July 15, 2010
Name:
Midterm
Exam
Math 2250-002
Summer 2010
1. Find the general solution for the following dierential equations.
(a)
dy
=p1
dx
x+2
Rewriting the equation in terms of fractional powers rather than
radicals
dy
= (x + 2) 1=2
dx
it is clear that the solution to this dierential equation can be found
by integrating both sides of the equation and applying the power rule
for integration with a u substitution u = (x + 2). We have
y (x) =
Z
(x + 2) 1=2 dx =
Z
u 1=2 du + C
= 2u1=2 + C
= 2(x + 2)1=2 + C:
(b)
x
dy
dx
y = 2x2 y
This dierential equation is seperable. To see this, notice that
x
Therefore
dy
dx
= 2x2 y + y = y(2x2 + 1):
dy
y
= 2x + x1
dx;
and integrating both sides of this equation gives
Z
Z
1
1
ln y =
dy = 2x + dx + C = x2 + ln x + C:
y
x
So we have
2
y (x) = ex +ln x+C
1
= ex eln x eC = Cxex
2
2
:
(c)
y 0 + 2xy
=x
This is a rst-order linear dierential equation. Multiplying each side
of this equation by the integrating factor
(x) = e 2x dx
R
gives
= ex
2
;
ex y 0 + 2xex y = xex :
2
2
2
An application of the product to the left-hand side of this equation
shows that
2
2 0
= xex :
yex
Integrating both sides of this equation (letting u = x2 and using a
u-substitution to evaluate the integral on the right-hand side) now
gives us the equation
Z
2
2
1 Z 2xex2 dx + C
yex = xex dx + C =
2Z
= 21 eu du + C
= 21 eu + C
2
= 21 ex + C:
Therefore, multiplying both sides of this equation by e x2 shows that
2
1
y (x) = + Ce x :
2
2
2. Solve the following systems of linear equations.
(a)
3x + 5y
z = 13
2x + 7y + z = 28
x + 7y + 2z = 32
2
3 5
4 2 7
1 7
2
7
( 2)R1 +R2 4 1
!
0 7
3 5
2
( 7)R2 ;( 3)R3 4 1 7
!
0 49
0 48
2
( 48)R2 +r3 4 1 7
0 1
!
0 0
(b)
1 13
1 28
2 32
2 32
3 36
1 13
2 32
21 252
21 249
2 32
0 3
21 105
3
2
3
SWAP(R1 ;R3 ) 4 1 7
5
2 32
2 7 1 28 5
!
3 5 1 13
3
2
7 2 32
( 3)R1 +R3 4 1
5
!
0
7 3 36
0 16 7 83
3
2
3
1
7
2
32
(
1)
R
+
R
5
!3 2 4 0 1 0 3 5
0 48 21 249
3
2
3
1 7 2 32
(1
=
21)
5 ! 4 0 1 0
3 5:
0 0 1 5
So, we see that x = 1, y = 3 and z = 5 is the unique solution to this
linear system.
x
x
3x
2
1 3
4 1
1
3 1
2
3
( 3)R1 +R3 4 1
0 4
!
0 8
2
1 3
R2 +!R3 4 0 4
0 0
+ 3y + 2z = 5
y + 3z = 3
+ y + 8z = 10
2 5
3 3
8 10
2 5
1 2
2 5
2 5
1 2
0 1
3
2
( 1)R1 +R2 4 1
5
3
3 2 5
!
0 4 1 25
3 1 8 10
3
2
3
1 3 2 5
(
1)
R
2
5
! 40 4 1 25
0 8 2 5
3
5
Because the last row of this transformed matrix implies that
0x + 0y + 0z = 1;
we see that the system is incosistent, and that there is no solution.
3
3
5
3. Suppose that the logistic equation dx=dt = kx(M x) models a population
x(t) of sh in a lake after t months during which no shing occurs. Now
suppose that, because of shing, sh are removed from the lake at the rate
of hx sh per month (with h a positive constant).
(a) Write a dierential equation that describes how the population size
x(t) of sh changes over time in terms of the parameter h.
dx
dt
= kx(M
x)
hx = (kM
h)x
kx2
= kx
M
h
k
x
(b) Identify the critical points of the dierential equation from part (a).
The critical points are c1 = 0 and c2 = M (h=k).
(c) If 0 < h < kM , describe the behavior of x(t) as t ! 1.
x(t) will approach the stable equalibrium solution x = M (h=k ) as
t ! 1.
(d) If h kM , describe the behavior of x(t) as t ! 1.
x(t) will approach the stable equalibrium solution of x = 0 as t ! 1.
(e) Construct a bifurcation diagram for the dierential equation from
part (a).
4
4. Determine whether or not the following matrices are invertible. If so,
compute the inverse of the given matrix.
(a)
5
7
A= 4 6
Because
jAj = 54 76 = 2;
we see that A is invertible. To nd A 1 we observe that
5 7 1 0 ( 1)R!2 +R1 1 1 1 1
4 6 0 1
4 6 0 1
( 4)R1 +R2 1 1
1 1 (1=2)!R2 1 1 1 1
!
0 2 4 5
0 1 2 5=2
( 1)R2 +R1 1 0
3 7=2
!
0 1 2 5=2 :
So, A 1 is given by
A
1=
5
3
2
7=2
5=2
:
(b)
2
B=4
Because
3 5 6
2 4 3
2 3 5
3
5
jBj = 3 43 35 2 53 65 + 2 54 63
= 3 11 2 7 + 2 ( 5)
= 33 14 10
= 9;
we see that B is invertible. To nd B 1 we observe that
2
3
2
3 5 6 1 0 0 ( 1)R2 +R1 1
4 2 4 3 0 1 0 5
! 42
2 3 5 0 0 1
2
2
3
2
1 1 3 1 1 0
1
! 4 0 2 3 2 3 0 5 ( 1)R!3 +R2 4 0
0 1 1 2 2 1
0
2
3
2
3 1 1 0
1 1 0
( 1)R2 +R3 4 1 1
0 1 2 0 1 1 5 !4 0 1 0
!
0 0 1 2 1 2
0 0 1
2
3
9
( 1)R2 +R1 4 1 0 0 11 7
0 1 0 4 4 35
!
0 0 1 2 1 2
So B 1 is given by
2
B
1=4
6
11
4
2
7
4
1
9
3
2
3
5:
1 3 1 1
4 3 0 1
3 5 0 0
1 3 1
1 2 0
1 1 2
7 4 6
4 4 3
2 1 2
0
0
1
1
1
2
3
5
3
5
0
1
1
3
5
5. The acceleration of a Maserati is proportional to the dierence between
250 kilometers per hour and the velocity of this sports car. If this machine
can accelerate from rest to 100 kilometers per hour in 10 seconds, how long
will it take for the car to accelerate from rest to 200 kilometers per hour?
The velocity of the car is governed by the dierential equation
dv
dt
= k(250
v ):
Seperating variables gives
dv
250 v = k dt:
Integrating both sides of this equation gives
ln(250
Therefore,
v ) = kt + C:
Ce kt
v (t) = 250
Because the car starts from rest, we have v(0) = 0. So,
0 = 250
Ce k0
= 250
C:
Therefore, C = 250, and
250e kt:
v (t) = 250
Because v(10) = 100, we see that
100 = 250 250e 10k :
Therefore,
k
So, we need to solve
v (t) = 250
for t. This gives
1 ln(3=5):
= 10
1 ln(3=5)t = 200
250 exp 10
t = 10
7
ln(1=5)
ln(3=5) :
6. Determine whether or not the following sets of vectors are linearly dependent or independent. If they are linearly dependent, nd a nontrivial
linear combination of them that is equal to the zero vector.
(a) v1 = (1; 0; 0; 1), v2 = (7; 6; 4; 5), v3 = (3; 3; 2; 3)
Notice that
2v3 + v1 = v2 :
Therefore,
2v3 + v1 v2 = 0;
and the vectors are linearly dependent.
(b) v1 = (2; 0; 3; 0), v2 = (5; 4; 2; 1), v3 = (2; 1; 1; 1)
Consider the matrix
2
3
2 5 2
A = v1 v2 v3 = 664 03 42 11 775 :
0 1 1
We know that if there exists a 3 3 submatrix of A with a nonzero
determinant, then the vectors are linearly independent. Let B be
the matrix made up of the last three rows of A. Then, computing
the determinant of B by computing the cofactor expansion along the
rst column of B we have
0 4 1
jBj = 3 2 1 = 3 41 11 = 3( 4 ( 1)) = 9:
0 1 1
Therefore, the vectors are linearly independent.
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7. Let V R3 be the intersection of the planes dened by the equations
x+y
x + 2y
2z = 0
7z = 0
That is, V = f(x; y; z ): x + y 2z = 0 and x + 2y 7z = 0g.
(a) Find a basis for V .
V is the solution space to the homogeneous matrix equation
Ax = 0
where
Because
A = 11 12
2
7
1 1 2 ! 1 1 2 ;
(1)
1 2 7
0 1 5
we see that any vector of the form x = ( 3t; 5t; t) is a solution of this
equation. Therefore, the vector v = ( 3; 5; 1) constitutes a basis of
V.
(b) Find a basis for V ? , the orthogonal complement of V .
Because the space V ? is the setof all vectors orthogonal to v, we can
conclude that the solution to the matrix equation
vx = [
351]x = 0
is a basis for V ? . Any vector of the form
x = 35 s + 13 t; s; t
is a solution. Therefore the vectors
v1 = (5=3; 1; 0)
constitute a basis of V ? .
9
and
v2 = (1=3; 0; 1)
8. A tank initially containing 60 gallons of pure water. Brine containing 1
pound of salt per gallon enters the tank at 2 gallons per minute, and the
(perfectly mixed) solution leaves the tank at 3 gallons per minute; thus
the tank is empty after exactly one hour.
(a) Find the amount of salt (in pounds) in the tank after t minutes.
The amount of salt in the tank at time t is governed by the dierential
equation
dx
= 2 3 60x t :
dt
Because this is a rst-order linear dierential equation, we will solve
using the method of integrating factors. First rewrite the equation
as
dx
+ 603 t x = 2:
dt
We can then see that by letting
(t) = exp
Z
60
3
t
dt
= (60 1 t)3 ;
and multiplying both sides of the dierential equation by (t), we
have
0
1
2
x
(60 t)3 = (60 t)3 :
Integrating both sides gives
1 = 1 + C:
x
(60 t)3 (60 t)2
Therefore,
x(t) = (60 t) + C (60 t)3 :
Because we know that x(0) = 0, we have
0 = (60 0) + C (60 0)3 ;
and C = 1=3600. Therefore,
x(t) = (60
10
t)
1 (60
3600
t)3 :
(b) What is the maximum amount of salt ever in the tank?
We rst need to nd the time tmax at which the tank contains the
maximum amount of salt. Because there is no salt at the tank at
time t = 0 or t = 60, we know that at 0 < tmax < 60 and that
x0 (tmax ) = 0. Therefore, tmax is the solution to
2 603 t x = 0:
Substituting the answer from part (a) into this equation gives,
3
2 60 t
Solving for t, we see that
1 (60
(60 t) 3600
tmax
t)3 :
p
= 60 20 3:
Therefore, the maximum amount of salt in the tank is
p
40 3 :
x(tmax ) =
3
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