Be sure this exam has 6 pages including the cover
The University of British Columbia
MATH 215/255, Section 104
Midterm Exam I – October 2014
Name
Signature
Student Number
This exam consists of 4 questions worth 10 marks each. No notes nor calculators.
Problem
1.
max score
10
2.
10
3.
10
4.
10
total
40
score
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2. Read and observe the following rules:
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during the first half hour of the examination.
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in examination questions.
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examination and shall be liable to disciplinary action.
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(b) Speaking or communicating with other candidates.
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October 2014
Math 215/255 Midterm 1, Section 104
Page 2 of 6
(10 points) 1. Consider a fish population in a lake. Suppose some toxic substance flows into the lake so that
the rate of growth of fish is slowing down. We model such a population dynamics by
dp
= (12 − 3t2 )p.
dt
a) Find the general solution to the equation.
b) What is the maximum fish population and when does it occur if p(0) = 500?
a) This equation is separable. p(x) ≡ 0 is a constant solution. To find other solutions, we
compute
dp
= (12 − 3t2 )p
dt
1 dp
⇒
= 12 − 3t2
p dt
d
⇒
(ln |p|) = 12 − 3t2
dt
⇒ ln |p| = 12t − t3 + c1
3
⇒ p = ce12t−t ,
where c1 is a constant, and c = ec1 or −ec1 .
3
b) If p(0) = 500, then c = 500. Thus p(t) = 500e12t−t is the solution. Note that
3
p′ (t) = 500(12 − 3t2 )e12t−t and p′ (t) = 0 if and only if t = ±2. As p′ (t) > 0 for t ∈ (−2, 2),
and p′ (t) < 0 for t ∈ (2, ∞), we obtain p(t) ≤ p(2) for all t ∈ [0, ∞). So the maximum fish
population occurs at t = 2, with the maximal value p(2) = 500e16 .
October 2014
Math 215/255 Midterm 1, Section 104
(10 points) 2. a) Sketch the slope field and some typical solutions for
b) Draw the phase diagram for the equation
equilibrium points.
dx
dt
dx
dt
Page 3 of 6
= −x2 + x + 2.
= −x2 + x + 2 and indicate the stability of
2
c) For dx
dt = −x + x + 2 with initial condition x(0) = 0, use Euler’s method with step size
h = 0.5 to approximate x(1).
d) Find all possible numbers a so that the equation
dx
= ax2 + x + 2
dt
has two equilibrium points and the larger equilibrium is stable. Hint: Sketch parabola for
different values of a.
a)
5
4
3
2
1
0
−1
−2
−3
−4
−5
0
2
4
6
8
b)
x
2
-1
Stable
unstable
10
12
October 2014
Math 215/255 Midterm 1, Section 104
Page 4 of 6
c) Euler’s formula: xi+1 = xi + hf (ti , xi ). Thus, x1 = 0 + 0.5 · 2 = 1, t1 = 0 + 0.5.
x2 = 1 + 0.5 · 2 = 2.
d) ax2 + x + 2 has two real roots if and only if a < 1/8, and a ̸= 0. The graph of ax2 + x + 2
is concave up if a > 0, and concave down if a < 0. Suppose a < 0 and the two roots of
ax2 + x + 2 are denoted by r1 , r2 with r1 < r2 , then ax2 + x + 2 > 0 if r1 < x < r2 , and
ax2 + x + 2 < 0 if x > r2 . Accordingly, the equation has two equilibrium points r1 and r2 , and
the larger equilibrium r2 is stable. If 0 < a < 1/8, the larger equilibrium r2 is unstable.
October 2014
Math 215/255 Midterm 1, Section 104
Page 5 of 6
(10 points) 3. Show that r(x, y) = (xy)−1 is an integrating factor for
(3xy 3 ex +x)y ′ + 2x2 y 4 ex +y = 0,
2
2
and solve the equation for y(1) = 1.
The functions N (x, y) and M (x, y) are given respectively by
2
2
N (x, y) = 3xy 3 ex +x,
M (x, y) = 2x2 y 4 ex +y.
To show that r(x, y) = (xy)−1 is an integration factor, we need to calculate
)
)
∂rN
∂ (
∂ ( 2 x2
2
2
=
r(x, y)(3xy 3 ex +x) =
3y e +y −1 = 6xy 2 ex ,
∂x
∂x
∂x
and
)
)
∂ (
∂rM
∂ (
2
2
2
r(x, y)(2x2 y 4 ex +y) =
2xy 3 ex +x−1 = 6xy 2 ex .
=
∂y
∂y
∂y
So, since the above derivatives are equal, equation
(3y 2 ex +y −1 )y ′ + 2xy 3 ex +x−1 = 0,
2
2
is exact. To find function ϕ(x, y), we need to integrate its derivatives
∂ϕ
2
= rN = 3y 2 ex +y −1
∂y
⇒
∂ϕ
2
= rM = 2xy 3 ex +x−1
∂x
It is clear that
⇒
2
ϕ = y 3 ex + ln |y| + f (x),
2
ϕ = y 3 ex + ln |x| + g(y).
2
ϕ = y 3 ex + ln |xy| = C,
is a general solution. The initial condition y(1) = 1 allows to find unknown constant as
C = e.
Since both x and y are positive, the solution is given by an implicit formula
2
y 3 ex + ln(xy) = e.
October 2014
Math 215/255 Midterm 1, Section 104
(10 points) 4. a) Find a general solution to
y ′′ + 2y ′ − 3y = 0.
This second order differential equation with constant coefficients has the following
characteristic equation
r2 + 2r − 3 = 0.
The roots of this equation are r1 = 1, r2 = −3. The general solution is then
y = C1 ex + C2 e−3x ,
where C1 and C2 are constants.
b) Solve
9y ′′ − 12y ′ + 4y = 0,
y(0) = 1,
The characteristic equation for this equation is
9r2 − 12r + 4 = 0.
It has a repeated root r = 2/3. The general solution is
y = C1 e2x/3 + C2 xe2x/3 .
From the initial conditions, we can find that
C1 = 1,
2
C1 + C2 = 0.
3
The constants are C1 = 1, C2 = − 32 , and the solution is
2
y = e2x/3 − xe2x/3 .
3
y ′ (0) = 0.
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