Exam 3
Math 216
Fall 2012
Name: ___________________ This is a closed book exam. You may use a calculator and the
formulas handed out with the exam. Show all work and explain any reasoning which is not clear
from the computations. (This is particularly important if I am to be able to give part credit.) Turn in
this exam with your answers.
1.
2.
3.
Consider the mass-spring system at the right. Assume the following.
i. m is the mass.
ii. The spring is linear with spring constant k.
k
iii. x = x(t) is the displacement from equilibrium at time t.
iv. The force of air resistance is proportional to the velocity with
proportionality constant b and is directed in the direction
opposite to the velocity.
m
v. The only forces that have to be taken into account are the
force of the spring and the force of air resistance. The force
of gravity can be neglected.
vi. k = 2 and b = 4.
a. (8 points) What is the differential equation that x(t) satisfies? Show how to derive it
from basic principles of physics.
b. (9 points) For what range of values of m is the system overdamped.
c. (8 points) Find a formula for the general solution to the equation in part a. when m is in
range in your answer to part b. Your answer will be a formula for x in terms of t and m.
Express your answer in terms of real valued functions.
d2u
du
(25 points) One solution to the differential equation t dt2 - t(t + 2) dt + (t + 2)u = 0 is
u1 = t. Find a formula for the general solution. You should be able to do the integrals that
arise.
2
(25 points) Find the general solution to the equation
d2u
du
2 = sin(2t) - 2
dt
dt - 10 u.
Express your answer in terms of real valued functions.
4.
d2x
(25 points) Find the general solution to the equation t2 dt2 - 6x = t2. You should be able to
do the integrals that arise. Hint: the homogeneous equation is an Euler equation.
Solutions
1.
a. Newton’s 2 law says (mass)(acceleration) = (Total force).
nd
In this case it becomes m
one gets m
d 2x
dt2
+4
dx
d2
d 2x
dt2
= Fspring + Fresistance = - kx – b
dx
d2
} 2 points
or m
d 2x
dt2
+b
dx
d2
+ kx = 0. Putting in k = 2 and b = 4
+ 2x =0.
} 6 points
b. To solve we try x = ert. Plugging in we get mr2 + 4r + 2 = 0.
Using the quadratic formula one obtains r =
-4
16 - (4)m(2)
2m
} 2 points
=
-4
16 - 8m
2m
=
-2
4 - 2m
.
m
} 3 points
The overdamped case is when the roots are real and distinct. This is when 4 – 2m > 0 or m < 2.
c. x = Ae-(2 +
2.
4 - 2m)t/m
+ Be-(2 -
4 - 2m)t/m
e-(q/p)dt
dt
(u1)2
} 4 points
e-(-(t+2)/t)dt
dt
t2
} 4 points
A second solution is u2 = u1 

Plugging in we get u2 = t 

Doing the integral in the exponent we get u2 = t 

Simplifying and doing the integral we get u2 = t 
e(1+2/t)dt
e(t+2 log t)dt
dt = t 
dt

2
t
t2
} 5 points
2
ete(log t )
ett2
dt = t  2 dt = t  et dt = tet.
2
t
t
} 8 points
t
So the general solution is u = At + Bte .
3.
} 4 points
} 4 points
rt
2
First solve the homogeneous equation u + 2u + 10 u = 0. Try u = e . Plug in to get r + 2r + 10 = 0. Using the
quadratic formula one obtains r =
-2
4 - (4)(1)(10)
2
=
-2
4 - 40
2
-t
=
-2
- 36
2
=
- 2  6i
2
= - 1  3i.
} 5 points
-t
So the solution to the homogeneous equation is u = Ae cos(3t) + Be sin(3t).
} 5 points
Next find a particular solution. The equation can be written as u'' + 2u' + 10u = Im[e2it]. So u = Im[v] where v is
the solution to v'' + 2v' + 10v = e2it. Try v = Ce2it. Plug in to get – 4Ce2it + 4iCe2it + 10Ce2it = e2it. So (6 + 4i)C = 1
1
or C =
.
} 6 points
6 + 4i
e2it
e2it
cos(2t) + i sin(2t)
cos(2t) + i sin(2t) 6 - 4i
So v =
. So u = Im[
] = Im[
] = Im[
6 + 4i
6 + 4i
6 + 4i
6 + 4i

 6 - 4i] =
6 cos(2t) - 4i cos(2t) + 6i sin(2t) + 4 sin(2t) - 4 cos(2t) + 6 sin(2t) - 2 cos(2t) + 3 sin(2t)
Im[
]=
=
.
} 5 points
52
52
26
- 2 cos(2t) + 3 sin(2t)
Add the general solution to the homogeneous to get u =
+ Ae-tcos(3t) + Be-tsin(3t).
} 4 points
26
4.
First solve the homogeneous equation. Try x = tr. Plug in to get r(r – 1) – 6 = 0 or r2 – r – 6 = 0 or (r –3)(r + 2) =0.
So r = 3 and r = - 2.
} 5 points
1
So two solutions to the homogeneous equation are u1 = t3 and u2 = t-2 = 2 and the general solution to the
t
B
homogenous equation is u = At3 + 2.
} 4 points
t
To find a particular solution use the variation of parameters formula u = - u1 
u2 f
 Wp
dt + u2 
u2  t3
t-2 
u1
W=
= 
 = - 2 – 3 = - 5.
(u1)' (u2)' 3t2 - 2t-3
t-2 t2
t3 t2
-2 
dt.
2 dt + t
 (-5)t
 (-5)t2
Plug into the formula above to get u = - t3 
t3
t-2
t3 t-1 t-2 t4
t2 t2
t2
t-2 dt - 
t3 dt =   -   = - =- .



5
5
5 -1 5  4 
5 20
4
t2
B
Add the general solution to the homogeneous to get u = - + At3 + 2.
4
t
Doing the integrals we get u =
u1 f
 Wp
dt.
} 3 points
} 3 points
} 3 points
} 3 points
} 4 points