MATH 2250—Midterm 1
SOLUTIONS
Thursday, October 2, 2008, 12:25–2:10PM
• Write your name and ID number at the top of this page.
• Show all your work.
• You may refer to one double-sided sheet of notes during the exam and
nothing else.
• Calculators are not allowed.
—Begin Midterm—
1. Find a general solution y(x) to the following differential equation,
y + xy ′ = 2e2x
SOLUTION:
The equation is not separable, but it is linear. In fact, the left-hand
side is already the derivative of the product xy. If you didn’t notice
that, you would start by putting the equation in standard linear form
by multiplying both sides by x−1 :
y ′ + x−1 y = 2x−1 e2x
Now the integrating factor is
R
x−1
e
= eln x = x
Multiply both sides of the d.e. by the integrating factor:
xy ′ + y = 2e2x
Now you see we’re back to the original equation! We proceed as follows:
d
(xy) = 2e2x
dx
xy =
Z
2e2x dx = e2x + C
y=
e2x + C
x
2. Find the solution y(x) to the following initial value problem:
y′
e sin x +
= e2y ,
cos x
2y
y(0) = 3
SOLUTION:
The equation is not linear, but it is separable:
y′
= e2y − e2y sin x
cos x
y′
= e2y (1 − sin x)
cos x
y ′ = e2y cos x(1 − sin x)
e−2y y ′ = cos x(1 − sin x)
Z
Z
−2y
e
dy =
e−2y dy =
Z
Z
(cos x − cos x sin x) dx
cos x dx −
Z
cos x sin x dx
There are a few ways to do the last integral (product of sine and cosine),
all of which should lead to equivalent answers in the end. One way is to
use
the substitution u = sin x, du = cos x dx, which makes the integral
R
u du = 21 u2 + C.
1
1
− e−2y = sin x − sin2 x + C1
2
2
e−2y = −2 sin x + sin2 x + C2
(where C2 = −2C1 )
−2y = ln(−2 sin x + sin2 x + C2 )
1
y = − ln(−2 sin x + sin2 x + C2 )
2
1
y(0) = − ln(0 + 0 + C2 ) = 3
2
ln(C2 ) = −6
C2 = e−6
1
y = − ln(−2 sin x + sin2 x + e−6 )
2
3. Find the solution y(x) to each of the following initial value problems:
(a) x3 y ′′′ = 1, y(1) = 2,
SOLUTION:
y ′ (1) = 2,
y ′′ (1) = 0
y ′′′ = x−3
1
x−3 dx = − x−2 + C1
2
1
1
C1 =
y ′′ (1) = − + C1 = 0
2
2
1
1
y ′′ = − x−2 +
2
2
Z 1
1
1
1
y′ =
− x−2 +
dx = x−1 + x + C2
2
2
2
2
1 1
C2 = 1
y ′ (1) = + + C2 = 2
2 2
1
1
y ′ = x−1 + x + 1
2
2
Z 1
1
1 −1 1
x + x + 1 dx = ln x + x2 + x + C3
y=
2
2
2
4
3
1
C3 =
y(1) = 0 + + 1 + C3 = 2
4
4
1
1 2
3
y = ln x + x + x +
2
4
4
a+c
dy
y, y(0) = d (a, b, c, and d are constants > 0)
=
(b)
dx
b
′′
y =
Z
SOLUTION:
) is just a positive constant. If we call that exThe expression ( a+c
b
dy
pression k, we are left with dx
= ky, which is the common exponential
equation that we’ve solved many times. The general solution is
y = Cekx ,
So our solution is
where y(0) = C
a+c
y = de( b )x
4. Given:


0 0 3


A= 0 2 0 
1 0 0


−1


b =  −1 
2
Do each of the following calculations, if possible. If the calculation is
not possible, explain why.
(a) Calculate the product Ab
SOLUTION:




6
0 · −1 + 0 · −1 + 3 · 2



Ab = 
 0 · −1 + 2 · −1 + 0 · 2  =  −2 
−1
1 · −1 + 0 · −1 + 0 · 2
(b) Calculate the product bA
SOLUTION:
The product bA is not defined, because the number of columns of
b (1) is not equal to the number of rows of A (3).
(c) Calculate the inverse A−1
SOLUTION:
We place the 3 × 3 identity matrix I next to A to make a 3 × 6
matrix, and then we perform row operations to convert the left
half of the new matrix to I. Then the right half of the matrix will
be A−1 . In the steps below, I first swapped row 1 with row 3, and
then multiplied row 2 and row 3 by 12 and 13 , respectively.






1 0 0 0 0 1
1 0 0 0 0 1
0 0 3 1 0 0






1
 0 2 0 0 1 0 → 0 2 0 0 1 0 → 0 1 0 0 2 0 
0 0 1 13 0 0
0 0 3 1 0 0
1 0 0 0 0 1
So we have
A−1


0 0 1


=  0 12 0 
1
0 0
3
5. (a) Show that the following system of linear equations,





1
x1
2 −1 7





A =  −2 1 −7   x2  =  −1 
−2
x3
−4 2 −14
has an infinite number of solutions, and express the entire set of
solutions in vector form.
SOLUTION:
We perform Gaussian elimination on the augmented matrix to put
it into echelon form. We add rows 1 and 2, and also add twice
row 1 to row 3:




2 −1 7 1
2 −1 7
1



−7 −1 
→ 0 0 0 0 
 −2 1
0 0 0 0
−4 2 −14 −2
In this form, x2 and x3 are free variables, which means they can
be any number, so there are infinite solutions. We can let x2 = r
and x3 = t, and then use the remaining top row to solve for x1 ,
and we get x1 = 12 + 12 r − 27 t. In vector form,



x1



 x2  = 
x3
1
2
1
1
− 72
+ 21 r − 27 t

 2 


 2 
r
 =  0  + r 1  + t 0 
0
0
1
t







(b) Find a way to change the third column of the matrix in the above
problem so that the system has a unique solution, or explain why
it is not possible to do so.
SOLUTION:
If we change the elements of the third column to some numbers
a, b, and c, then Gaussian elimination gives us




2 −1
a
1
2 −1 a 1



b −1  →  0 0 a + b 0 

 −2 1
0 0 2a + c 0
−4 2 c −2
If a + b 6= 0 or 2a + c 6= 0, then x3 = 0 will be the unique solution
for x3 . However, x2 would still be a free variable, so the entire
system will still have infinite solutions for any a, b, c.
6. Newton’s Law of Cooling says that the rate of change of the temperature of an object at time t is proportional to the difference between the
ambient temperature at time t and the object’s temperature at time t.
Let’s call the constant of proportionality k. Consider an object sitting
in a broken freezer, where both the object and the (ambient) freezer
temperatures are 100 degrees. At time zero, the freezer is fixed, and the
ambient temperature begins dropping at a constant rate of one degree
per minute. Find a formula for the temperature of the object at time
t. Leave k unspecified in your answer. The following integral-table
formula may be useful:
Z
1
1
ueau du = ueau − 2 eau + C
a
a
SOLUTION:
Let u(t) be the temperature of the object and A(t) be the (ambient)
temperature of the freezer, t minutes after the freezer is fixed. Newton’s
law of cooling says
du
= k(A(t) − u)
dt
The freezer temperature is 100 degrees at time zero and drops one
degree per minute, so A(t) = 100 − t. Now we have the linear equation
du
du
= k(100 − t − u) →
+ ku = k(100 − t)
dt
dt
du
+ kekt u = 100kekt − ktekt
ekt
dt
Z
Z
ekt u = 100k
ekt dt − k
tekt dt
1 kt
1
1 kt
e u = 100k
−k
e
te − 2 ekt + C
k
k
k
1
ekt u = 100ekt − tekt + ekt + C
k
1
u(t) = 100 − t + + Ce−kt
k
1
1
u(0) = 100 − 0 + + C = 100
C=−
k
k
1 1 −kt
u(t) = 100 − t + − e
k k
kt
7. In population dynamics, an Allee effect occurs when populations struggle to recover from small numbers. For example, if the population is
too small, it may be exceedingly difficult to find a mate, or species
that rely on group defense may be more vulnerable to predation at low
levels. The following model is an alteration to the logistic model that
captures the Allee effect:
dP
= P (P − 10)(1000 − P )
dt
For this equation, find all of the critical points (equilibria), determine
their stability, and draw a phase diagram. Finally, explain how this
model is consistent with the Allee effect described above.
SOLUTION:
Equlibrium solutions occur when dP
= 0, which is true when P = 0,
dt
P = 10, or P = 100. If we check the sign of dP
for different ranges of
dt
P we can construct a phase diagram:
→ 0 ← 10 →→→ 1000 ←
which demonstrates that P = 0 and P = 1000 are stable equilibria and
P = 10 is an unstable equlibrium.
The Allee effect is exhibited when 0 < P < 10. In this range of
small population levels, the population decreases toward extinction.
For P > 10, the population is large enough to avoid the disadvantages
of being small and the dynamics behave like the logistic equation we
studied in class.