Math 2280
Summer 2009
Test 1
You will have one hour to complete this test. You may not use a calculator, computer, or any other electronic
device on this test. Please be sure to show as much of your work as possible, as I will determine your score on a
given problem from the work and not necessarily by the answer. Please be neat so that I can grade your answers
easily.
1. Solve the initial value problem
dy
= xe−x , y(0) = 1.
dx
In this equation, the right hand side depends only on x, so we can simply integrate
Z
Z
dy
dx = xe−x dx.
y=
dx
In order to evaluate the integral on the right, we must use integration by parts
Z
Z
−x
−x
xe dx = −xe + e−x dx = −xe−x − e−x + C = −e−x (x + 1) + C.
The initial condition implies that
1 = y(0) = −1 + C ⇒ C = 2.
Therefore,
y(x) = −e−x (x + 1) + 2.
2. Find the general solution of the equation
dy
√
= 3 xy.
dx
This equation is separable, so we separate and integrate to get
Z
Z
√
dy
√ = 3 x dx.
y
Upon performing the integration, we get
1
3
3
2y 2 = 2x 2 + C ⇒ y = (x 2 + C)2 .
3. A 400 gallon tank initially contains 100 gallons of brine containing 50 pounds of salt. Brine containing 1 pound
of salt per gallon enters the tank at a rate of 5 gallons per second, and the well-mixed brine in the tank flows out
at the rate of 3 gallons per second. Find the amount of salt in the tank as a function of time.
If x(t) is the amount of salt in the tank at time t, then
x(t)
3
dx
= rate salt flows in − rate salt flows out = 1 · 5 −
·3 = 5−
x.
dt
V (t)
100 + 2t
If we rewrite this as
dx
3
+
x = 5,
dt
100 + 2t
then it can easily be seen that the proper integrating factor is
Z
3
3
3
dt = exp
ln(100 + 2t) = (100 + 2t) 2 .
ρ(t) = exp
100 + 2t
2
1
Multiplying through by the integrating factor gives us
3
(100 + 2t) 2
3
1
dx
+ 3(100 = 2t) 2 x = 5(100 + 2t) 2 .
dt
The left hand side is now recognizable as a derivative, so we have
i
h
3
3
Dt (100 + 2t) 2 x = 5(100 + 2t) 2 ,
and when we integrate we obtain
3
3
5
(100 + 2t) 2 x = (100 + 2t) 2 + C ⇒ x(t) = (100 + 2t) + C(100 + 2t)− 2 .
Our initial condition is x(0) = 50, so
3
3
50 = 100 + C(100)− 2 ⇒ C = −50(100)− 2 = −50, 000.
The particular solution to the equation is then
3
x(t) = (100 + 2t) − 50000(100 + 2t)− 2 .
4. Solve the logistic initial value problem
dx
= 7x(x − 13), x(0) = 17.
dt
As with all logistic equations, this one is separable, and when we rewrite it and integrate it becomes
Z
Z
dx
= 7 dt.
x(x − 13)
The integral on the left requires the use of a partial fraction decomposition. We write
1
A
B
= +
x(x − 13)
x
x − 13
⇒ 1 = A(x − 13) + Bx = (A + B)x − 13A ⇒ −13A = 1, A + B = 0 ⇒ A = −
1
1
, B=
.
13
13
The integral on the left becomes
Z
Z
1
1
1
1
x − 13
1
,
dx +
dx = − ln x +
ln(x − 13) =
ln
−
13x
13(x − 13)
13
13
13
x
so
1
ln
13
x − 13
x
= 7t + C.
This is a good time to apply the initial condition x(0) = 17. Upon doing so, we find
1
4
C=
.
ln
13
17
We can solve the implicit equation for x as follows:
x − 13
x − 13
13
4 91t
ln
= 91t + 13C ⇒
= e91t e13C ⇒ 1 −
=
e
x
x
x
17
⇒ −
4 91t
x
13
=
e −1 ⇒
=
x
17
−13
2
1
4 91t
17 e
−1
⇒ x=
−13
.
−1
4 91t
17 e
5. Find and classify the equilibrium solutions to the equation
dx
= 7x − x2 − 10.
dt
The equilibrium solutions are solutions to the equation
2
9
49
7
7
3
7 3
49
=
= −10 +
⇒ x−
⇒ x − = ± ⇒ x = ± = 5 or 2.
−x2 + 7x − 10 = 0 ⇒ x2 − 7x +
4
4
2
4
2
2
2 2
The graph of f (x) = −x2 + 7x − 10 looks like
2
4
6
8
-5
-10
-15
-20
By analyzing the sign of f (x) near the critical points, we and see that x = 2 is unstable and x = 5 is stable.
6. Suppose that a motorboat is moving at 40 ft/sec when its motor suddenly quits, and that 10 sec later the boat
has slowed to 20 ft/sec. Assume that the resistance it encounters while coasting is proportional to its velocity.
Find an equation for the distance the boat coasts as a function of time.
The only force acting on the boat is the resistance, so Newton’s second law says
m
dv
= −kv.
dt
The solution to this equation is
k
v(t) = Ce− m t
for the appropriate choice of the constant C. From the infromation provided in the problem, we have that
v(0) = 40 ft/sec, which implies that
40 = v(0) = C.
The other condition we are given implies that v(10) = 20ft/sec. Therefore,
k
20 = v(10) = 40e− m (10) ⇒
The position function is
x(t) = −
1
1
k
1
k
= e−10 m ⇒ − ln = .
2
10 2
m
40m − k t
e m + C1 ,
k
and the initial condition x(0) = 0 implies that C1 = 40m
k . Therefore, the distance the boat has coasted as a
function of time is
40m − k t 40m
e m +
.
x(t) = −
k
k
7. Apply Euler’s method to approximate the solution to the initial value problem
y ′ = x − y, y(0) = 1
on the interval [0, 1/2] with stepsize h = 1/4.
The initial conditions imply that
x0 = 0, y0 = 1.
3
The formula for Euler’s method is yn+1 = yn + hf (xn , yn ), xn+1 = xn + h, so
1
3
1
y1 = 1 + (0 − 1) =
4
4
4
5
3 1 1 3
1
= .
y2 = +
−
x2 =
2
4 4 4 4
8
x1 =
8. Find the solution to the initial value problem
y ′′ + y ′ = 0, y(0) = −2, y ′ (0) = 8.
The characteristic equation for this second order linear equation is
r2 = r = 0 ⇒ r(r + 1) = 0 ⇒ r = 0 or r = 1,
so two linearly independent solutions are
y1 (x) = e0x = 1 and y2 (x) = e−x .
The general solution to the equation is
y(x) = c1 + c2 e−x .
To solve the IVP, we must choose c1 and c2 so that y(0) = −2 and y ′ (0) = 8. We have that y ′ (x) = −c2 e−x , so
c1 and c2 must satisfy
c1 + c2 = −2
⇒ c2 = −8, c1 = 6.
−c2 = 8
Thus, our particular solution is y(x) = 6 − 8e−x .
9. Find the general solution to the equation
y ′′ − 10y ′ + 25y = 0.
The characteristic equation is
r2 − 10r + 25 = 0 ⇒ r2 − 10r + 25 = −25 + 25 ⇒ (r − 5)2 = 0 ⇒ r = 5.
We have only one root, so
y1 (x) = e5x
is one solution to our equation. As seen in class, another solution which is linearly independent with the first is
y2 (x) = xe5x . Therefore, our general solution is
y(x) = c1 e5x + c2 xe5x .
10. Show that
f (x) = 0, g(x) = sin x, and h(x) = ex
are linearly dependent by finding a linear combination that vanishes identically.
This is acutally easier than it sounds. Choose c1 to be any real number, for example, c1 = π. Let c2 = c3 = 0.
Then
c1 f (x) + c2 sin x + c3 ex = π(0) + (0) sin x + (0)ex = 0
for each x. This shows that the set {f, g, h} is linearly dependent.
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