Problem 1
edit
Problem Statement
edit
ODE:
y
″
−
3
y
′
−
10
y
=
3
c
o
s
7
x
{\displaystyle y''-3y'-10y=3cos7x}
Part 3: Find the overall solution y(x) that corresponds to the initial conditions:
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0}
Plot the solution over 3 periods
Solution
edit
Wronskian: Function is linearly independent if
W
(
f
,
g
)
≠
0
{\displaystyle W(f,g)\neq 0}
W
(
f
,
g
)
:=
[
f
g
f
′
g
′
]
=
f
g
′
−
g
f
′
{\displaystyle W(f,g):={\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'}
f
=
c
o
s
7
x
,
f
′
=
−
7
s
i
n
7
x
{\displaystyle f=cos7x,f'=-7sin7x}
g
=
s
i
n
7
x
,
g
′
=
7
c
o
s
7
x
{\displaystyle g=sin7x,g'=7cos7x}
W
(
c
o
s
7
x
,
s
i
n
7
x
)
:=
[
c
o
s
7
x
s
i
n
7
x
−
7
s
i
n
7
x
7
c
o
s
7
x
]
=
7
c
o
s
2
7
x
+
7
s
i
n
2
7
x
=
1
{\displaystyle W(cos7x,sin7x):={\begin{bmatrix}cos7x&sin7x\\-7sin7x&7cos7x\end{bmatrix}}=7cos^{2}7x+7sin^{2}7x=1}
g(x) and f(x) are linearly independent
Gramian: Function is linearly independent if
Γ
(
f
,
g
)
≠
0
{\displaystyle \Gamma (f,g)\neq 0}
Γ
(
f
,
g
)
:=
[
<
f
,
f
>
<
f
,
g
>
<
g
,
f
>
<
g
,
g
>
]
{\displaystyle \Gamma (f,g):={\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}
f
(
x
)
=
c
o
s
7
x
,
g
(
x
)
=
s
i
n
7
x
{\displaystyle f(x)=cos7x,g(x)=sin7x}
<
f
,
f
>=
∫
−
1
1
c
o
s
7
x
∗
c
o
s
7
x
d
x
=
1.07
{\displaystyle <f,f>=\int _{-1}^{1}cos7x*cos7xdx=1.07}
<
f
,
g
>=
∫
−
1
1
c
o
s
7
x
∗
s
i
n
7
x
d
x
=
0
{\displaystyle <f,g>=\int _{-1}^{1}cos7x*sin7xdx=0}
<
g
,
f
>=
∫
−
1
1
s
i
n
7
x
∗
c
o
s
7
x
d
x
=
0
{\displaystyle <g,f>=\int _{-1}^{1}sin7x*cos7xdx=0}
<
g
,
g
>=
∫
−
1
1
s
i
n
7
x
∗
s
i
n
7
x
d
x
=
0.93
{\displaystyle <g,g>=\int _{-1}^{1}sin7x*sin7xdx=0.93}
Γ
(
f
,
g
)
:=
[
1.07
0
0
0.93
]
=
0.9951
{\displaystyle \Gamma (f,g):={\begin{bmatrix}1.07&0\\0&0.93\end{bmatrix}}=0.9951}
g(x) and f(x) are linearly independent
The particular solution for a
r
(
x
)
=
3
c
o
s
7
x
{\displaystyle r(x)=3cos7x}
y
p
(
x
)
=
M
c
o
s
7
x
+
N
s
i
n
7
x
{\displaystyle y_{p}(x)=Mcos7x+Nsin7x}
Differentiate to get:
y
p
′
(
x
)
=
−
M
7
s
i
n
7
x
+
N
7
c
o
s
7
x
{\displaystyle y_{p}'(x)=-M7sin7x+N7cos7x}
y
p
″
(
x
)
=
−
M
7
2
c
o
s
7
x
−
N
7
2
s
i
n
7
x
{\displaystyle y_{p}''(x)=-M7^{2}cos7x-N7^{2}sin7x}
Plug the derivatives into the equation:
y
p
″
(
x
)
−
3
y
p
′
(
x
)
−
10
y
p
(
x
)
=
3
c
o
s
7
x
{\displaystyle y_{p}''(x)-3y_{p}'(x)-10y_{p}(x)=3cos7x}
(
−
M
7
2
c
o
s
7
x
−
N
7
2
s
i
n
7
x
)
−
3
(
−
M
7
s
i
n
7
x
+
N
7
c
o
s
7
x
)
−
10
(
M
c
o
s
7
x
+
N
s
i
n
7
x
)
{\displaystyle (-M7^{2}cos7x-N7^{2}sin7x)-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)}
Separate the sin and cos terms to get 2 equations in order to solve for M and N
−
M
7
2
c
o
s
7
x
−
3
N
7
c
o
s
7
x
−
10
M
c
o
s
7
x
=
3
c
o
s
7
x
{\displaystyle -M7^{2}cos7x-3N7cos7x-10Mcos7x=3cos7x}
−
N
7
2
s
i
n
7
x
+
3
M
7
s
i
n
7
x
−
10
N
s
i
n
7
x
=
0
{\displaystyle -N7^{2}sin7x+3M7sin7x-10Nsin7x=0}
dividing each equation by cos7x and sin7x respectively:
−
49
M
−
21
N
−
10
M
=
3
{\displaystyle -49M-21N-10M=3}
−
49
N
+
21
M
−
10
N
=
0
{\displaystyle -49N+21M-10N=0}
M
=
−
0.0454
{\displaystyle M=-0.0454}
N
=
−
0.0154
{\displaystyle N=-0.0154}
So the particular solution is:
y
p
(
x
)
=
−
0.0454
c
o
s
7
x
−
0.0154
s
i
n
7
x
{\displaystyle y_{p}(x)=-0.0454cos7x-0.0154sin7x}
The overall solution can be found by:
y
(
x
)
=
y
h
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{h}(x)+y_{p}(x)}
The roots given in the problem statement
λ
1
=
−
2
,
λ
2
=
+
5
{\displaystyle \lambda _{1}=-2,\ \lambda _{2}=+5}
Lead to the homogeneous solution of:
y
h
(
x
)
=
C
1
e
−
2
x
+
C
2
e
5
x
{\displaystyle y_{h}(x)=C_{1}e^{-2x}+C_{2}e^{5x}}
Combining the homogeneous and particular solution gives us:
y
(
x
)
=
C
1
e
−
2
x
+
C
2
e
5
x
−
0.0454
c
o
s
7
x
−
0.0154
s
i
n
7
x
{\displaystyle y(x)=C_{1}e^{-2x}+C_{2}e^{5x}-0.0454cos7x-0.0154sin7x}
Solving for the constants by using the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0}
y
(
0
)
=
C
1
e
0
+
C
2
e
0
−
0.0454
c
o
s
(
0
)
−
0.0154
s
i
n
(
0
)
=
1
{\displaystyle y(0)=C_{1}e^{0}+C_{2}e^{0}-0.0454cos(0)-0.0154sin(0)=1}
y
′
(
0
)
=
−
2
C
1
e
0
+
5
C
2
e
0
+
0.3178
s
i
n
(
0
)
−
0.1078
c
o
s
(
0
)
=
0
{\displaystyle y'(0)=-2C_{1}e^{0}+5C_{2}e^{0}+0.3178sin(0)-0.1078cos(0)=0}
C
1
=
0.73
{\displaystyle C_{1}=0.73}
C
2
=
0.31
{\displaystyle C_{2}=0.31}
The overall solution is:
y
(
x
)
=
0.73
e
−
2
x
+
0.31
e
5
x
−
0.0454
c
o
s
7
x
−
0.0154
s
i
n
7
x
{\displaystyle y(x)=0.73e^{-2x}+0.31e^{5x}-0.0454cos7x-0.0154sin7x}
Plot
y
(
x
)
=
0.73
e
−
2
x
+
0.31
e
5
x
−
0.0454
c
o
s
7
x
−
0.0154
s
i
n
7
x
{\displaystyle y(x)=0.73e^{-2x}+0.31e^{5x}-0.0454cos7x-0.0154sin7x}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 2
edit
Problem Statement
edit
Complete the solution to problem on p.8-6.
y
(
x
)
{\displaystyle y(x)}
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0}
Solution
edit
Given:
y
″
+
4
y
′
+
13
y
=
2
∗
e
−
2
x
∗
3
c
o
s
7
x
{\displaystyle y''+4y'+13y=2*e^{-2x}*3cos7x}
y
h
(
x
)
=
e
−
2
x
∗
(
A
c
o
s
(
3
x
)
+
B
N
s
i
n
(
3
x
)
)
{\displaystyle y_{h}(x)=e^{-2x}*(Acos(3x)+BNsin(3x))}
y
P
(
x
)
=
x
∗
e
−
2
x
∗
(
M
c
o
s
(
3
x
)
+
N
s
i
n
(
3
x
)
)
{\displaystyle y_{P}(x)=x*e^{-2x}*(Mcos(3x)+Nsin(3x))}
y
P
′
(
x
)
=
e
−
2
x
∗
c
o
s
(
3
x
)
(
x
(
−
2
M
+
N
)
+
M
)
+
e
−
2
x
∗
s
i
n
(
3
x
)
(
x
(
−
2
N
−
M
)
+
N
)
{\displaystyle y_{P}'(x)=e^{-2x}*cos(3x)(x(-2M+N)+M)+e^{-2x}*sin(3x)(x(-2N-M)+N)}
y
P
″
(
x
)
=
e
−
2
x
(
−
3
s
i
n
(
3
x
)
(
x
(
−
2
M
+
N
)
+
M
)
{\displaystyle y_{P}''(x)=e^{-2x}(-3sin(3x)(x(-2M+N)+M)}
+
c
o
s
(
3
x
)
(
−
2
M
+
N
)
)
−
2
e
−
2
x
c
o
s
(
3
x
)
(
x
(
−
2
M
+
N
)
+
M
)
+
e
−
2
x
(
3
c
o
s
(
3
x
)
(
x
(
−
2
N
−
M
)
{\displaystyle +cos(3x)(-2M+N))-2e^{-2x}cos(3x)(x(-2M+N)+M)+e^{-2x}(3cos(3x)(x(-2N-M)}
+
N
)
+
s
i
n
(
3
x
)
(
−
2
N
−
M
)
)
−
2
e
−
2
x
s
i
n
(
3
x
)
(
x
(
−
2
N
−
M
)
+
N
)
{\displaystyle +N)+sin(3x)(-2N-M))-2e^{-2x}sin(3x)(x(-2N-M)+N)}
y
p
″
+
4
y
p
′
+
13
y
p
=
2
∗
e
−
2
x
∗
3
c
o
s
7
x
{\displaystyle y_{p}''+4y_{p}'+13y_{p}=2*e^{-2x}*3cos7x}
4
N
=
2
,
−
2
M
+
4
N
{\displaystyle 4N=2,-2M+4N}
M
=
1
,
N
=
0.5
{\displaystyle M=1,N=0.5}
y
(
x
)
=
e
−
2
x
∗
(
A
c
o
s
(
3
x
)
+
B
N
s
i
n
(
3
x
)
)
+
x
∗
e
−
2
x
∗
(
c
o
s
(
3
x
)
+
.5
∗
s
i
n
(
3
x
)
)
{\displaystyle y(x)=e^{-2x}*(Acos(3x)+BNsin(3x))+x*e^{-2x}*(cos(3x)+.5*sin(3x))}
A
=
1
,
−
2
A
+
3
B
+
1
=
0
{\displaystyle A=1,-2A+3B+1=0}
A
=
1
,
B
=
1
/
3
{\displaystyle A=1,B=1/3}
y
(
x
)
=
e
−
2
x
∗
(
c
o
s
(
3
x
)
+
s
i
n
(
3
x
)
/
3
)
+
x
∗
e
−
2
x
∗
(
c
o
s
(
3
x
)
+
.5
∗
s
i
n
(
3
x
)
)
{\displaystyle y(x)=e^{-2x}*(cos(3x)+sin(3x)/3)+x*e^{-2x}*(cos(3x)+.5*sin(3x))}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 3
edit
Problem Statement
edit
Is the given function even or odd or neither even nor odd? Find its Fourier Series.
Solution
edit
f
(
x
)
=
x
2
,
(
−
1
<
x
<
1
)
,
p
=
2
{\displaystyle f(x)=x^{2},(-1<x<1),p=2}
f
(
−
x
)
=
f
(
x
)
{\displaystyle f(-x)=f(x)}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
even function.
f
(
x
)
=
a
0
+
∑
n
=
1
∞
[
a
n
c
o
s
(
n
w
x
)
+
b
n
s
i
n
(
n
w
x
)
]
{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(nwx)+b_{n}sin(nwx)]}
a
0
=
1
2
L
∫
−
L
L
f
(
x
)
d
x
{\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx}
n
=
1
,
2
,
.
.
.
{\displaystyle n=1,2,...}
a
n
=
1
L
∫
−
L
L
f
(
x
)
c
o
s
(
n
w
x
)
d
x
{\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos(nwx)dx}
b
n
=
1
L
∫
−
L
L
f
(
x
)
s
i
n
(
n
w
x
)
d
x
{\displaystyle b_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)sin(nwx)dx}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
a
0
=
1
2
(
1
)
∫
−
1
1
x
2
d
x
=
1
3
{\displaystyle a_{0}={\frac {1}{2(1)}}\int _{-1}^{1}x^{2}dx={\frac {1}{3}}}
a
n
=
1
1
∫
−
1
1
x
2
c
o
s
(
n
π
x
)
d
x
{\displaystyle a_{n}={\frac {1}{1}}\int _{-1}^{1}x^{2}cos(n\pi x)dx}
a
n
=
1
n
π
[
s
i
n
(
n
π
)
−
s
i
n
(
−
n
π
)
]
+
2
n
2
π
2
[
c
o
s
(
n
π
)
+
c
o
s
(
−
n
π
)
]
−
2
n
3
π
3
[
s
i
n
(
n
π
)
−
s
i
n
(
−
n
π
)
]
{\displaystyle a_{n}={\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]}
b
n
=
1
1
∫
−
1
1
x
2
s
i
n
(
n
π
x
)
d
x
{\displaystyle b_{n}={\frac {1}{1}}\int _{-1}^{1}x^{2}sin(n\pi x)dx}
b
n
=
−
1
n
π
[
c
o
s
(
n
π
)
−
c
o
s
(
−
n
π
)
]
+
2
n
2
π
2
[
s
i
n
(
n
π
)
+
s
i
n
(
−
n
π
)
]
+
2
n
3
π
3
[
c
o
s
(
n
π
)
−
c
o
s
(
−
n
π
)
]
{\displaystyle b_{n}={\frac {-1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
1
3
+
∑
n
=
1
∞
c
o
s
(
n
w
x
)
[
1
n
π
[
s
i
n
(
n
π
)
−
s
i
n
(
−
n
π
)
]
+
2
n
2
π
2
[
c
o
s
(
n
π
)
+
c
o
s
(
−
n
π
)
]
−
2
n
3
π
3
[
s
i
n
(
n
π
)
−
s
i
n
(
−
n
π
)
]
]
{\displaystyle {\frac {1}{3}}+\sum _{n=1}^{\infty }cos(nwx)[{\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]]}
+
s
i
n
(
n
w
x
)
[
−
1
n
π
[
c
o
s
(
n
π
)
−
c
o
s
(
−
n
π
)
]
+
2
n
2
π
2
[
s
i
n
(
n
π
)
+
s
i
n
(
−
n
π
)
]
+
2
n
3
π
3
[
c
o
s
(
n
π
)
−
c
o
s
(
−
n
π
)
]
]
{\displaystyle +sin(nwx)[-{\frac {1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]]}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 4
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Problem Statement
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1) Develop the Fourier series of
f
(
x
¯
)
{\displaystyle f({\bar {x}})}
f
(
x
¯
)
{\displaystyle f({\bar {x}})}
f
n
(
x
¯
)
{\displaystyle f_{n}({\bar {x}})}
f
n
(
x
¯
)
:=
a
¯
0
+
∑
k
=
1
n
[
a
¯
k
cos
k
ω
x
¯
+
b
¯
k
sin
k
ω
x
¯
]
{\displaystyle f_{n}({\bar {x}}):={\bar {a}}_{0}+\sum _{k=1}^{n}[{\bar {a}}_{k}\cos k\omega {\bar {x}}+{\bar {b}}_{k}\sin k\omega {\bar {x}}]}
f
(
x
~
)
{\displaystyle f({\tilde {x}})}
f
(
x
)
{\displaystyle f(x)}
Solution
edit
To begin, the function
f
(
x
¯
)
{\displaystyle f({\bar {x}})}
f
(
x
¯
)
{\displaystyle f({\bar {x}})}
a
0
=
1
L
∫
0
2
f
(
x
¯
)
d
x
¯
=
A
2
{\displaystyle a_{0}={\frac {1}{L}}\int _{0}^{2}f({\bar {x}})d{\bar {x}}={\frac {A}{2}}}
a
k
=
2
L
∫
0
2
f
(
x
¯
)
cos
(
k
π
x
¯
2
)
d
x
¯
{\displaystyle a_{k}={\frac {2}{L}}\int _{0}^{2}f({\bar {x}})\cos({\frac {k\pi {\bar {x}}}{2}})d{\bar {x}}}
a
k
=
2
A
k
π
sin
(
k
π
2
)
{\displaystyle a_{k}={\frac {2A}{k\pi }}\sin({\frac {k\pi }{2}})}
f
k
(
x
¯
)
=
a
0
+
∑
k
=
1
n
[
a
n
cos
n
π
x
¯
2
]
{\displaystyle f_{k}({\bar {x}})=a_{0}+\sum _{k=1}^{n}[a_{n}\cos {\frac {n\pi {\bar {x}}}{2}}]}
f
k
=
A
2
+
∑
k
=
1
n
[
2
A
k
π
sin
(
k
π
2
)
cos
k
π
x
¯
2
]
{\displaystyle f_{k}={\frac {A}{2}}+\sum _{k=1}^{n}[{{\frac {2A}{k\pi }}\sin({\frac {k\pi }{2}})}\cos {\frac {k\pi {\bar {x}}}{2}}]}
f
0
(
x
¯
)
=
A
2
{\displaystyle f_{0}({\bar {x}})={\frac {A}{2}}}
f
1
(
x
¯
)
=
A
2
+
2
A
π
cos
(
π
x
¯
2
)
{\displaystyle f_{1}({\bar {x}})={\frac {A}{2}}+{\frac {2A}{\pi }}\cos({\frac {\pi {\bar {x}}}{2}})}
x
¯
=
x
−
1.25
{\displaystyle {\bar {x}}=x-1.25}
f
k
(
x
)
=
A
2
+
A
π
cos
(
π
(
x
−
1.25
)
2
)
{\displaystyle f_{k}(x)={\frac {A}{2}}+{\frac {A}{\pi }}\cos({\frac {\pi (x-1.25)}{2}})}
To begin, the function
f
(
x
¯
)
{\displaystyle f({\bar {x}})}
f
(
x
¯
)
{\displaystyle f({\bar {x}})}
a
0
=
1
L
∫
0
2
f
(
x
¯
)
d
x
¯
=
A
2
{\displaystyle a_{0}={\frac {1}{L}}\int _{0}^{2}f({\bar {x}})d{\bar {x}}={\frac {A}{2}}}
a
0
=
1
2
L
∫
0
4
f
(
x
~
)
d
x
~
=
A
2
{\displaystyle a_{0}={\frac {1}{2L}}\int _{0}^{4}f({\tilde {x}})d{\tilde {x}}={\frac {A}{2}}}
a
n
=
1
2
∫
0
4
f
x
~
cos
(
n
π
x
~
2
d
x
~
{\displaystyle a_{n}={\frac {1}{2}}\int _{0}^{4}f{\tilde {x}}\cos({\frac {n\pi {\tilde {x}}}{2}}d{\tilde {x}}}
a
n
=
1
2
f
(
x
~
)
cos
(
n
π
x
~
2
{\displaystyle a_{n}={\frac {1}{2}}f({\tilde {x}})\cos({\frac {n\pi {\tilde {x}}}{2}}}
a
n
=
A
n
π
sin
(
π
k
)
{\displaystyle a_{n}={\frac {A}{n\pi }}\sin(\pi k)}
b
n
=
1
2
∫
0
4
f
x
~
sin
(
n
π
x
~
2
d
x
~
{\displaystyle b_{n}={\frac {1}{2}}\int _{0}^{4}f{\tilde {x}}\sin({\frac {n\pi {\tilde {x}}}{2}}d{\tilde {x}}}
b
n
=
1
2
f
(
x
~
)
sin
(
n
π
x
~
2
)
{\displaystyle b_{n}={\frac {1}{2}}f({\tilde {x}})\sin({\frac {n\pi {\tilde {x}}}{2}})}
b
n
=
A
n
π
(
1
−
cos
(
π
k
)
)
{\displaystyle b_{n}={\frac {A}{n\pi }}(1-\cos(\pi k))}
f
k
(
x
~
)
=
A
2
+
∑
k
=
1
n
[
A
k
π
(
1
−
cos
(
π
k
)
)
(
sin
(
k
π
x
~
2
)
)
]
{\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+\sum _{k=1}^{n}[{\frac {A}{k\pi }}(1-\cos(\pi k))(\sin({\frac {k\pi {\tilde {x}}}{2}}))]}
f
0
(
x
~
)
=
A
2
{\displaystyle f_{0}({\tilde {x}})={\frac {A}{2}}}
f
1
(
x
~
)
=
A
2
+
A
π
sin
(
π
x
~
2
)
{\displaystyle f_{1}({\tilde {x}})={\frac {A}{2}}+{\frac {A}{\pi }}\sin({\frac {\pi {\tilde {x}}}{2}})}
x
~
=
x
−
.25
{\displaystyle {\tilde {x}}=x-.25}
f
1
(
x
)
=
A
2
+
A
π
sin
(
π
(
x
−
.25
)
2
)
{\displaystyle f_{1}(x)={\frac {A}{2}}+{\frac {A}{\pi }}\sin({\frac {\pi (x-.25)}{2}})}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 5
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Problem Statement
edit
Find the separated ODE's for the Heat Equation:
∂
2
u
∂
t
2
=
k
∂
2
u
∂
x
2
{\displaystyle {\frac {\partial ^{2}u}{\partial t^{2}}}=k{\frac {\partial ^{2}u}{\partial x^{2}}}}
k
=
{\displaystyle k=}
Solution
edit
Separation of Variables:
u
(
x
,
t
)
=
F
(
x
)
⋅
G
(
t
)
{\displaystyle u(x,t)=F(x)\cdot G(t)}
∂
u
(
x
,
t
)
∂
x
=
F
′
(
x
)
⋅
G
(
t
)
{\displaystyle {\frac {\partial u(x,t)}{\partial x}}=F'(x)\cdot G(t)}
∂
2
u
(
x
,
t
)
∂
x
2
=
F
″
(
x
)
⋅
G
(
t
)
{\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}=F''(x)\cdot G(t)}
∂
u
(
x
,
t
)
∂
t
=
F
(
x
)
⋅
G
˙
(
t
)
{\displaystyle {\frac {\partial u(x,t)}{\partial t}}=F(x)\cdot {\dot {G}}(t)}
∂
2
u
(
x
,
t
)
∂
t
2
=
F
(
x
)
⋅
G
¨
(
t
)
{\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial t^{2}}}=F(x)\cdot {\ddot {G}}(t)}
F
(
x
)
⋅
G
˙
(
t
)
=
k
F
″
(
x
)
⋅
G
(
t
)
{\displaystyle F(x)\cdot {\dot {G}}(t)=kF''(x)\cdot G(t)}
G
˙
(
t
)
k
G
(
t
)
=
F
″
(
x
)
F
(
x
)
=
c
(constant)
{\displaystyle {\frac {{\dot {G}}(t)}{kG(t)}}={\frac {F''(x)}{F(x)}}=c{\text{ (constant)}}}
G
˙
(
t
)
k
G
(
t
)
=
c
{\displaystyle {\frac {{\dot {G}}(t)}{kG(t)}}=c}
G
˙
(
t
)
=
k
c
G
(
t
)
{\displaystyle {\dot {G}}(t)=k\,c\,G(t)}
G
˙
(
t
)
−
k
c
G
(
t
)
=
0
{\displaystyle {\dot {G}}(t)-k\,c\,G(t)=0}
F
″
(
x
)
F
(
x
)
=
c
{\displaystyle {\frac {F''(x)}{F(x)}}=c}
F
″
(
x
)
=
c
F
(
x
)
{\displaystyle F''(x)=c\,F(x)}
F
″
(
x
)
−
c
F
(
x
)
=
0
{\displaystyle F''(x)-c\,F(x)=0}
G
˙
(
t
)
−
k
c
G
(
t
)
=
0
{\displaystyle {\dot {G}}(t)-k\,c\,G(t)=0}
F
″
(
x
)
−
c
F
(
x
)
=
0
{\displaystyle F''(x)-c\,F(x)=0}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 6
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Problem Statement
edit
Verify (4)-(5) p.19-9
<
ϕ
i
,
ϕ
j
>=
0
{\displaystyle <\phi _{i},\phi _{j}>=0}
i
≠
j
{\displaystyle i\neq j}
<
ϕ
i
,
ϕ
j
>=
L
/
2
{\displaystyle <\phi _{i},\phi _{j}>=L/2}
i
=
j
{\displaystyle i=j}
Solution
edit
Verification of (4)
edit
Using the integral scalar product calculation,
∫
0
L
ϕ
i
(
x
)
ϕ
j
(
x
)
d
x
{\displaystyle \int _{0}^{L}\phi _{i}(x)\phi _{j}(x)dx}
∫
0
L
s
i
n
(
ω
i
(
x
)
)
s
i
n
(
ω
j
(
x
)
)
d
x
{\displaystyle \int _{0}^{L}sin(\omega _{i}(x))sin(\omega _{j}(x))dx}
z
=
π
x
L
{\displaystyle z={\frac {\pi x}{L}}}
d
z
=
π
L
d
x
{\displaystyle dz={\frac {\pi }{L}}dx}
∫
0
π
s
i
n
(
i
z
)
s
i
n
(
j
z
)
d
z
L
π
{\displaystyle \int _{0}^{\pi }sin(iz)sin(jz)dz{\frac {L}{\pi }}}
1
2
[
s
i
n
(
i
−
j
)
z
i
−
j
−
s
i
n
(
i
+
j
)
z
i
+
j
]
{\displaystyle {\frac {1}{2}}[{\frac {sin(i-j)z}{i-j}}-{\frac {sin(i+j)z}{i+j}}]}
z
=
0
{\displaystyle z=0}
z
=
π
{\displaystyle z=\pi }
i
≠
j
{\displaystyle i\neq j}
Verification of (5)
edit
You can use the same equation from the verification of (4) from this point:
1
2
[
s
i
n
(
i
−
j
)
z
i
−
j
−
s
i
n
(
i
+
j
)
z
i
+
j
]
{\displaystyle {\frac {1}{2}}[{\frac {sin(i-j)z}{i-j}}-{\frac {sin(i+j)z}{i+j}}]}
z
=
0
{\displaystyle z=0}
z
=
π
{\displaystyle z=\pi }
L
2
{\displaystyle {\frac {L}{2}}}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 7
edit
Problem Statement
edit
u
(
x
,
t
)
=
∑
j
=
1
∞
a
j
cos
(
c
w
j
t
)
sin
(
w
j
x
)
{\displaystyle u(x,t)=\sum _{j=1}^{\infty }a_{j}\cos(cw_{j}t)\sin(w_{j}x)}
t
=
α
p
1
=
α
2
L
c
{\displaystyle t=\alpha p_{1}={\frac {\alpha 2L}{c}}}
α
=
.5
,
1
,
1.5
,
2
{\displaystyle \alpha =.5,1,1.5,2}
Solution
edit
a
j
=
2
[
(
(
−
1
)
j
)
−
1
]
π
3
j
3
{\displaystyle a_{j}={\frac {2[((-1)^{j})-1]}{\pi ^{3}j^{3}}}}
w
j
=
j
π
L
{\displaystyle w_{j}={\frac {j\pi }{L}}}
u
(
x
,
2
/
3
)
{\displaystyle u(x,2/3)}
u
(
x
,
4
/
3
)
{\displaystyle u(x,4/3)}
u
(
x
,
2
)
{\displaystyle u(x,2)}
u
(
x
,
8
/
3
)
{\displaystyle u(x,8/3)}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
![{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(nwx)+b_{n}sin(nwx)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/572aa93560e3630683588a21c7ec56e16ddf8c5b)
![{\displaystyle a_{n}={\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d258d6131ac7db6db3bb6f452fefd88e4665da2a)
![{\displaystyle b_{n}={\frac {-1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1701f6016a8083df714cbc0dd71c9713fd9bcc6)
![{\displaystyle {\frac {1}{3}}+\sum _{n=1}^{\infty }cos(nwx)[{\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d1b3c785ae7c556a2ef2c7898b84e5b699ca899)
![{\displaystyle +sin(nwx)[-{\frac {1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d9a819ff8a5c1b538d2e0c1f188752c8061c7d1)
![{\displaystyle f_{n}({\bar {x}}):={\bar {a}}_{0}+\sum _{k=1}^{n}[{\bar {a}}_{k}\cos k\omega {\bar {x}}+{\bar {b}}_{k}\sin k\omega {\bar {x}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4148a2e3b7a30083ad8bd117fa313bd2b8f2204)
.JPG)
![{\displaystyle f_{k}({\bar {x}})=a_{0}+\sum _{k=1}^{n}[a_{n}\cos {\frac {n\pi {\bar {x}}}{2}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cc46d3cbd89239f72ac5880a5d2d543eb352c72)
![{\displaystyle f_{k}={\frac {A}{2}}+\sum _{k=1}^{n}[{{\frac {2A}{k\pi }}\sin({\frac {k\pi }{2}})}\cos {\frac {k\pi {\bar {x}}}{2}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e8c93229a59b63538e7e80387133e57164cd1b3)
.JPG)
![{\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+\sum _{k=1}^{n}[{\frac {A}{k\pi }}(1-\cos(\pi k))(\sin({\frac {k\pi {\tilde {x}}}{2}}))]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/933cdaee1e78aa31c54808efafc3dc19c3b120dc)
![{\displaystyle {\frac {1}{2}}[{\frac {sin(i-j)z}{i-j}}-{\frac {sin(i+j)z}{i+j}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ce36a3c8f876d2f7be13eb64e2a9a2867294867)
![{\displaystyle a_{j}={\frac {2[((-1)^{j})-1]}{\pi ^{3}j^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e26da390654d425ff6ffa2e7c79e2834e53977a0)
