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School
Texas A&M University *
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Course
308
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
14
Uploaded by CorporalRaccoonMaster1109
Test I Practice
September, 2023
MATH 308 Di
ff
erential 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 instructed.
⇤
Please turn o
ff
all mobile phones, calculators, and other electronic device, and stow those away in your bags.
⇤
Print
your
name and UIN below.
⇤
Your desk may only contain pens, pencils, erasers, and student IDs during the test. No calculators, no notes,
no formula sheets, no extra scratch paper are allowed.
⇤
In part I (Multiple choice questions), print your answer clearly and legibly in the box provided.
⇤
In part II (Short answer questions), ensure that your final response is typed clearly and legibly in the provided
answer box.
⇤
In part III (Work out questions), show your work to earn credits for work out problems: merely having final
answers does not su
ffi
ce.
⇤
There are
14
pages in this test set, including a scratch paper on the last page. Carefully separate the scratch
paper from the test booklet. Do not turn in this scratch paper. Any work shown on the scratch paper will
neither be accepted nor graded.
Name
UIN
Section
507
,
515
Last name
Fist name
THE AGGIE HONOR CODE
"An Aggie does not lie, cheat, or steal or tolerate those who do.”
Signature:
Test I Practice| MATH 308 Di
ff
erential Equations
[1/14]
Part I. Multiple choice
Print your answer clearly and legibly in the box
provided.
(
6 problems
◊
3 points
)
Problem
1.
Find di
ff
erential equation of the slope
field.
A .
y
Õ
= 1 +
y
B .
y
Õ
=
x
+
y
*
C .
y
Õ
=
x
y
D .
y
Õ
=
y
x
E . None of above
Problem 2.
Which statement is true for solutions of
the di
ff
erential equation?
A . For
x <
0
all solutions are decreasing
B .
lim
x
æŒ
f
(
x
) = 0
for all solutions
f
(
x
)
C . For
y <
0
all solutions are increasing
D .
y
Õ
= 0
for
y
= 0
*
E . None of above
Problem 3.
Find the order of
x
3
y
ÕÕ
+
x
4
(
y
Õ
)
3
+
xy
5
= 0
A . 1
B . 2 *
C . 3
D . 4
Problem 4.
Which of the following initial value prob-
lems may not have the unique solution?
A .
y
Õ
=
x
2
y
2
;
y
(0) = 0
B .
y
Õ
=
xy
2
;
y
(0) = 0
C .
y
Õ
=
xy
;
y
(0) = 0
D .
y
Õ
=
x
Ô
x
;
y
(0) = 0
*
E . None of above
Problem 5.
Which of the following di
ff
erential equa-
tions is nonlinear PDE?
A .
y
Õ
=
x
2
+
x
2
y
2
B .
yy
Õ
= sin(
x
2
+ 1)
C .
cos(
x
2
+
y
2
)
u
xx
=
u
xy
+
u
D .
uu
t
+
tu
x
=
u
E . None of above
Test I Practice| MATH 308 Di
ff
erential Equations
[2/14]
Problem 6.
Solve
y
Õ
=
x
y
A .
y
=
x
+
C
B .
y
=
Ô
x
2
+
C
*
C .
y
=
Cx
2
D .
y
=
x
ln
|
y
|
+
C
E . Doesn’t not exist
Problem 7.
Find a separable first-order di
ff
erential
equation
A .
y
Õ
+
e
x
y
2
=
x
2
y
2
*
B .
y
+
e
x
sin
x
=
x
3
y
Õ
C .
y
ln
x
≠
x
2
y
=
xy
Õ
*
D .
y
Õ
+ cos
y
= tan
x
E . Doesn’t not exist
Problem 8.
Which of followings is an exact di
ff
erential
equation?
A .
ydx
≠
xdy
= 0
B .
(2
x
≠
y
)
dx
+ (
y
≠
2
x
)
dy
= 0
C .
(2
x
≠
y
)
dx
+ (
x
≠
2
y
)
dy
= 0
D .
(2
x
≠
y
)
dx
+ (2
y
≠
x
)
dy
= 0
*
E . None of above
Problem
9.
Which of followings can be solved by
y
=
ux
substitution?
A .
(
x
+
y
2
)
dx
+ (
x
+
y
2
)
dy
= 0
B .
(
x
2
≠
1)
dx
+ (
y
2
≠
2
x
)
dy
= 0
C .
(
x
2
≠
xy
)
dx
+ (2
xy
≠
y
2
)
dy
= 0
*
D .
(
xy
≠
y
2
)
dx
+ (
y
2
≠
1)
dy
= 0
E . None of above
Problem 10.
Which of followings equations has a so-
lution
x
sin
y
= 2
?
A .
sin
ydy
+
xdx
= 0
B .
sin
ydy
+
xdx
= 2
C .
sin
y dx
+
x
cos
ydy
= 0
*
D .
sin
y dx
+
x
cos
ydy
= 2
E . None of above
Problem 11.
What is a possible substitution to solve
y
Õ
=
Ô
2
x
+
y
+ 1
A .
y
=
ux
B .
u
=
y
≠
1
2
C .
u
= 2
x
+
y
*
D .
u
=
Ô
2
x
+
y
E . None of above
Test I Practice| MATH 308 Di
ff
erential Equations
[3/14]
M
m
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Problem 12.
What is a possible substitution to solve
2
yy
Õ
+ 3 =
y
2
+ 3
x
A .
y
=
ux
B .
u
=
y
≠
2
C .
u
=
y
2
*
D .
u
=
y
2
+ 3
x
E . None of above
Problem 13.
Find an integrating factor that converts
the following di
ff
erential equation to exact form
dy
≠
!
x
+
y
x
"
dx
= 0
A .
e
x
B .
e
≠
x
C .
x
D .
1
/x
*
E . None of above
Problem 14.
Find the Wronskian of
1
≠
t
and
1 +
t
A . 1
B .
t
C . 2 *
D .
2
t
E . None of above
Part II. Short answer
Write your answer clearly and legibly in the box
provided.
(
7 problems
◊
8 points
)
Problem 15.
If
y
=
e
2
t
is a solution to the di
ff
erential
equation
y
ÕÕ
≠
5
y
Õ
+
ky
= 0
,
find the value of the constant
k
.
[Answer]
k
=
6
y
ÕÕ
≠
5
y
Õ
+
ky
=4
e
2
t
≠
10
e
2
t
+
ke
2
t
=(
k
≠
6)
e
2
t
=0
Therefore,
k
= 6
Problem 16.
The function
y
=
x
2
+
x
≠
2
is a solution
of equation:
xy
Õ
+ 2
y
=
kx
2
, (
k
œ
R
,
x >
0
). Find the
constant
k
.
[Answer]
k
=
4
y
Õ
+
2
x
y
=
kx
I.F.
=
x
2
(
x
2
y
)
Õ
=
kx
3
x
2
y
=
k
4
x
4
+
C
y
=
k
4
x
2
+
Cx
≠
2
Test I Practice| MATH 308 Di
ff
erential Equations
[4/14]
Mo
g
ae
see
Problem 17.
In problems below, (a) identify the in-
dependent variable and the dependent variable of each
equation (b) give the order of each di
ff
erential equation
and (c) state whether the equation is linear or nonlinear.
If your answer to (c) is nonlinear, make sure that you
can explain why this is true.
1.
y
ÕÕ
≠
y
+
t
2
= 0
2.
t
2
y
ÕÕ
+
ty
Õ
+ 2
y
= sin
y
3.
sin
xy
ÕÕÕ
+ (
y
Õ
)
4
+
e
x
= 1
[Answer]
1
2
3
independent
variable
dependent
variable
order
linear/
nonlinear
why?
[Show your work here.]
1.
y
ÕÕ
≠
y
+
t
2
= 0
Linear 2nd order. Indep. var =
t
, Dep. var. =
y
2.
t
2
y
ÕÕ
+
ty
Õ
+ 2
y
= sin
y
Nonlinear 2nd order. Indep. var =
t
, Dep. var. =
y
3.
sin
xy
ÕÕÕ
+ (
y
Õ
)
4
+
e
x
= 1
Nonlinear 3rd order. Indep. var =
x
, Dep. var. =
y
For the next problems on 1st order ODE, choose
the equation category from
⇤
2nd order linear w/ constant coe
ffi
cients
⇤
Bernoulli’s equation
⇤
Cauchy-Euler equation
⇤
Exact (w/ or w/o integration factor)
⇤
Homogeneous with 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
⇤
Substitution
u
=
ax
+
by
+
c
⇤
Substitution
u
=
y
k
⇤
Substitution
y
=
ux
Test I Practice| MATH 308 Di
ff
erential Equations
[5/14]
Mid
Problem 18.
Solve
(1 +
x
)
dy
≠
ydx
= 0
,
y
(1) = 2
Category:
Method:
Solution:
Separable DE
Separation of variables
dy
dx
=
y
1 +
x
(
x
”
= 0)
1
y
dy
=
1
1 +
x
dx
⁄
1
y
dy
=
⁄
1
1 +
x
dx
ln
|
y
|
= ln
|
x
+ 1
|
+
C
ln
|
y
|
|
x
+ 1
|
=
C
y
=
C
(
x
+ 1)
From
y
(1) = 2
,
C
= 1
.
y
= (
x
+ 1)
Problem 19.
Solve
(
e
2
y
≠
y
) cos
x
dy
dx
= 2
e
y
sin 2
x, y
(0) = 0
.
Category:
Method:
Solution:
[Show your work here.]
Separable DE
Separation of variables
By a double angle formula,
sin 2
x
= 2 sin
x
cos
x
. Hence,
(
e
2
y
≠
y
) cos
x
dy
dx
=4
e
y
sin
x
cos
x
e
2
y
≠
y
e
y
dy
=4 sin
xdx
(cos
x
”
= 0)
⁄
(
e
y
≠
ye
≠
y
)
dy
=4
⁄
sin
xdx
e
y
≠
(
≠
ye
≠
y
≠
e
≠
y
) =
≠
4 cos
x
+
C
e
y
+ (
y
+ 1)
e
≠
y
=
≠
4 cos
x
+
C
From
y
(0) = 0
,
e
0
+ (0 + 1)
e
≠
0
=
≠
4 cos 0 +
C
2 =
≠
4 +
C
C
=6
Therefore, the solution is
e
y
+ (
y
+ 1)
e
≠
y
=
≠
4 cos
x
+ 6
where
x
”
=
(2
n
+1)
2
fi
for
n
œ
Z
.
Test I Practice| MATH 308 Di
ff
erential Equations
[6/14]
messed
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Problem 20.
Find the general solution of
(
x
2
≠
9)
y
Õ
+
xy
= 0
.
Category:
Method:
Solution:
[Show your work here.]
Separable DE
Separation of variables
(
x
2
≠
9)
dy
dx
=
≠
xy
1
y
dy
=
≠
x
x
2
≠
9
dx
(
x
”
=
±
3)
⁄
1
y
dy
=
≠
⁄
x
x
2
≠
9
dx
u
≠
sub:
u
=
x
2
≠
9
xdx
=
1
2
du
ln
|
y
|
=
≠
1
2
⁄
1
u
du
ln
|
y
|
=
≠
1
2
ln
|
u
|
+
C
ln
|
y
|
= ln
|
x
2
≠
9
|
≠
1
2
+
C
|
y
|
=
e
c
Ô
x
2
≠
9
(
>
0)
y
=
C
Ô
x
2
≠
9
Problem 21.
Solve
x
dy
dx
≠
4
y
=
x
6
e
x
Category:
Method:
Solution:
[Show your work here.]
Linear 1st order DE.
In normal form,
dy
dx
≠
4
x
y
=
x
5
e
x
(
x
”
= 0)
Integration factor (I.F.) is
exp
3⁄
≠
4
x
dx
4
= exp(
≠
4 ln
|
x
|
)
= exp
3
ln
1
|
x
|
4
4
=
1
x
4
(
*
x
4
Ø
0)
By multiplying the I.F. on both sides
3
1
x
4
y
4
Õ
=
xe
x
1
x
4
y
=
xe
x
≠
e
x
+
C
(
By IBP
)
y
=
x
4
(
x
≠
1)
e
x
+
Cx
4
Test I Practice| MATH 308 Di
ff
erential Equations
[7/14]
today
Problem 22.
Solve the initial-value problem
y
Õ
≠
2
xy
= 2
, y
(0) = 1
.
Category:
Method:
Solution:
[Show your work here.]
Linear 1st order DE.
Integrating factor
An integration factor (I.F.) is
exp
3⁄
≠
2
xdx
4
= exp(
≠
x
2
)
By multiplying the I.F. on both sides
1
e
≠
x
2
y
2
Õ
=2
e
≠
x
2
e
≠
x
2
y
=2
erf
(
x
) +
C
y
=2
e
x
2
erf
(
x
) +
Ce
x
2
From
y
(0) = 1
,
1 =2
e
0
erf
(0) +
Ce
0
2 =0 +
C
C
=1
Therefore,
y
= 2
e
x
2
erf
(
x
) +
e
x
2
Problem 23.
Solve
2
xydx
+ (
x
2
≠
1)
dy
= 0
Category:
Method:
Solution:
[Show your work here.]
Separable DE
Separation of variables
2
xydx
=
≠
(
x
2
≠
1)
dy
1
y
dy
=
≠
2
x
x
2
≠
1
dx
(
x
”
= 1)
ln
|
y
|
=
≠
ln
|
x
2
≠
1
|
+
C
|
y
|
=
e
C
|
x
2
≠
1
|
(
Ø
0)
y
=
C
|
x
2
≠
1
|
y
=
±
C
x
2
≠
1
y
=
C
x
2
≠
1
because
±
C
is another arbitrary constant.
Test I Practice| MATH 308 Di
ff
erential Equations
[8/14]
see
Problem 24.
Solve
dy
dx
=
2
x
3
≠
4
y
4
x
+ 3
y
4
Category:
Method:
Solution:
[Show your work here.]
Exact
Partial integration
By writing in di
ff
erential form,
(2
x
3
≠
4
y
)
dx
≠
(4
x
+ 3
y
4
)
dy
= 0
(2
x
3
≠
4
y
)
y
=
≠
4
≠
(4
x
+ 3
y
4
)
x
=
≠
4
f
x
= 2
x
3
≠
4
y
f
=
1
2
x
4
≠
4
xy
+
C
(
y
)
f
y
=
≠
4
x
+
C
Õ
(
y
) =
≠
4
x
≠
3
y
4
C
Õ
(
y
) =
≠
3
y
4
;
C
(
y
) =
≠
3
5
y
5
+
C
Therefore,
f
(
x, y
) =
1
2
x
4
≠
4
xy
≠
3
5
y
5
+
C
A solution to the given di
ff
erential equation is
1
2
x
4
≠
4
xy
≠
3
5
y
5
=
C
Problem 25.
Solve
3
1
≠
2
y
+
x
4
y
Õ
+
y
=
2
x
≠
1
[Show your work here.]
By writing in di
ff
erential form,
Partial integration
(1
≠
2
y
+
x
)
dy
+(
y
≠
2
x
+ 1)
dx
= 0
Since
ˆ
ˆ
x
(1
≠
2
y
+
x
) =1
ˆ
ˆ
y
(
y
≠
2
x
+ 1) =1
this equation is an exact 1st order di
ff
erential equation
with solution
f
(
x, y
) =
C
that satisfies
I
f
x
=
y
≠
2
x
+ 1
f
y
=
1
≠
2
y
+
x
By integrating
f
x
,
f
(
x, y
) =
yx
≠
2 ln
|
x
|
+
x
+
C
(
y
)
.
By partially di
ff
erentiating with respect to
y
,
f
y
=
x
+
C
Õ
(
y
) = 1
≠
2
y
+
x
C
Õ
(
y
) =1
≠
2
y
C
(
y
) =
y
≠
2 ln
|
y
|
+
C
Therefore,
f
(
x, y
) =
yx
≠
2 ln
|
x
|
+
x
+
y
≠
2 ln
|
y
|
+
C
=
x
+
y
+
xy
≠
2 ln
|
xy
|
+
C
A solution to the given di
ff
erential equation is
x
+
y
+
xy
≠
2 ln
|
xy
|
=
C
Test I Practice| MATH 308 Di
ff
erential Equations
[9/14]
i
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Problem 26.
Solve the given di
ff
erential equations
(
y
2
+
yx
)
dx
≠
x
2
dy
= 0
,
y
(1) = 1
Category:
Method:
Solution:
Homogeneous equation of degree zero.
Substitute
y
=
ux
.
0 =(
u
2
x
2
+
ux
2
)
dx
≠
x
2
(
udx
+
xdu
)
=
x
2
(
u
2
+
u
)
dx
≠
x
2
(
udx
+
xdu
)
0 =(
u
2
+
u
)
dx
≠
(
udx
+
xdu
)
, x
”
= 0
=
u
2
dx
≠
xdu
xdu
=
u
2
dx
1
u
2
du
=
1
x
dx
⁄
1
u
2
du
=
⁄
1
x
dx
≠
1
u
= ln
|
x
|
+
C
x
y
=
≠
ln
|
x
|
+
C
y
=
x
C
≠
ln
|
x
|
For
y
(1) = 1
,
C
= 1
.
y
=
x
1
≠
ln
|
x
|
.
Problem 27.
Solve the given di
ff
erential equations
x
2
y
Õ
≠
2
xy
= 3
y
4
,
y
(1) =
3
Ô
5
.
Category:
Method:
Solution:
y
Õ
≠
2
x
y
=
3
x
2
y
4
Bernoulli’s equation with
n
= 4
Substitute
u
=
y
≠
3
(i.e.
y
=
u
≠
1
/
3
)
Since
y
Õ
=
≠
1
3
u
≠
4
3
u
Õ
≠
1
3
u
≠
4
3
u
Õ
≠
2
x
u
≠
1
3
=
3
x
2
u
≠
4
3
u
Õ
+
6
x
u
=
≠
9
x
2
!
x
6
u
"
Õ
=
≠
9
x
4
x
6
u
=
≠
9
5
x
5
+
C
u
=
≠
9
5
x
+
C
x
6
y
≠
3
=
≠
9
5
x
+
C
x
6
y
3
=
5
x
6
C
≠
9
x
5
y
=
3
5
x
6
C
≠
9
x
5
4
1
/
3
For
y
(1) =
3
Ô
5
,
C
= 10
.
y
=
1
5
x
6
10
≠
9
x
5
2
1
/
3
Test I Practice| MATH 308 Di
ff
erential Equations
[10/14]
Problem 28.
Solve the given di
ff
erential equations
y
Õ
= (
x
+
y
+ 1)
2
,
y
(0) = 0
.
Category:
Method:
Solution:
Linear composite Equation.
Substitute
u
=
x
+
y
+ 1
i.e.
y
=
u
≠
x
≠
1
Now that
dy
dx
=
du
dx
≠
1
,
du
dx
≠
1 =
u
2
du
dx
=
u
2
+ 1
⁄
1
u
2
+ 1
du
=
⁄
dx
arctan
u
=
x
+
C
arctan(
x
+
y
+ 1) =
x
+
C
For
y
(0) = 0
,
C
=
fi
4
.
Problem 29.
Solve the given initial-value problem.
y
ÕÕ
+ 16
y
= 0
,
y
(0) = 2
,
y
Õ
(0) =
≠
2
[Answer]
Characteristic Eq. is
⁄
2
+ 16 = 0
.
(
⁄
+ 4
i
)(
⁄
≠
4
i
) = 0
The root of the characteristic Eq.
⁄
=
±
4
i
The fundamental set of solutions is
I
y
1
=
cos 4
x
y
2
=
sin 4
x
,
I
y
Õ
1
=
≠
4 sin 4
x
y
Õ
2
=
4 cos 4
x
W
=
-
-
-
-
y
1
(0)
y
2
(0)
y
Õ
1
(0)
y
Õ
2
(0)
-
-
-
-
=
-
-
-
-
1
0
0
4
-
-
-
-
= 4
W
1
=
-
-
-
-
y
0
y
2
(0)
y
Õ
0
y
Õ
2
(0)
-
-
-
-
=
-
-
-
-
2
0
≠
2
4
-
-
-
-
= 8
W
2
=
-
-
-
-
y
1
(0)
y
0
y
Õ
1
(0)
y
Õ
0
-
-
-
-
=
-
-
-
-
1
2
0
≠
2
-
-
-
-
≠
2
I
c
1
=
W
1
W
= 2
c
1
=
W
2
W
=
≠
1
2
Therefore, the solution is
y
=2 cos 4
x
≠
1
2
sin 4
x
Test I Practice| MATH 308 Di
ff
erential Equations
[11/14]
Woody
Problem 30.
Solve the given initial-value problem.
4
y
ÕÕ
≠
4
y
Õ
≠
3
y
= 0
,
y
(0) = 1
,
y
Õ
(0) = 5
[Answer]
Characteristic Eq. is
⁄
2
≠
4
⁄
≠
3 = 0
.
(2
⁄
+ 1)(2
⁄
≠
3) = 0
The root of the characteristic Eq.
⁄
=
≠
1
2
or
⁄
=
3
2
The fundamental set of solutions is
I
y
1
=
e
≠
1
2
x
y
2
=
e
3
2
x
,
I
y
Õ
1
=
≠
1
2
e
≠
1
2
x
y
Õ
2
=
3
2
e
3
2
x
W
=
-
-
-
-
y
1
(0)
y
2
(0)
y
Õ
1
(0)
y
Õ
2
(0)
-
-
-
-
=
-
-
-
-
1
1
≠
1
2
3
2
-
-
-
-
=
3
2
+
1
2
= 2
W
1
=
-
-
-
-
y
0
y
2
(0)
y
Õ
0
y
Õ
2
(0)
-
-
-
-
=
-
-
-
-
1
1
5
3
2
-
-
-
-
=
3
2
≠
5
=
≠
7
2
W
2
=
-
-
-
-
y
1
(0)
y
0
y
Õ
1
(0)
y
Õ
0
-
-
-
-
=
-
-
-
-
1
1
≠
1
2
5
-
-
-
-
= 5 +
1
2
=
11
2
I
c
1
=
W
1
W
=
≠
7
4
c
1
=
W
2
W
=
11
4
Therefore, the solution is
y
=
≠
7
4
e
≠
1
2
x
+
11
4
e
3
2
x
III. Workout problems
Show your complete work to earn credits: merely
having final answers does not su
ffi
ce. In accurate
statements will be marked incorrect even if they
achieve correct results.
(
2 problems
◊
13 points
)
Problem 31.
The indicated function
y
1
(
x
)
is a solu-
tion of the given Euler’s di
ff
erential equation. Use reduc-
tion of order to find a second solution
y
2
(
x
)
.
x
2
y
ÕÕ
≠
7
xy
Õ
+ 16
y
= 0
;
y
1
=
x
4
[Show your work here.]
1.
x
2
y
ÕÕ
≠
7
xy
Õ
+ 16
y
= 0
;
y
1
=
x
4
In a normal form,
y
ÕÕ
≠
7
x
y
Õ
+
16
x
2
y
= 0
From the order reduction formula for
y
ÕÕ
+
p
(
x
)
y
Õ
+
q
(
x
)
y
= 0
,
y
2
=
y
1
⁄
e
≠
s
p
(
x
)
y
2
1
dx
Since
e
≠
s
p
(
x
)
=
e
s
7
x
dx
=
e
ln
|
x
|
7
=
|
x
|
7
,
y
2
=
±
x
4
⁄
x
7
x
8
dx
=
±
x
4
ln
|
x
|
Since
±
1
can be included in a arbitrary constant,
y
2
=
x
4
ln
|
x
|
Test I Practice| MATH 308 Di
ff
erential Equations
[12/14]
staggered
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Problem 32.
Determine a particular solution to
y
ÕÕ
≠
4
y
Õ
≠
12
y
= 3
e
5
t
by the method of undetermined coe
ffi
cients
[Solution]
Y
P
(
t
) =
Ae
5
t
Plugging into the di
ff
erential equation,
25
Ae
5
t
≠
4(5
Ae
5
t
)
≠
12(
Ae
5
t
) = 3
e
5
t
≠
7
Ae
5
t
= 3
e
5
t
A
=
≠
3
7
)
Y
P
(
t
) =
≠
3
7
e
5
t
Problem 33.
Determine a particular solution to
y
ÕÕ
≠
4
y
Õ
≠
12
y
=
e
6
t
by the method of undetermined coe
ffi
cients
[Solution]
Y
P
(
t
) =
Ate
6
t
Plugging into the di
ff
erential equation,
(12
Ae
6
t
+ 36
Ate
6
t
)
≠
4(
Ae
6
t
+ 6
Ate
6
t
)
≠
12
Ate
6
t
=
e
6
t
(36
A
≠
24
A
≠
12
A
)
te
6
t
+ (12
A
≠
4
A
)
e
6
t
=
e
6
t
8
Ae
6
t
=
e
6
t
A
=
1
8
)
Y
P
(
t
) =
1
8
te
6
t
MC
SA
LA
Total
Score
Test I Practice| MATH 308 Di
ff
erential Equations
[13/14]
MY
Scratch paper
Carefully separate the scratch paper from the test booklet. Do not turn in this scratch paper. Any work shown
on the scratch paper will neither be accepted nor graded.
Test I Practice| MATH 308 Di
ff
erential Equations
[14/14]
IF
Exact