311sp23final_sol
pdf
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School
Rumson Fair Haven Reg H *
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Course
311
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
7
Uploaded by CoachRiverTiger30
Math 311
Final Exam
Spring 2023
Fri May 5
No electronic devices are allowed.
No collaboration is allowed.
There are 10 pages and
each page is worth 6 points, for a total of 60 points.
1. Direct Integration.
Consider the equation
x
00
(
t
) =
t
.
(a) Find the general solution.
Integrating twice gives
x
00
(
t
) =
t
x
0
(
t
) =
1
2
t
2
+
c
1
x
(
t
) =
1
6
t
3
+
c
1
t
+
c
2
for some constants
c
1
, c
2
.
(b) Find the solution with
x
(0) = 2 and
x
0
(0) = 3.
Substituting
t
= 0 in
x
0
(
t
) gives
x
0
(0) =
1
2
0
2
+
c
1
3 =
c
1
,
and substituting
t
= 0 in
x
(0) gives
x
(0) =
1
6
t
3
+
c
1
(0) +
c
2
2 =
c
2
,
hence
x
(
t
) =
1
6
t
3
+ 3
t
+ 2
.
2. Separation of Variables.
Consider the equation
dy/dx
=
x/y
.
(a) Use separation of variables to find the general solution.
We have
ydy
=
xdx
Z
y dy
=
Z
x dx
+
C
1
2
y
2
=
1
2
x
2
+
C
y
2
=
x
2
+
D
y
=
±
p
x
2
+
D
for some constant
D
.
1
2
(b) Find the specific solution with
y
(2) = 4.
Substituting
x
= 2 and
y
= 4 gives
(4)
2
= (2)
2
+
D
16 = 4 +
D
12 =
D,
hence
y
(
x
) =
±
p
x
2
+ 12
.
3. Logistic Growth.
The logistic equation
dy/dx
=
y
(1
-
y
) has general solution
y
(
x
) =
1 +
e
-
x
y
(0)
1
-
y
(0)
-
1
.
(a) Sketch the slope field of the equation.
(b) Sketch the solutions with initial conditions
y
(0) = 0
.
1,
y
(0) = 1 and
y
(0) = 2.
4. Damped Oscillations.
Consider the equation
x
00
(
t
) + 2
x
0
(
t
) + 2
x
(
t
) = 0.
(a) Find the general solution.
We look for basic solutions of the form
x
(
t
) =
e
λt
. Substituting gives
λ
2
e
λt
+ 2
λe
λt
+ 2
e
λt
= 0
λ
2
+ 2
λ
+ 2 = 0
λ
=
-
2
±
√
4
-
8
2
=
-
1
±
i.
Hence the general solution is
x
(
t
) =
c
1
e
(
-
1+
i
)
t
+
c
2
e
(
-
1
-
i
)
t
=
e
-
t
(
c
1
e
it
+
c
2
e
-
it
)
3
=
e
-
t
(
c
3
cos
t
+
c
4
sin
t
)
for some constants
c
3
, c
4
.
(b) Find the specific solution with
x
(0) = 0 and
x
0
(0) = 1.
Substituting
t
= 0 in
x
(
t
) gives
x
(0) =
e
0
(
c
3
1 +
c
4
0)
0 =
c
3
.
Computing
x
0
(
t
) and substituting
t
= 0 gives
x
0
(
t
) =
-
e
-
t
(
c
3
cos
t
+
c
4
sin
t
) +
e
-
t
(
-
c
3
sin
t
+
c
4
cos
t
)
x
0
(0) =
-
e
0
(
c
3
1 +
c
4
0) +
e
0
(
-
c
3
0 +
c
4
1)
1 =
-
c
3
+
c
4
1 =
-
0 +
c
4
1 =
c
4
,
hence
x
(
t
) =
e
-
t
(0 cos
t
+ 1 sin
t
) =
e
-
t
sin
t.
5. Undetermined Coefficients.
Consider the equation
x
00
(
t
) +
x
0
(
t
) =
t
.
(a) Find the general solution of the homogeneous equation
x
00
(
t
) +
x
0
(
t
) = 0.
Substituting the guess
x
(
t
) =
e
λt
gives
λ
2
e
λt
+
λe
λt
= 0
λ
2
+
λ
= 0
λ
(
λ
+ 1) = 0
λ
= 0
,
-
1
.
Hence the general solution of
x
00
(
t
) +
x
0
(
t
) = 0 is
x
(
t
) =
c
1
e
0
t
+
c
2
e
-
1
t
=
c
1
+
c
2
e
-
t
.
(b) Find the general solution of the non-homogeneous equation
x
00
(
t
) +
x
0
(
t
) =
t
. Use
the guess
x
p
(
t
) =
A
+
Bt
+
Ct
2
for the particular solution.
To find a particular solution we substitute the guess
x
p
(
t
) =
A
+
Bt
+
Ct
2
:
(
A
+
Bt
+
Ct
2
)
00
+ (
A
+
Bt
+
Ct
2
)
0
=
t
2
C
+
B
+ 2
Ct
=
t
(2
C
)
t
+ (
B
+ 2
C
) = 1
t
+ 0
.
Comparing coefficients gives 2
C
= 1 and
B
+ 2
C
= 0, hence
C
= 1
/
2 and
B
=
-
2
C
=
-
1.
The value of
A
is arbitrary.
Let’s just take
A
= 0 to obtain the
particular solution
x
p
(
t
) = 0
-
t
+
t
2
/
2
.
Combining this with the homogeneous solution from part (a) gives the general
solution
x
(
t
) =
x
c
(
t
) +
x
p
(
t
) =
c
1
+
c
2
e
-
t
-
t
+
t
2
/
2
.
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4
6. Rules for Laplace Transforms.
Use the rules in the attached table to compute the
following Laplace transforms.
(a)
L
[
t
·
e
2
t
]
The table says that
L
[
t
] = 1
/s
2
and hence
L
[
t
·
e
2
t
] =
L
[
t
]
s
→
s
-
2
=
1
(
s
-
2)
2
.
(b)
L
[
e
2
t
·
sin
t
]
The table says that
L
[sin
t
] = 1
/
(
s
2
+ 1) and hence
L
[
e
2
t
·
sin] =
L
[sin
t
]
s
→
s
-
2
=
1
(
s
-
2)
2
+ 1
.
(c)
L
[
t
·
sin
t
]
The table says that
L
[sin
t
] = 1
/
(
s
2
+ 1) and hence
L
[
t
·
sin] =
-
d
ds
L
[sin
t
] =
-
d
ds
1
(
s
2
+ 1)
=
2
s
(
s
2
+ 1)
2
.
7. Solve Using Laplace Transforms.
Consider the equation
x
0
(
t
) +
x
(
t
) = 2 sin
t
.
(a) Find the partial fraction decomposition of
2
(
s
+1)(
s
2
+1)
.
We are looking for
A, B, C
so that
2
(
s
+ 1)(
s
2
+ 1)
=
A
s
+ 1
+
Bs
+
C
s
2
+ 1
2
(
s
+ 1)(
s
2
+ 1)
=
A
(
s
2
+ 1) + (
Bs
+
C
)(
s
+ 1)
(
s
+ 1)(
s
2
+ 1)
2 =
A
(
s
2
+ 1) + (
Bs
+
C
)(
s
+ 1)
0
s
2
+ 0
s
+ 2 = (
A
+
B
)
s
2
+ (
B
+
C
)
s
+ (
A
+
C
)
.
Comparing coefficients gives
A
+
B
= 0,
B
+
C
= 0 and
A
+
C
= 2, which implies
that
A
=
-
B
=
C
= 1. We conclude that
2
(
s
+ 1)(
s
2
+ 1)
=
1
s
+ 1
+
-
s
+ 1
s
2
+ 1
=
1
s
+ 1
-
s
s
2
+ 1
+
1
s
2
+ 1
.
(b) Use Laplace transforms to solve the equation with
x
(0) = 0.
Applying Laplace transforms gives
x
0
(
t
) +
x
(
t
) = 2 sin
t
sX
-
x
(0) +
X
=
2
s
2
+ 1
(
s
+ 1)
X
=
2
s
2
+ 1
X
=
2
(
s
+ 1)(
s
2
+ 1)
5
X
=
1
s
+ 1
-
s
s
2
+ 1
+
1
s
2
+ 1
x
(
t
) =
L
-
1
1
s
+ 1
-
L
-
1
s
s
2
+ 1
+
L
-
1
1
s
2
+ 1
x
(
t
) =
e
-
t
-
cos
t
+ sin
t.
8. Discontinuous Input.
Consider the step function
H
(
t
) and the delta function
δ
(
t
).
(a) Compute the partial fraction decomposition of
1
s
(
s
-
1)
.
We are looking for
A, B
such that
1
s
(
s
-
1)
=
A
s
+
B
s
-
1
1
s
(
s
-
1)
=
A
(
s
-
1) +
Bs
s
(
s
-
1)
1 =
A
(
s
-
1) +
Bs
Substituting
s
= 0 gives 1 =
A
(0
-
1) =
-
A
and substituting
s
= 1 gives 1 =
B
,
hence
1
s
(
s
-
1)
=
-
1
s
+
1
s
-
1
.
(b) Use your answer from (a) to evaluate
L
-
1
h
e
-
3
s
s
(
s
-
1)
i
If
F
(
s
) =
L
[
f
(
t
)] then
L
-
1
[
e
-
as
·
f
(
t
)] =
H
(
t
-
a
)
f
(
t
-
a
). In our case we have
F
(
s
) =
1
s
(
s
-
1)
, which implies that
f
(
t
) =
L
-
1
1
s
(
s
-
1)
=
L
-
1
-
1
s
+
1
s
-
1
=
-
L
-
1
1
s
+
L
-
1
1
s
-
1
=
-
1 +
e
t
,
and hence
L
-
1
e
-
3
s
s
(
s
-
1)
=
H
(
t
-
3)
f
(
t
-
3) =
H
(
t
-
3)
(
-
1 +
e
t
-
3
)
.
(c) Use your answers from (a) and (b) to solve the equation
x
00
(
t
)
-
x
0
(
t
) =
δ
(
t
-
3)
with initial conditions
x
(0) =
x
0
(0) = 0.
Applying Laplace transforms gives
x
00
(
t
)
-
x
0
(
t
) =
δ
(
t
-
3)
s
2
X
-
sx
(0)
-
x
0
(0)
-
(
sX
-
x
(0)) =
e
-
3
s
(
s
2
-
s
)
X
=
e
-
3
s
6
X
=
e
-
3
s
s
(
s
-
1)
x
(
t
) =
L
-
1
e
-
3
s
s
(
s
-
1)
x
(
t
) =
H
(
t
-
3)
(
-
1 +
e
t
-
3
)
.
9. First Order Linear System.
Consider the linear system
x
0
(
t
)
=
-
x
(
t
) + 2
y
(
t
)
,
y
0
(
t
)
=
x
(
t
)
.
(a) Find the eigenvalues and eigenvectors of the matrix
-
1
2
1
0
.
The eigenvalues are the roots of the characteristic equation:
-
1
-
λ
2
1
0
-
λ
= 0
(
-
1
-
λ
)(0
-
λ
)
-
(1)(2) = 0
λ
2
+
λ
-
2 = 0
(
λ
+ 2)(
λ
-
1) = 0
λ
= 1
,
-
2
.
The eigenvectors (
u, v
) for
λ
= 1 satisfy
-
1
-
1
2
1
0
-
1
u
v
=
0
0
-
2
2
1
-
1
u
v
=
0
0
u
v
= any multiple of
1
1
The eigenvectors (
u, v
) for
λ
=
-
2 satisfy
-
1 + 2
2
1
0 + 2
u
v
=
0
0
1
2
1
2
u
v
=
0
0
u
v
= any multiple of
2
-
1
(b) Find the general solution of the linear system.
The general solution is
x
(
t
)
y
(
t
)
=
c
1
1
1
e
t
+
c
2
2
-
1
e
-
2
t
.
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7
10. Second Order Linear System.
Consider the linear system
x
00
(
t
)
=
-
3
x
(
t
) + 2
y
(
t
)
,
y
00
(
t
)
=
x
(
t
)
-
2
y
(
t
)
.
Here are the eigenvalues and eigenvectors of the coefficient matrix:
-
3
2
1
-
2
1
1
=
-
1
2
1
1
and
-
3
2
1
-
2
2
-
1
=
-
2
2
2
-
1
.
(a) Use the given information to find the general solution of the system.
The general solution is
x
(
t
)
y
(
t
)
=
a
1
1
1
cos
t
+
b
1
1
1
sin
t
+
a
2
2
-
1
cos(2
t
) +
b
2
2
-
1
sin(2
t
)
.
(b) Find the specific solution with
x
(0) =
x
0
(0) =
y
(0) = 0 and
y
0
(0) = 1.
Substituting
t
= 0 into
x
(
t
) and
y
(
t
) gives
0
0
=
x
(0)
y
(0)
=
a
1
1
1
+
a
2
2
-
1
.
This implies that 0 =
a
1
+ 2
a
2
and 0 =
a
1
-
a
2
, which has solution
a
1
=
a
2
= 0.
Now we know that
x
(
t
)
y
(
t
)
=
b
1
1
1
sin
t
+
b
2
2
-
1
sin(2
t
)
.
Taking derivatives and substituting
t
= 0 gives
x
0
(
t
)
y
0
(
t
)
=
b
1
1
1
cos
t
+ 2
b
2
2
-
1
cos(2
t
)
0
1
=
b
1
1
1
+ 2
b
2
2
-
1
This implies that 0 =
b
1
+ 4
b
2
and 1 =
b
1
-
2
b
2
, which has solution
b
1
= 2
/
3 and
b
2
=
-
1
/
6. Hence the solution is
x
(
t
)
y
(
t
)
=
2
3
1
1
sin
t
-
1
6
2
-
1
sin(2
t
)
.