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
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}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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}
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))}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Is the given function even or odd or neither even nor odd? Find its Fourier Series.
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 )]]}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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)}
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}})}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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=}
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}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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}
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}
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}}}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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}
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)}
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)
