 and hence 5 sin(c) = 0. This means c = 2πn
where n can be any integer (note that there are an infinite number of
solutions).
(b) [5 points] What is the slope of the unique solution through (t, y) =
(1, π/2)? What about the slope of the unique solution through (t, y) =
(1, −π/2)?
At (t, y) = (1, π/2) the derivative y ′ (1) is equal to 5 sin(π/2) = 5 so the
slope is 5.
At (t, y) = (1, −π/2) the derivative y ′ (1) is equal to 5 sin(−π/2) = −5 so
the slope is −5.
(c) [5 points] Sketch the two solutions y1 (t) and y2 (t) satisfying the initial
value conditions y1 (0) = π/2 and y2 (0) = −π/2.
See the figure above.
2
3. Consider the differential ω = ( yx − y)dx + xdy = 0.
(a)[5 points] Show that ω is not exact.
We have
∂ y2
2y
( − y) =
−1
∂y x
x
while
∂
(x) = 1
∂x
Since these two are not equal ω is not exact.
(b)[10 points] Find an integrating factor µ for ω assuming you know it
only depends on the variable y (i.e. µ = µ(y)).
We want µ = µ(y) such that
y2
∂
∂
(µ − µy) =
(xµ)
∂y x
∂x
The left side is equal to
LS = µ′
y2
y
y(y − x)
2y − x
+ 2µ − µ − µ′ y = µ′ (
) + µ(
)
x
x
x
x
while the right side equals RS = µ. Equating the two gives
µ′ (
2y − x
y(y − x)
) + µ(
− 1) = 0
x
x
or equivalently
y(y − x)
y−x
= −2µ
x
x
Cancelling on both sides we
µ′
µ′ y = −2µ
This separates to give
dµ
µ
= − y2 dy which gives ln(µ) = −2 ln(y) or
µ=
as an integrating factor.
1
y2
4. [10 points]
Solve the differential equation ω = (xy − 2)dx + (x2 − xy 2 )dy = 0 by using
the integrating factor µ = 1/x.
Notice
that
ω
x
= (y − x2 )dx + (x − y 2 )dy is exact. We want to find F (x, y) such
dF =
From
∂F
∂x
=y−
∂F
∂F
2
dx +
dy = (y − )dx + (x − y 2 )dy
∂x
∂y
x
2
x
we get
Z
2
F = (y − )dx + φ(y) = xy − 2 ln |x| + φ(y)
x
where φ is any function of y. Then
∂F
= x + φ′ (y) = x − y 2
∂y
3
Solving for φ we get φ(y) = − y3 + C. Thus
F = xy − 2 ln |x| −
y3
+C
3
and the general solution is
xy − 2 ln |x| −
y3
+C =0
3
5. A large tub having volume 6, 000 liters contains 1, 000 liters of salt water
which contains 0.2 kg/liter of dissolved salt. Salt water containing 0.1
kg/liter of salt is pouring into the tub at 20 liters/min. Meanwhile, water
pours out at 5 liters/min. Assume the mixing is instantaneous.
(a)[10 points] Show that the amount of salt A(t) at time t in the tub
satisfies
A(t)
dA
=2−
.
dt
200 + 3t
We have
salt in = 20 × 0, 1 = 2kg/min
salt out =
Therefore
A(t)
A(t)
×5=
kg/min
1000 + (20 − 5)t
200 + 3t
A(t)
dA
= salt in − salt out = 2 −
dt
200 + 3t
(b)[7 points] Solve for A(t).
A
This is a linear ODE which we rewrite as A′ + 200+3t
= 2. The integrating
factor is
R
1
µ = e 200+3t dt = eln(200+3t)/3 = (200 + 3t)1/3
Multiplying by µ and integrating we get
Z
1
4
1
4
3 1
(200+3t) 3 A = 2(200+3t)1/3 = 2(200+3t) 3 × × +C = (200+3t) 3 +C
4 3
2
Solving for A we get
A=
C
1
(200 + 3t) +
1
2
(200 + 3t) 3
To find C note that A(0) = 1000 so
1
C
= 1000
(200) +
2
200/f rac13
1
which gives C = 900 × 200 3 .
(c)[5 points] Find the amount of salt in the tub at the moment it becomes
full.
The tub becomes full when 1000 + (20 − 5)t = 6000 or t = 1000/3. Thus
we want
A(1000/3) =
=
=
1
1
1
(1200) + 900 × 200 3 ×
1
2
1200 3
200 1
)3
600 + 900 × (
1200
900
600 + 1
63
6. [10 points]
M. Poppins drops her umbrella from a balloon at height 80 meters. Suppose her umbrella has mass m and that the wind resistance to the umbrella
is proportional (with constant c) to the square of its speed. Write down
an ODE with initial value conditions which describes the motion of the
umbrella as it falls. (Note: you do not need to solve the initial value
problem.) hint: make sure you get your signs right.
What are the forces on the umbrella? There’s gravity acting with a force
−mg. There’s air resistance acting with a force −cv|v| where v is the
velocity of the umbrella. You might be tempted to write −cv 2 instead but
then the force would always be acting downward, regardless of whether
the umbrella is moving up or down (as it happens the umbrella is always
moving down in this example but that need not be the case in general).
So to get the signs right the force due to air resistance is −cv|v|. Adding
the two forces we get
ma = −mg − cv|v|
which gives the ODE governing motion
mx′′ = −mg − cx′ |x′ |
where x(t) is the height of the umbrella at time t. The initial conditions
are x(0) = 80 and x′ (0) = 0 since the umbrella is dropped from a height
of 80 meters.
7. Consider the differential equation x′ = xet − 2et .
(a)[3 points] Find a constant solution without solving.
Constant solution x(t) = c implies 0 = cet − 2et . This means et (c − 2) = 0
so c = 2 and the constant solution is x(t) = 2.
(b)[10 points] Find a solution satisfying x(0) = 1.
This ODE is linear so we write it as x′ − xet = −2et . The integrating
factor is
R
t
t
µ = e −e dt = e−e
Multiplying by µ and integrating we get
Z
t
t
t
e−e x = −2et e−e = 2e−e + C
t
0
This gives x(t) = 2 + Cee . The condition x(0) = 1 implies 2 + Cee =
2 + Ce1 = 1 so that C = − 1e .