Name________________________
Student I.D.___________________
Math 2250-010
Super Quiz 3
April 18, 2014 SOLUTIONS
1) Solve one of the two following Laplace transform problems. There is a Laplace transform table at the
end of the superquiz. If you try both problems, indicate clearly which one you wish to have graded.
(5 points)
1a) Use Laplace transform to solve this forced oscillation initial value problem.
2
0%t!1
x## t C 5 x# t C 6 x t =
0
tR1
x 0 =0
x# 0 = 0
Solution: Notice that we can rewrite the piecewise forcing function as f t = 2 K 2 u t K 1 . Then,
since the solution to this IVP makes both sides of the differential equation equal, their Laplace transforms
will be too, i.e.
2
2
s2 X s K 0 s K 0 C 5 sX s K 0 C 6 X s =
K eKs
s
s
Ks
1Ke 2
X s s2 C 5 s C 6 =
s
2
X s = 1KeKs
.
2
s C5 sC6 s
partial fractions:
2
A
B
C
=
C
C
.
sC3 sC2 s
sC2
sC3
s
Equating what the numerators would be if the RHS was written as a fraction using the denominator on the
LHS yields
2 = A s sC3 CB s sC2 CC sC2 sC3
@ s = 0,K2,K3 yields
1
2=6C0C=
3
2 =K2 A 0 A =K1
2
2=3B0B=
.
3
1
2
1
1 1
0 X s = 1KeKs K
C
C
sC2
3 sC3
3 s
The inverse Laplace transform of
1
2
1
1 1
K
C
C
sC2
3 sC3
3 s
2
1
is g t =KeK2 t C eK3 t C . Therefore, using the "translation" table entry, the inverse Laplace
3
3
transform of
1
2
1
1 1
C
C
sC2
3 sC3
3 s
eKs K
is u t K 1 g t K 1 , i.e.
u tK1
KeK2
tK1
C
2 K3
e
3
tK1
C
1
3
.
Thus the solution to the IVP is
2
1
2
1
x t =KeK2 t C eK3 t C
C u t K 1 KeK2 t K 1 C eK3 t K 1 C
3
3
3
3
> with DEtools : #tech check
dsolve x## t C 5$x# t C 6$x t = 2 K 2$Heaviside t K 1 , x 0 = 0, x# 0 = 0
2 K3 t
1
1
x t = KeK2 t C
e C
K
Heaviside t K 1 C Heaviside t K 1 eK2 t C 2
3
3
3
2
K
Heaviside t K 1 eK3 t C 3
3
> plot KeK2 t C
K
.
;
2 K3 t
1
1
e C
K
Heaviside t K 1 C Heaviside t K 1 eK2 t C 2
3
3
3
2
Heaviside t K 1 eK3 t C 3 , t = 0 ..5, color = black ; #solution plot
3
0.20
0
0
1
>
2
3
t
4
5
1b) Find an integral convolution formula that will work for any forcing function f t , for the solution
x t to
x## t C 5 x# t C 6 x t = f t
x 0 =0
x# 0 = 0
Solution: proceeding as in 1a, one finds that
1
1
X s = 2
F s =
F s =F s G s .
sC3 sC2
s C5 sC6
Since
1
1
1
G s =
=
K
sC3 sC2
sC2
sC3
the inverse Laplace transform
g t = eK2 t KeK3 t .
Therefore the solution to the IVP is given by
t
x t = f)g t =
0
f t g t K t dt =
t
f t
eK2
tKt
0
Since convolution is commutative, you could also write x t = g)f t .
KeK3
tKt
dt
(1)
2a) Find the eigenvalues and eigenvectors (eigenspace bases) for the following matrix
K3
1
2 K4
Hint: if you compute your characteristic polynomial correctly you will find that the eigenvalues are
negative integers.
Solution: Start by finding the characterstic polynomial and its roots, the eigenvalues:
K3Kl
1
2
K4Kl
= lC3
2
l C 4 K 2 = l C 7 l C 10 = l C 5
lC2 .
Thus the eigenvalues are l =K5,K2
El =K5 is the solution space to the homogeneous system with matrix
2 1
/
2 1
and a basis for this eigenspace is v = K1, 2
T
2 1
0 0
.
El =K2 is the solution space to the homogeneous system with matrix
K1
1
2 K2
and a basis for this eigenspace is v = 1, 1
T
/
K1 1
0 0
.
(3 points)
2b) Set m1 = 1, k1 = 2. Find values for the other mass m2 and for the other two spring constants k2 , k3 ,
so that the displacements x1 t , x2 t of the two masses from equilibrium in the configuration below
satisfy
x1 ## t =K3 x1 C x2
x2 ## t = 2 x1 K 4 x2
solution: From Newton's laws, the system of DE's for x1 t , x2 t are
m1 x1 ## t =Kk1 x1 C k2 x2 K x1 =K k1 C k2 x1 C k2 x2
m2 x2 ## t =Kk2 x2 K x1 K k3 x2 = k2 x1 K k2 C k3 x2
i.e.
k1 C k2
k2
x1 ## t =K
x1 C
x
m1
m1 2
k2
k2 C k3
x2 ## t =
x1 K
x2 .
m2
m2
Since m1 = 1 and k1 = 2 the first equation simplifies to
x1 ## t =K 2 C k2 x1 C k2 x2
which is supposed to be
x1 ## t =K3 x1 C x2
so k2 = 1. Thus the second DE simplifies to
1 C k3
1
x2 ## t =
x1 K
x2
m2
m2
which is supposed to be
x2 ## t = 2 x1 K 4 x2
1
so m2 = , k3 = 1.
2
(2 points)
2c) Find the general solution to
x1 ## t =K3 x1 C x2
x2 ## t = 2 x1 K 4 x2
(3 points)
solution: In matrix-vector form this is the mass-spring system
x1 ## t
x2 ## t
=
1
x1
2 K4
x2
K3
and we know the eigenvalues and eigenvectors from 2a)
K1
1
El =K5 = span
, El =K2 = span
2
1
For this second order DE describing a conservative mass-spring system, we get solutions
cos w t v, sin w t v where v is an eigenvector of the acceleration matrix A, with eigenvalue l and w
=
Kl . Thus the general solution is
x1 t
= c1 cos 2 t C c2 sin 2 t
x2 t
1
1
C c3 cos 5 t C c4 sin 5 t
K1
2
(Notice the slow mode has both masses in phase and with equal amplitude, and angular frequency
w = 2 ; the fast mode has both masses out of phase, with the second mass oscillating at twice the
amplitude of the first mass, and angular frequency w = 5 .)
3) Consider a general input-output model with two compartments as indicated below. The compartments
contain volumes V1 , V2 and solute amounts x1 t , x2 t respectively. The flow rates (volume per time)
are indicated by ri , i = 1 ..6 . The two input concentrations (solute amount per volume) are c1 , c5 .
m3
. Explain why the volumes
hour
3a) Suppose r2 = r3 = 100, r1 = r4 = r5 = 200, r6 = 300
V1 t , V2 t remain constant.
(2 points)
We have
V1 # t = r1 C r3 K r2 K r4 = 200 C 100 K 100 K 200 = 0
V2 # t = r4 C r5 K r3 K r6 = 200 C 200 K 100 K 300 = 0 .
Since V1 # t , V2 # t are identically zero, V1 , V2 are constant.
3b) Using the flow rates above, incoming concentrations c1 = 0.05, c5 = 0
kg
, volumes
m3
V1 = V2 = 100 m3 , show that the amounts of solute x1 t in tank 1 and x2 t in tank 2 satisfy
x1 # t
=
x2 # t
x1 # t = r1 c1 C r3
x2
V2
x2 # t = r5 c5 C r4
K r2 C r4
x1
x1
V1
K r3 C r6
1
x1
2 K4
x2
K3
= 200$.05 C 100
x2
+
x2
100
x1
10
0
.
K 100 C 200
x1
100
x2
(4 points)
= 10 K 3 x1 C x2
= 200$0 C 200
K 100 C 300
= 2 x1 K 4 x2 .
V1
V2
100
100
The two differential equations are equivalent to the displayed system that is written in vector-matrix form.
3c) Verify that a particular solution to this system of differential equations is given by the constant vector
function
4
xP =
.
2
(2 points)
solution: (One should be able to find such a particular solution as well, but here it was given to us so we
only need to check that this constant vector function makes the differential equation true. For
4
xP t =
2
We have xP # t = 0 (the left side of the differential equation). On the other hand the right side of the
differential equation in this case is
K3
1
4
2 K4
2
+
10
0
=
K12 C 2 C 10
8K8
=
0
0
.
Thus, since the differential equation is true for the constant vector function xP t , this constant is a
particular solution.
3d) Find the general solution to the system of differential equations in 3b. Hints: Use the particular
solution from 3c as part of your solution, and notice that the matrix in this system is the same as the one
in problem 2, so you can use the eigendata you found in that problem
(4 points)
solution: By the fundamental theorem for linear operators, the general solution is
4
K2
1
x t = xP t C xH t =
C c1 eK5 t
C c2 eK2 t
.
2
1
1