Math 2280
Summer 2007
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. Find the position function x(t) of a moving particle with acceleration a(t) = 50 sin 5t, initial velocity v(0) =
−10, and initial position x(0) = 8.
Since v ′ (t) = a(t), we have that
v(t) =
Z
50 sin 5t dt + C = −10 cos 5t + C.
The initial condition implies that −10 = −10 + C, so C = 0. Since x(t) = v ′ (t), we have
Z
x(t) = −10 cos 5t dt + C = −2 sin 5t + C.
The initial condition for the position function gives us that C = 8. Therefore, the position function is given by
x(t) = −2 sin 5t + 8.
2. Find a general solution of the ODE y ′ = 2x sec y.
First note that the problem becomes a little easier when it is rewritten as
2x
dy
=
.
dx
cos y
We now seperate the variables and integrate to arrive at
Z
Z
cos y dy = 2x dx ⇒ sin y = x2 + C.
3. A tank initially contains 60 gal. of pure water. Brine containing 1 lb. of salt per gallon enters the tank at 2
gal./min. and the perfectly mixed solution leaves the tank at 3 gal./min; thus the tank is empty after exactly 1
h. Find the amount of salt in the tank after t minutes.
First, we have to set up an ODE from the information given in the problem. Recall from the lecture that the
aprropriate equation is
dx
x
= rate salt flows in − rate salt flows out = ri ci − ro ,
dt
V
where ri is the rate at which brine flows in, ci is the salt concentration in the incoming brine, and ro is the rate
at which brine leaves the tank. Substituting the correct values from the problem, we have
dx
3
=2−
x.
dt
60 − t
We know that dV /dt = ri − ro = −1, so V (t) = −t + V0 = 60 − t. Next, we recognize this as a first order linear
equation, so we rearrange it and identify P (t) = 3/(60 − t). This gives us an integrating factor of
Z
1
3
,
dt = exp(−3 ln(60 − t)) =
ρ(t) = exp
60 − t
(60 − t)3
1
Multiplying the equation through by the integrating factor gives us
1
3
2
x′ +
x=
.
(60 − t)3
(60 − t)4
(60 − t)3
We recognize the left hand side as a derivative:
1
2
Dt
x
=
.
(60 − t)3
(60 − t)3
Integrating both sides gives us
1
1
x=
+ C ⇒ x = (60 − t) + C(60 − t)3 .
(60 − t)3
(60 − t)2
The appropriate initial condition is x(0) = 0, which tells us that
0 = 60 + 603 C ⇒ C = −
Therefore, the solution is
x(t) = 60 − t −
60
1
=−
.
603
3600
(60 − t)3
.
3600
4. Given the equation
dx
= −x2 + 7x − 10,
dt
find the equilibrium solutions and determine whether or not each is stable. Then sketch a graph of several solution
curves with different initial conditions.
The equilibrium solutions are the solutions to the equation −x2 + 7x − 10 = 0. The quadratic formula tells us
that
p
−7 ± 49 − 4(−1)(−10)
7±3
=
= 5 or 2.
x=
2(−1)
2
A graph of dx/dt versus x is shown in Figure 1.
An analysis of the signs of dx/dt tells us that the solution x = 2 is unstable and the solution x = 5 is stable. A
sketch of some solution curves might look something like Figure 2.
5. Using Euler’s method, find an approximation to the solution y = y(x) of the initial value problem
y ′ = 2xy 2 , y(0) = 1
at the points x =
1
2
and x = 1 using step size h = 21 .
Euler’s method consists in applying the formula
yn+1 = yn + hf (xn , yn ).
The initial data gives us that x0 = 0 and y0 = 1. It then follows that x1 = x0 + h = 1/2 and
1
y1 = 1 + (2(0)(1)2 ) = 1.
2
We apply the same process one more time to find that x2 = 1 and
1
3
1
2
(1)2 = .
y2 = 1 +
2
2
2
6. Find the general solution of the ODE
2y ′′ − 12y ′ + 18y = 0.
2
2
1
x
3
2
4
5
6
0
–2
–4
–6
–8
–10
Figure 1: A graph of dx/dt versus x.
6
4
x(t)
2
t
0.1
0.05
0.15
–2
–4
–6
Figure 2: Some solutions of the equation.
3
0.2
Primes denote differentiation with respect to x.
First we note that this equation is second order linear with constant coefficients. This means that to solve the
given equation, we really need to solve the corresponding characteristic equation, which is
p
12 ± 144 − 4(2)(18)
12 ± 0
2
=
= 3.
2r − 12r + 18 = 0 ⇒ r =
2(2)
4
We have here a repeated root, so the general solution must be
y(x) = c1 e3x + c2 xe3x .
7. Prove or disprove: the functions
f (x) = 1, g(x) = cos 2x, h(x) = sin2 x
are linearly independent. If they are linearly dependent, find a nontrivial linear combination of f, g, and h that
vanishes identically on R.
To determine whether these functions
1
cos 2x
0 −2 sin 2x
0 −4 cos 2x
are linearly independent or not, we evaluate their Wronskian:
sin2 x sin 2x = 1(−4 sin 2x cos 2x + 4 sin 2x cos 2x) = 0.
2 cos 2x Since the Wronskian is identically zero, the functions are linearly dependent. This means that there must be
some linear combination of them that is identically zero. What this linear combination is might be clear, but in
case it’s not, here is how you can find it. We seek constants c1 , c2 , and c3 such that c1 + c2 cos 2x + c3 sin2 x = 0
for all x. Another way of thinking of this is that one of the functions can be written as a linear combination of
the others. For example,
cos 2x = cos2 x − sin2 x = (1 − sin2 x) − sin2 x = 1 − 2 sin2 x.
In other words, cos 2x + 2 sin2 x − 1 = 0, so c1 = −1, c2 = 1, and c3 = 2.
8. Given that y(x) = e3x is a solution of the ODE
y (3) − 6y ′′ + 11y ′ − 6y = 0,
find the general solution.
The characteristic polynomial for this equation is r3 − 6r2 + 11r − 6. This is a cubic equation and could possibly
be hard to factor, exept for the fact that we know that r = 3 is one of it’s roots. This means that (r − 3) divides
the characteristic polynomial evenly. After performing long division, we find that the quotient is
r2 − 3r + 2.
The quadratic formula tells us that
r=
3±
p
9 − 4(1)(2)
3±1
=
= 1 or 2.
2(1)
2
We now know all the roots of the characteristic polynomial, so the general solution is
y(x) = c1 e3x + c2 e2x + c1 ex .
9. Find the general solution of the equation
y (3) + 27y = 0.
4
The corresponding characteristic equaiton is
r3 + 27 = 0 ⇒ r3 = −27.
Therefore,
1
1
iπ
2
r = (−27) 3 = (27ei(π+2nπ) ) 3 = 3e 3 + 3 iπn .
For n = 0, we get
π
3ei 3 = 3(cos
√
π
π
3
3 3
+ i sin ) = + i
.
3
3
2
2
√
For n = 1 we have 3eiπ = −3. The other root must be the complex conjugate of the first, 3/2 − 3i 3/2. Thus,
the general solution of the equation is
√
√
3 3
3 3
3
3
y(x) = c1 e−3x + c2 e 2 x cos
x + c3 e 2 x sin
x.
2
2
5