C Roettger, Fall 13
Math 267 - Practice exam 1 - solutions
Problem 1 Have a look at the whole exam paper. Which of the numbered
differential equations is/are linear? (just list their numbers).
Solution. Only (3), (5), (17) (new consecutive numbering, because there
are now more equations). The numbers on the exam itself read (3), (5),
(13).
Problem 2 For which of the following initial value problems does Theorem
1.2.1 (section 1.2 of Chapter 1) guarantee existence and uniqueness of a
solution, at least in a small rectangle? Justify your answers.
√
y(0) = 2
(1)
y 0 = √4 − x2 + |y|
0
2
y = √4 − x + |y|
y(2) = 0
(2)
3
0
2
y(2) = 0
(3)
y = p4 − x
0
2
4 − y + |x|
y(0) = 1
(4)
y =
Solution. Theorem 1 applies to (1), (3), and (4). Checks for (1): f (x, y) =
√
4 − x2 +|y| is continuous near x = 0, y = 2. The same is true for the partial
derivative (note that |y| = √
y near y = 2, having derivative 1).
Checks for (2): f (x, y) = 4 − x2 + |y| is not even defined on the right of
x = 2, so not continuous there. Also, ∂f /∂y is not defined near y = 0 because
of the absolute value.
√
Checks for (3): f (x, y) = 3 4 − x2 is continuous everywhere. The derivative
∂f /∂x is not, but that does not matter, we only need ∂f /∂y which is constant
zero, so continuous.
p
Checks for (4): f (x, y) = 4 − y 2 + |x| is continuouspnear x = 0, y = 1. The
same is true for the partial derivative ∂f /∂y = −y/ 4 − y 2 .
Problem 3 Consider the differential equation
2
xy 0 + 2x2 y = (x + 5)e−x .
(5)
a) Find the general solution. Write down intermediate steps explaining the
method you are using - the result alone is not enough.
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b) Solve the initial value problem consisting of (5) with initial condition
y(1) = 2e−1 .
c) What can you say about existence and uniqueness of solutions of an initial
value problem (IVP) involving (5) and a condition y(a) = b? What is the
biggest interval where you can be sure that solutions of such an IVP are
defined?
Solution. a) This equation is linear. But to apply the recipe, you have to
divide by x first (y 0 has to be isolated)! You get
x + 5 −x2
e .
x
y 0 + 2xy =
(6)
Then the recipe tells us to put
µ(x) := e
R
2x dx
2
= ex ,
to multiply equation (6) by µ(x) and then integrate. We get
Z
x+5
µ(x)y =
dx = x + 5 ln |x| + C.
x
(7)
Solve for y to get
2
y(x) = e−x (x + 5 ln |x| + C).
(8)
b) Substitute x = 1, y = 2e−1 into equation (8) to get 2/e = e−1 (1 + C), so
2
C = 1 and y(x) = e−x (x + ln |x| + 1).
c) To apply theorem 2.4.1, we already need to divide the original equation
2
by x. Looking at e−x (x + 5)/x, the right-hand side of equation (6), we have
a discontinuity only at x = 0. So a solution to the equation exists for every
initial condition y(a) = b provided a 6= 0 and is unique. Because the equation
is linear, these solutions can be defined either on (−∞, 0) or on (0, ∞).
Problem 4 Consider the differential equation
y 3 + cos(x)y 4 + [4y 3 + 4 sin(x)y 3 + 3xy 2 ]
dy
= 0.
dx
(9)
a) Show that this equation is exact by performing the test for exactness.
b) Find an implicit solution of equation (9).
Solution.
a) Define M (x, y) := y 3 + cos(x)y 4 and N (x, y) := 4y 3 +
3
4 sin(x)y + 3xy 2 . Then
∂
∂
M = 3y 2 + 4 cos(x)y 3 =
N,
∂y
∂x
2
(10)
so the equation is exact
by theorem 2.4.2.
R
b) Put F (x, y) := M dx = xy 3 + sin(x)y 4 + g(y) with some function g.
Compare Fy to N :
d
g
dy
N = 4y 3 + 4 sin(x)y 3 + 3xy 2
Fy = 3xy 2 + 4 sin(x)y 3 +
So g 0 (y) = 4y 3 , and g(y) = y 4 + D. The implicit solution is F (x, y) = C
where
F (x, y) = xy 3 + sin(x)y 4 + y 4 .
Problem 5 Which of the following differential equations belongs to the direction field plotted below?
p0 = p(2 − p),
p0 = p2 − 3p,
p0 = p2 − 4p.
(11)
(12)
(13)
For initial conditions P (0) = 0, 1, 2, 3, 3.9, 4, 4.1, make a rough sketch of
solution curves.
Solution.
It is p0 = p2 − 4p. You can tell from the horizontal arrows at
level p = 4, corresponding to p0 = 0. Note that the t-axis (p = 0) would also
have horizontal arrows, if the sketch had drawn them - and indeed, there is
a constant solution p = 0 and another p = 4.
I will not ask you to sketch better than what I can do by hand. But it’s nice
to see what the solutions really look like!
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Note that the solution for initial condition y(0) = 4.1 seems to tend to infinity
at about t = 0.9. The vertical line is a sign that the computer approximation
failed at this point.
Some biological populations follow a model like this. It is called explosionextinction model: for all initial values up to a threshold, the population
collapses. Above the threshold, the population ’explodes’ in finite time.
Problem 6 Consider the differential equations
p
a) y 0 = x + y − 1
b) xy 0 + 3y = 6xy 4
c) (x − y)y 0 = 2x + y
(14)
(15)
(16)
Give substitutions v = v(x, y) for each of them transforming them into separable or linear equations. Write down the definition of v and the differential
equation for v which is separable/linear - do NOT solve them.
Solution. a) Here, the right-hand side is a function of v = x + y − 1, so
y = v − x + 1 and y 0 = v 0 − 1. The given equation becomes
√
v 0 = v + 1,
which is separable.
b) This is a Bernoulli equation with n = 4, so use v = y −3 . You get
dv
9v
= −3y −4 (6y 4 − 3y/x) = −18 +
dx
x
4
which is linear.
c) This is a homogeneous equation, so use v = y/x. Then
2+v
y0 =
1−v
and consequently,
dv
y0 − v
2 + v − v(1 − v)
2 + v2
=
=
=
dx
x
x(1 − v)
x(1 − v)
which is separable.
Problem 7 A leaf floats gently down from a tree. Its vertical velocity v
obeys the equation
dv
= −δv − g
(17)
dt
where −g = −10m/s2 is the acceleration due to gravity and the drag coefficient δ = 5 models air resistance. The tree is 20 meters high and the leaf
has velocity zero initially.
a) Find the solution of this initial value problem for v(t).
b) Integrate once more to find the position at time t.
c) Bonus: estimate how long it takes for the leaf to reach the ground. Hint
– compute the height above ground at t = 10 and t = 10.5 seconds.
Solution. a) solve the given linear equation for v: the integrating factor
is
R
µ(t) = e δ dt = eδt
and therefore
Z
g
gµ dt = − eδt + C
δ
g
v = Ce−δt −
δ
g
0 = C−
δ
where in the last step, we have substituted t = 0 and v = 0, so C = g/δ = 2
and v = 2e−5t − 2.
b) Integrate once more to get the vertical position function s,
Z t
s(t) =
v(τ ) dτ + 20
vµ = −
0
= 20 −
2 −5t
e − 1 − 2t
5
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c) The formula above gives s(10) > 0, so we can be sure the leaf reaches the
ground only after t = 10 (of course, the model is not valid anymore when
the leaf reaches the ground). But t = 10.5 gives s(t) = −0.6 < 0, so the leaf
reaches the ground between these times. Actually, t = 10.2 is for all intents
and purposes an exact solution, because e−50 is so small.
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