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Texas A&M University *
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Course
308
Subject
Mathematics
Date
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Type
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Test1 23F a sol - 1st midterm Prof. Kang
Differential Equations (Texas A&M University)
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Test1 23F a sol - 1st midterm Prof. Kang
Differential Equations (Texas A&M University)
Scan to open on Studocu
Studocu is not sponsored or endorsed by any college or university
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Test I A
Fall Semester 2023
MATH 308 Differential Equations
Dr. Minchul Kang
Texas A&M University | Department of Mathematics | 335B Blocker Building | College Station | TX 77840 | grcorg@tamu.edu
Direction
□
Do not turn to the next page until you are told to do so.
□
Print
your
name and UIN below as well as on the Scantron form.
□
Please turn off all mobile phones, calculators, and other electronic device, and stow those away in your bags.
□
Your desk may only contain pens, pencils, erasers, and student IDs during the test.
□
In part I (Multiple choice questions), mark your answer on the scantron from using No. 2 pencil.
□
In part II (Short answer questions), ensure that your final response is typed clearly and legibly in the provided
answer box. You don’t have to show your work.
□
In part III (Work out questions), show your work to earn credits for work out problems: merely having final
answers does not suffice.
□
There are
8
pages in this test set.
Name:
,
UIN:
Last Name
First Name
THE AGGIE HONOR CODE
"An Aggie does not lie, cheat, or steal or tolerate those who do.”
Signature:
Test I A| MATH 308 Differential Equations
[1/8]
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Part I. Multiple choice (4 points each)
Print your answer in the box provided.
Problem 1.
Find differential equation of the direction
fields.
A .
y
′
=
y
x
*
B .
y
′
=
x
y
C .
y
′
=
-
y
x
D .
y
′
=
-
x
y
E . None of above
[Answer]
Problem 2.
Find the order of
x
(
y
′′′
)
2
+
x
2
(
y
′′
)
3
+
x
3
(
y
′
)
4
+
x
4
y
5
= 0
A . 2
B . 3 *
C . 4
D . 5
[Answer]
Problem 3.
Which of the following is a separable dif-
ferential equation?
A .
y
2
ln
x
-
x
2
y
=
xy
′
B .
y
ln
x
-
xy
2
=
xy
′
C .
y
ln
x
-
x
2
y
=
xy
′
*
D .
y
ln
x
-
xy
+
x
=
xy
′
E . None of above
[Answer]
Problem 4.
Which of followings is an exact differential
equation?
A .
(3
x
-
2
y
)
dx
+ (3
y
+ 2
x
)
dy
= 0
B .
(2
x
-
3
y
)
dx
+ (3
x
-
2
y
)
dy
= 0
C .
(3
x
-
2
y
)
dx
+ (2
y
-
3
x
)
dy
= 0
D .
(2
x
-
3
y
)
dx
+ (2
y
-
3
x
)
dy
= 0
*
E . None of above
[Answer]
Test I A| MATH 308 Differential Equations
[2/8]
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Problem 5.
For
y
′
=
y
2
-
4
, which of followings is
NOT true?
A .
y
= 2
is a steady state solution.
B .
y
=
-
2
is a a steady state solution.
C . The equation has a unique solution for
y
(0) = 0
.
D . The equation has two solution for
y
(0) = 2
.*
E . None of above
[Answer]
Problem 6.
Which statement is true for solutions for
the direction fields?
A . For
x <
0
all solutions are decreasing.
B .
lim
x
→∞
f
(
x
) = 1
for all solutions
f
(
x
)
. *
C . For
y >
1
all solutions are increasing.
D .
y
′
= 0
for
x
= 0
.
E . None of above.
[Answer]
Problem 7.
To solve
y
′
-
(4
x
-
y
+ 1)
2
= 0
, what is a
proper substitution?
A .
y
=
ux
B .
u
=
y
−
1
C .
u
= (4
x
-
y
+ 1)
2
D .
u
= 4
x
-
y
+ 1
*
E . None of above
[Answer]
Problem 8.
To solve
yy
′
+
x
2
y
2
-
x
3
y
3
= 0
, what is a
proper substitution?
A .
y
=
ux
B .
u
=
y
−
1
*
C .
u
=
y
−
2
D .
u
=
y
2
E . None of above
[Answer]
Test I A| MATH 308 Differential Equations
[3/8]
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Part II. Short answer (8 points each)
Print your answer in the box provided. You don’t
have to show your work.
Choose the equation category from
□
2nd order linear w/ constant coefficients
□
Bernoulli’s equation
□
Cauchy-Euler equation
□
Exact (w/ or w/o integration factor)
□
Homogeneous degree zero
□
Integrable
□
Linear 1st order
□
Linear composite
□
Separable
Choose the solution method from
□
Characteristic equation
□
Direct integration
□
Integrating factor
□
Partial integrations
□
Separation of variables
□
u
=
ax
+
by
+
c
substitution
□
u
=
y
k
substitution
□
y
=
ux
substitution
Problem 9.
Identify the equation category, outline so-
lution methods, and solve the equation.
y
′
= 2
e
−
y
x
-
2
e
−
y
,
y
(1) = 0
Category:
Separable
Method:
Separation of variables
Solution:
e
y
=
x
2
-
2
x
+ 2
Problem 10.
Identify the equation category, outline
solution methods, and solve the equation.
ty
′
=
-
3
y
+ 6
t
3
,
y
(1) = 2
.
Category:
1st order linear
Method:
Integrating factor
Solution:
y
=
t
3
+
t
−
3
Test I A| MATH 308 Differential Equations
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Problem 11.
Identify the equation category, outline
solution methods, and solve the equation.
x
(1
-
sin
y
)
dy
= (cos
x
-
cos
y
-
y
)
dx
.
Category:
Exact
Method:
Partial integration
Solution:
x
cos
y
-
sin
x
+
xy
=
C
Problem 12.
Identify the equation category, outline
solution methods, and solve the equation.
y
′′
-
4
y
′
+ 4
y
= 0
,
y
(0) = 1
,
y
′
(0) = 1
Category:
2nd order linear w/ constant
coefficients
Method:
Characteristic equation
Solution:
y
=
e
2
t
-
te
2
t
Problem 13.
Identify the equation category, outline
solution methods, and solve the equation.
y
′′
+ 2
y
′
+ 2
y
= 0
Category:
2nd order linear w/ constant
coefficients
Method:
Characteristic equation
Solution:
y
=
e
−
t
(
C
1
cos
t
+
C
2
sin
t
)
Test I A| MATH 308 Differential Equations
[5/8]
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Part III. Work out questions (14 points
each)
Show your complete work to earn credits: merely
having final answers does not suffice. In accurate
statements will be marked incorrect even if they
achive correct results.
Problem 14.
Find the general solution to Cauchy Eu-
ler’s equation following suggested steps
x
2
y
′′
-
xy
′
+
y
= 0
for
x >
0
.
Step 1:
By making a guess
y
(
x
) =
x
r
, find a solution
y
1
(
x
)
to the equation.
[Show your work here.]
0 =
r
(
r
-
1)
-
r
+ 1
=
r
2
-
2
r
+ 1
= (
r
-
1)
2
Step 2:
Use reduction of order to find a second solution
y
2
(
x
)
.
From the order reduction formula for
y
′′
+
p
(
x
)
y
′
+
q
(
x
)
y
= 0
, or
y
′′
-
1
x
y
′
+
1
x
2
y
= 0
.
y
2
=
y
1
integraldisplay
e
−
integraltext
p
(
x
)
y
2
1
dx
=
x
integraldisplay
x
2
e
−
integraltext
1
/tdt
dx
=
x
ln
x
[Show your work here.]
Test I A| MATH 308 Differential Equations
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Step 3:
Show that
y
1
and
y
2
form the complete set of
solution (i.e. independent).
[Show your work here.]
W
(
y
1
, y
2
) =
vextendsingle
vextendsingle
vextendsingle
vextendsingle
x
x
ln
x
1
ln
x
+ 1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= ln
x
+ 1
-
x
ln
x
= 1
>
0
Problem 15.
Following suggested steps to determine
the general solution to
y
′′
+ 2
y
′
-
3
y
= 4
e
t
Step 1:
Solve the associated homogeneous equation to
find
y
c
[Show your work here.]
y
′′
+ 2
y
′
-
3
y
= 0
λ
2
+ 2
λ
-
3 = 0
(
λ
+ 3)(
λ
-
1) = 0
y
c
=
C
1
e
t
+
C
2
e
−
3
t
Test I A| MATH 308 Differential Equations
[7/8]
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Step 2:
Find
y
p
by the method of undetermined coeffi-
cients
[Show your work here.]
y
p
(
t
) =
Ate
t
(
∵
e
t
is part of
y
c
)
y
′
p
(
t
) =
A
(1 +
t
)
e
t
y
′′
p
(
t
) =
A
(2 +
t
)
e
t
Plugging into the differential equation,
A
(2 +
t
)
e
t
+ 2
A
(1 +
t
)
e
t
-
3
Ate
t
= 4
e
t
A
(1 + 2
-
3)
te
t
+
A
(2 + 2)
e
t
= 4
e
t
4
Ae
t
= 4
e
t
A
= 1
∴
y
p
(
t
) =
te
t
Step 3:
Find the general solution to
y
′′
+2
y
′
-
3
y
= 4
e
t
[Show your work here.]
y
c
=
C
1
e
t
+
C
2
e
−
3
t
+
te
t
Test I A| MATH 308 Differential Equations
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