Report 6
Intermediate Engineering Analysis
Problem Statement
edit
Solve the problems 11 and 12 from Kreyszig p.491
Solution
edit
First, for problem 11:
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}\!}
<
−
1
<
x
<
1
>
{\displaystyle <-1<x<1>\!}
p
=
2
{\displaystyle p=2\!}
p
=
2
=
2
L
{\displaystyle p=2=2L\!}
L
=
1
{\displaystyle L=1\!}
We check is the function is even, odd or neither.
f
(
x
)
{\displaystyle f(x)\!}
f
(
x
)
=
f
(
−
x
)
{\displaystyle f(x)=f(-x)\!}
b
n
=
0
{\displaystyle b_{n}=0\!}
b
n
{\displaystyle b_{n}\!}
a
0
{\displaystyle a_{0}\!}
a
0
=
1
2
L
∫
−
L
L
f
(
x
)
d
x
{\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}
a
0
=
1
2
∫
−
1
1
x
2
d
x
=
1
2
[
1
3
x
3
]
−
1
1
=
1
2
[
1
3
+
1
3
]
=
1
3
{\displaystyle a_{0}={\frac {1}{2}}\int _{-1}^{1}x^{2}dx={\frac {1}{2}}\left[{\frac {1}{3}}x^{3}\right]_{-1}^{1}={\frac {1}{2}}\left[{\frac {1}{3}}+{\frac {1}{3}}\right]={\frac {1}{3}}\!}
a
n
{\displaystyle a_{n}\!}
a
n
=
1
L
∫
−
L
L
f
(
x
)
c
o
s
(
n
π
x
L
)
d
x
{\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos({\frac {n\pi x}{L}})dx\!}
a
n
=
1
∫
−
1
1
x
2
c
o
s
(
n
π
x
)
d
x
{\displaystyle a_{n}=1\int _{-1}^{1}x^{2}cos(n\pi x)dx\!}
a
n
=
2
∫
0
1
x
2
c
o
s
(
n
π
x
)
d
x
{\displaystyle a_{n}=2\int _{0}^{1}x^{2}cos(n\pi x)dx\!}
∫
f
d
g
=
f
g
−
∫
g
d
f
{\displaystyle \int fdg=fg-\int gdf\!}
f
=
x
2
d
g
=
c
o
s
(
π
n
x
)
d
x
d
f
=
2
x
d
x
g
=
s
i
n
(
π
n
x
)
π
n
{\displaystyle {\begin{matrix}f=x^{2}&dg=cos(\pi nx)dx\\df=2xdx&g={\frac {sin(\pi nx)}{\pi n}}\end{matrix}}\!}
[
x
2
s
i
n
(
π
n
x
)
π
n
]
0
1
−
2
π
n
∫
0
1
x
s
i
n
(
π
n
x
)
d
x
{\displaystyle \left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\int _{0}^{1}xsin(\pi nx)dx\!}
f
=
x
d
g
=
s
i
n
(
π
n
x
)
d
f
=
d
x
g
=
−
c
o
s
(
π
n
x
)
π
n
{\displaystyle {\begin{matrix}f=x&dg=sin(\pi nx)\\df=dx&g={\frac {-cos(\pi nx)}{\pi n}}\end{matrix}}\!}
[
−
x
c
o
s
(
π
n
x
)
π
n
]
0
1
−
1
π
n
∫
0
1
c
o
s
(
π
n
x
)
d
x
{\displaystyle \left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\!}
2
[
[
x
2
s
i
n
(
π
n
x
)
π
n
]
0
1
−
2
π
n
[
−
x
c
o
s
(
π
n
x
)
π
n
]
0
1
−
1
π
n
∫
0
1
c
o
s
(
π
n
x
)
d
x
]
{\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\right]\!}
2
[
[
x
2
s
i
n
(
π
n
x
)
π
n
]
0
1
−
2
π
n
[
−
x
c
o
s
(
π
n
x
)
π
n
]
0
1
−
1
π
n
[
s
i
n
(
π
n
x
)
π
n
]
0
1
]
{\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\left[{\frac {sin(\pi nx)}{\pi n}}\right]_{0}^{1}\right]\!}
2
[
[
0
−
0
]
+
2
π
2
n
2
[
c
o
s
(
π
n
)
]
−
1
π
n
[
0
−
0
]
]
{\displaystyle 2\left[\left[0-0\right]+{\frac {2}{\pi ^{2}n^{2}}}\left[cos(\pi n)\right]-{\frac {1}{\pi n}}\left[0-0\right]\right]\!}
4
π
2
n
2
c
o
s
(
π
n
)
{\displaystyle {\frac {4}{\pi ^{2}n^{2}}}cos(\pi n)\!}
c
o
s
(
π
n
)
=
(
−
1
)
n
{\displaystyle cos(\pi n)=(-1)^{n}\!}
∴
a
n
=
4
π
2
n
2
(
−
1
)
n
{\displaystyle \therefore a_{n}={\frac {4}{\pi ^{2}n^{2}}}(-1)^{n}\!}
f
(
x
)
=
a
0
+
∑
n
=
1
∞
a
n
c
o
s
(
π
n
L
)
x
{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {\pi n}{L}})x\!}
f
(
x
)
=
1
3
+
∑
n
=
1
∞
4
π
2
n
2
(
−
1
)
n
c
o
s
(
π
n
)
x
{\displaystyle f(x)={\frac {1}{3}}+\sum _{n=1}^{\infty }{\frac {4}{\pi ^{2}n^{2}}}(-1)^{n}cos(\pi n)x\!}
Now problem 12:
f
(
x
)
=
1
−
x
2
4
{\displaystyle f(x)=1-{\frac {x^{2}}{4}}\!}
<
−
2
<
x
<
2
>
{\displaystyle <-2<x<2>\!}
p
=
2
{\displaystyle p=2\!}
p
=
4
=
2
L
{\displaystyle p=4=2L\!}
L
=
2
{\displaystyle L=2\!}
We check is the function is even, odd or neither.
f
(
x
)
{\displaystyle f(x)\!}
f
(
x
)
=
f
(
−
x
)
{\displaystyle f(x)=f(-x)\!}
b
n
=
0
{\displaystyle b_{n}=0\!}
b
n
{\displaystyle b_{n}\!}
a
0
{\displaystyle a_{0}\!}
a
0
=
1
2
L
∫
−
L
L
f
(
x
)
d
x
{\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}
a
0
=
1
4
∫
−
2
2
1
−
x
2
4
d
x
=
1
4
[
x
−
1
12
x
3
]
−
2
2
=
1
4
[
24
−
8
12
−
24
+
8
12
]
=
2
3
{\displaystyle a_{0}={\frac {1}{4}}\int _{-2}^{2}1-{\frac {x^{2}}{4}}dx={\frac {1}{4}}\left[x-{\frac {1}{12}}x^{3}\right]_{-2}^{2}={\frac {1}{4}}\left[{\frac {24-8}{12}}-{\frac {24+8}{12}}\right]={\frac {2}{3}}\!}
a
n
{\displaystyle a_{n}\!}
a
n
=
1
L
∫
−
L
L
f
(
x
)
c
o
s
(
n
π
x
L
)
d
x
{\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos({\frac {n\pi x}{L}})dx\!}
a
n
=
1
2
∫
−
2
2
(
1
−
x
2
4
c
o
s
(
n
π
x
)
d
x
{\displaystyle a_{n}={\frac {1}{2}}\int _{-2}^{2}(1-{\frac {x^{2}}{4}}cos(n\pi x)dx\!}
2
π
2
n
2
{\displaystyle {\frac {2}{\pi ^{2}n^{2}}}\!}
Finally,
f
(
x
)
=
a
0
+
∑
n
=
1
∞
a
n
c
o
s
(
π
n
L
)
x
{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {\pi n}{L}})x\!}
f
(
x
)
=
2
3
+
∑
n
=
1
∞
−
2
π
2
n
2
c
o
s
(
π
n
2
)
x
{\displaystyle f(x)={\frac {2}{3}}+\sum _{n=1}^{\infty }{\frac {-2}{\pi ^{2}n^{2}}}cos({\frac {\pi n}{2}})x\!}
Solved by Francisco Arrieta
Problem Statement
edit
Find the Fourier series expansion for
f
(
x
)
{\displaystyle f(x)\!}
Develop the Fourier series expansion of
f
(
x
¯
)
{\displaystyle f({\bar {x}})\!}
Plot
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}}]\!}
for n=0,1,2,4,8. Observe the values of
f
n
(
x
¯
)
{\displaystyle f_{n}({\bar {x}})\!}
f
(
x
)
{\displaystyle f(x)\!}
Solution
edit
p
=
2
L
=
4
{\displaystyle p=2L=4\!}
L
=
2
{\displaystyle L=2\!}
ω
=
2
π
p
{\displaystyle \omega ={\frac {2\pi }{p}}\!}
a
¯
0
=
1
2
L
∫
−
L
L
f
(
x
¯
)
d
x
¯
{\displaystyle {\bar {a}}_{0}={\frac {1}{2L}}\int _{-L}^{L}f({\bar {x}})d{\bar {x}}\!}
a
¯
0
=
1
4
∫
−
2
2
f
(
x
¯
)
d
x
¯
{\displaystyle {\bar {a}}_{0}={\frac {1}{4}}\int _{-2}^{2}f({\bar {x}})d{\bar {x}}\!}
a
¯
0
=
A
2
{\displaystyle {\bar {a}}_{0}={\frac {A}{2}}\!}
a
¯
k
=
1
L
∫
−
L
L
f
(
x
¯
)
cos
k
ω
x
¯
d
x
¯
{\displaystyle {\bar {a}}_{k}={\frac {1}{L}}\int _{-L}^{L}f({\bar {x}})\cos k\omega {\bar {x}}d{\bar {x}}\!}
a
¯
k
=
1
2
∫
−
2
2
f
(
x
¯
)
cos
k
π
x
¯
2
d
x
¯
{\displaystyle {\bar {a}}_{k}={\frac {1}{2}}\int _{-2}^{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 {\bar {a}}_{k}={\frac {2A}{k\pi }}\sin {\frac {k\pi }{2}}\!}
Note:
If k is even
a
¯
k
=
0
{\displaystyle {\bar {a}}_{k}=0\!}
If k=1, 5, 9...
a
¯
k
=
2
A
k
π
{\displaystyle {\bar {a}}_{k}={\frac {2A}{k\pi }}\!}
If k=3, 7, 11...
a
¯
k
=
−
2
A
k
π
{\displaystyle {\bar {a}}_{k}={\frac {-2A}{k\pi }}\!}
b
¯
k
=
1
L
∫
−
L
L
f
(
x
¯
)
sin
k
ω
x
¯
d
x
¯
{\displaystyle {\bar {b}}_{k}={\frac {1}{L}}\int _{-L}^{L}f({\bar {x}})\sin k\omega {\bar {x}}d{\bar {x}}\!}
Note:
b
¯
k
=
0
{\displaystyle {\bar {b}}_{k}=0\!}
Then Fourier series becomes a Fourier cosine series:
f
k
(
x
¯
)
=
A
2
+
∑
k
=
1
n
(
2
A
k
π
cos
k
π
2
x
¯
)
{\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\cos {\frac {k\pi }{2}}{\bar {x}})\!}
For n=0
f
k
(
x
¯
)
=
A
2
{\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}\!}
For n=1
f
k
(
x
¯
)
=
A
2
+
(
2
A
π
cos
π
2
x
¯
)
{\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}+({\frac {2A}{\pi }}\cos {\frac {\pi }{2}}{\bar {x}})\!}
For n=5
f
k
(
x
¯
)
=
A
2
+
2
A
π
(
cos
π
2
x
¯
−
1
3
cos
3
π
2
x
¯
+
1
5
cos
5
π
2
x
¯
)
{\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}+{\frac {2A}{\pi }}(\cos {\frac {\pi }{2}}{\bar {x}}-{\frac {1}{3}}\cos {\frac {3\pi }{2}}{\bar {x}}+{\frac {1}{5}}\cos {\frac {5\pi }{2}}{\bar {x}})\!}
After transforming the variable of
f
(
x
¯
)
{\displaystyle f({\bar {x}})\!}
f
k
(
x
)
=
A
2
+
∑
k
=
1
n
(
2
A
k
π
cos
k
π
2
(
x
−
1.25
)
)
{\displaystyle f_{k}(x)={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\cos {\frac {k\pi }{2}}(x-1.25))\!}
Do the same as above, but using
f
(
x
~
)
{\displaystyle f({\tilde {x}})\!}
f
(
x
)
{\displaystyle f(x)\!}
Solution
edit
p
=
2
L
=
4
{\displaystyle p=2L=4\!}
L
=
2
{\displaystyle L=2\!}
ω
=
2
π
p
{\displaystyle \omega ={\frac {2\pi }{p}}\!}
a
~
0
=
1
2
L
∫
0
2
L
f
(
x
~
)
d
x
~
{\displaystyle {\tilde {a}}_{0}={\frac {1}{2L}}\int _{0}^{2L}f({\tilde {x}})d{\tilde {x}}\!}
a
~
0
=
1
4
∫
0
4
f
(
x
~
)
d
x
~
{\displaystyle {\tilde {a}}_{0}={\frac {1}{4}}\int _{0}^{4}f({\tilde {x}})d{\tilde {x}}\!}
a
~
0
=
A
2
{\displaystyle {\tilde {a}}_{0}={\frac {A}{2}}\!}
a
~
k
=
1
L
∫
0
2
L
f
(
x
~
)
cos
k
ω
x
~
d
x
~
{\displaystyle {\tilde {a}}_{k}={\frac {1}{L}}\int _{0}^{2L}f({\tilde {x}})\cos k\omega {\tilde {x}}d{\tilde {x}}\!}
a
~
k
=
1
2
∫
0
4
f
(
x
~
)
cos
k
ω
x
~
d
x
~
{\displaystyle {\tilde {a}}_{k}={\frac {1}{2}}\int _{0}^{4}f({\tilde {x}})\cos k\omega {\tilde {x}}d{\tilde {x}}\!}
a
~
k
=
A
k
π
sin
π
k
{\displaystyle {\tilde {a}}_{k}={\frac {A}{k\pi }}\sin \pi k\!}
Note:
b
¯
k
=
0
{\displaystyle {\bar {b}}_{k}=0\!}
b
~
k
=
1
L
∫
0
2
L
f
(
x
~
)
sin
k
ω
x
~
d
x
~
{\displaystyle {\tilde {b}}_{k}={\frac {1}{L}}\int _{0}^{2L}f({\tilde {x}})\sin k\omega {\tilde {x}}d{\tilde {x}}\!}
b
~
k
=
1
2
∫
0
4
f
(
x
~
)
sin
k
π
2
x
~
d
x
~
{\displaystyle {\tilde {b}}_{k}={\frac {1}{2}}\int _{0}^{4}f({\tilde {x}})\sin k{\frac {\pi }{2}}{\tilde {x}}d{\tilde {x}}\!}
b
~
k
=
−
A
π
k
(
cos
π
k
−
1
)
{\displaystyle {\tilde {b}}_{k}={\frac {-A}{\pi k}}(\cos \pi k-1)\!}
Note:
If k is even
a
¯
k
=
0
{\displaystyle {\bar {a}}_{k}=0\!}
If k=1, 3, 5...
a
¯
k
=
2
A
k
π
{\displaystyle {\bar {a}}_{k}={\frac {2A}{k\pi }}\!}
Then Fourier series becomes a Fourier sine series:
f
k
(
x
~
)
=
A
2
+
∑
k
=
1
n
(
2
A
k
π
sin
k
π
2
x
~
)
{\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\sin {\frac {k\pi }{2}}{\tilde {x}})\!}
For n=0:
f
k
(
x
~
)
=
A
2
{\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}\!}
For n=1:
f
k
(
x
~
)
=
A
2
+
(
2
A
π
sin
π
2
x
~
)
{\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+({\frac {2A}{\pi }}\sin {\frac {\pi }{2}}{\tilde {x}})\!}
For n=5:
f
k
(
x
~
)
=
A
2
+
2
A
π
(
sin
π
2
x
~
+
1
3
sin
3
π
2
x
~
+
1
5
sin
5
π
2
x
~
)
{\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+{\frac {2A}{\pi }}(\sin {\frac {\pi }{2}}{\tilde {x}}+{\frac {1}{3}}\sin {\frac {3\pi }{2}}{\tilde {x}}+{\frac {1}{5}}\sin {\frac {5\pi }{2}}{\tilde {x}})\!}
After transforming the variable of
f
(
x
¯
)
{\displaystyle f({\bar {x}})\!}
f
k
(
x
)
=
A
2
+
∑
k
=
1
n
(
2
A
k
π
sin
k
π
2
(
x
−
.25
)
)
{\displaystyle f_{k}(x)={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\sin {\frac {k\pi }{2}}(x-.25))\!}
Problem Statement
edit
Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.
Solution
edit
15.
f
(
x
)
=
{
−
x
−
π
,
−
π
<
x
<
−
π
/
2
x
,
−
π
/
2
<
x
<
π
/
2
π
−
x
,
π
/
2
<
x
<
π
{\displaystyle f(x)=\left\{{\begin{matrix}-x-\pi ,-\pi <x<-\pi /2\\x,-\pi /2<x<\pi /2\\\pi -x,\pi /2<x<\pi \end{matrix}}\right.}
Since
f
(
−
x
)
=
−
f
(
x
)
{\displaystyle f(-x)=-f(x)}
a
0
=
(
1
/
π
)
∗
[
−
1
∫
−
π
−
π
/
2
(
x
+
π
)
d
x
+
∫
−
π
/
2
π
/
2
(
x
)
d
x
+
∫
π
/
2
π
(
π
−
x
)
d
x
]
{\displaystyle a_{0}=(1/\pi )*[-1\int _{-\pi }^{-\pi /2}(x+\pi )dx+\int _{-\pi /2}^{\pi /2}(x)dx+\int _{\pi /2}^{\pi }(\pi -x)dx]}
a
0
=
0
{\displaystyle a_{0}=0}
f
(
x
)
=
8
k
/
π
2
[
s
i
n
(
π
∗
x
/
L
)
−
s
i
n
(
3
π
∗
x
/
L
)
/
9...
)
{\displaystyle f(x)=8k/\pi ^{2}[sin(\pi *x/L)-sin(3\pi *x/L)/9...)}
L
=
π
,
k
=
π
/
2
{\displaystyle L=\pi ,k=\pi /2}
f
(
x
)
=
4
/
π
(
s
i
n
x
−
s
i
n
(
3
x
)
/
9
+
s
i
n
(
5
x
)
/
25
−
s
i
n
(
6
x
)
/
36....
)
{\displaystyle f(x)=4/\pi (sinx-sin(3x)/9+sin(5x)/25-sin(6x)/36....)}
f
(
x
)
=
0
{\displaystyle f(x)=0}
f
(
x
)
=
1
−
|
x
|
,
−
1
≤
x
≤
1
{\displaystyle f(x)=1-|x|,-1\leq x\leq 1}
f
(
x
)
=
{
1
+
x
−
π
,
−
1
<
x
<
0
1
−
x
,
0
<
x
<
1
{\displaystyle f(x)=\left\{{\begin{matrix}1+x-\pi ,-1<x<0\\1-x,0<x<1\\\end{matrix}}\right.}
a
0
=
1
/
2
[
∫
−
1
0
(
1
+
x
)
d
x
+
∫
0
1
(
1
−
x
)
d
x
]
{\displaystyle a_{0}=1/2[\int _{-1}^{0}(1+x)dx+\int _{0}^{1}(1-x)dx]}
a
0
=
1
/
2
[
1
−
1
/
2
+
1
−
1
/
2
]
{\displaystyle a_{0}=1/2[1-1/2+1-1/2]}
a
0
=
1
/
2
{\displaystyle a_{0}=1/2}
a
n
=
1
/
1
[
∫
−
1
0
(
1
+
x
)
c
o
s
(
n
π
∗
x
/
1
)
d
x
+
∫
0
1
(
1
−
x
)
c
o
s
(
n
π
∗
x
/
1
)
d
x
]
{\displaystyle a_{n}=1/1[\int _{-1}^{0}(1+x)cos(n\pi *x/1)dx+\int _{0}^{1}(1-x)cos(n\pi *x/1)dx]}
a
n
=
2
[
1
/
(
n
2
π
2
)
(
−
1
)
n
/
(
n
2
π
2
)
]
{\displaystyle a_{n}=2[1/(n^{2}\pi ^{2})_{(}-1)^{n}/(n^{2}\pi ^{2})]}
a
n
=
4
/
(
n
2
π
2
)
{\displaystyle a_{n}=4/(n^{2}\pi ^{2})}
when n is odd.
b
n
=
1
/
1
[
∫
−
1
0
(
1
+
x
)
s
i
n
(
n
π
∗
x
/
1
)
d
x
+
∫
0
1
(
1
−
x
)
s
i
n
(
n
π
∗
x
/
1
)
d
x
]
{\displaystyle b_{n}=1/1[\int _{-1}^{0}(1+x)sin(n\pi *x/1)dx+\int _{0}^{1}(1-x)sin(n\pi *x/1)dx]}
b
n
=
0
{\displaystyle b_{n}=0}
f
(
x
)
=
1
/
2
+
∑
n
=
0
∞
4
/
(
n
2
π
2
)
∗
c
o
s
(
n
π
∗
x
)
{\displaystyle f(x)=1/2+\sum _{n=0}^{\infty }4/(n^{2}\pi ^{2})*cos(n\pi *x)}
Since
a
n
=
0
{\displaystyle a_{n}=0}
f
(
x
)
{\displaystyle f(x)}
f
(
x
)
=
1
/
2
{\displaystyle f(x)=1/2}
solved by Luca Imponenti
Problem Statement
edit
Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:
y
″
−
3
y
′
+
2
y
=
f
(
x
)
{\displaystyle y''-3y'+2y=f(x)\!}
and the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1\ ,\ y'(0)=0\!}
1. Find
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
y
n
″
+
a
y
n
′
+
b
y
n
=
r
n
(
x
)
{\displaystyle y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!}
with the same initial conditions as above.
Plot
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
n
=
2
,
4
,
8
{\displaystyle n=2,4,8\!}
[
0
,
10
]
{\displaystyle [0,10]\!}
2. Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.
Level 1:
n
=
0
,
1
{\displaystyle n=0,1\!}
Fourier Series
edit
One period of the window function p9.8 is described as follows
f
(
x
)
=
{
0
−
1.75
<
x
<
l
0.25
A
0.25
<
x
<
2.25
{\displaystyle f(x)={\begin{cases}0&\ \ -1.75<x<l0.25\\A&\ \ 0.25<x<2.25\\\end{cases}}}
From the above intervals one can see that the period,
p
=
4
{\displaystyle p=4\!}
L
=
2
{\displaystyle L=2\!}
−
1.75
{\displaystyle -1.75}
2.25
{\displaystyle 2.25\!}
a
0
=
1
2
L
∫
−
1.75
2.25
f
(
x
)
d
x
{\displaystyle a_{0}={\frac {1}{2L}}\int _{-1.75}^{2.25}f(x)\ dx\!}
a
0
=
1
4
(
∫
−
1.75
0.25
0
d
x
+
∫
0.25
2.25
A
d
x
)
{\displaystyle a_{0}={\frac {1}{4}}(\int _{-1.75}^{0.25}0\ dx+\int _{0.25}^{2.25}A\ dx)\!}
a
0
=
+
1
4
(
0
+
∫
0.25
2.25
A
d
x
)
{\displaystyle a_{0}=+{\frac {1}{4}}(0+\int _{0.25}^{2.25}A\ dx)\!}
a
0
=
1
4
(
2.25
A
−
0.25
A
)
{\displaystyle a_{0}={\frac {1}{4}}(2.25A-0.25A)\!}
a
0
=
A
2
{\displaystyle a_{0}={\frac {A}{2}}\!}
The integral from
−
1.75
{\displaystyle -1.75\!}
0.25
{\displaystyle 0.25\!}
a
n
=
1
L
∫
0.25
2.25
f
(
x
)
c
o
s
(
n
π
x
L
)
d
x
{\displaystyle a_{n}={\frac {1}{L}}\int _{0.25}^{2.25}f(x)cos({\frac {n\pi x}{L}})\ dx\!}
a
n
=
1
2
∫
0.25
2.25
A
c
o
s
(
n
π
x
2
)
d
x
{\displaystyle a_{n}={\frac {1}{2}}\int _{0.25}^{2.25}Acos({\frac {n\pi x}{2}})\ dx\!}
a
n
=
2
A
2
n
π
(
s
i
n
(
2.25
n
π
2
)
−
s
i
n
(
0.25
n
π
2
)
{\displaystyle a_{n}={\frac {2A}{2n\pi }}(sin({\frac {2.25n\pi }{2}})-sin({\frac {0.25n\pi }{2}})\!}
a
n
=
A
n
π
(
s
i
n
(
9
n
π
8
)
−
s
i
n
(
n
π
8
)
{\displaystyle a_{n}={\frac {A}{n\pi }}(sin({\frac {9n\pi }{8}})-sin({\frac {n\pi }{8}})\!}
and
b
n
=
1
L
∫
0.25
2.25
f
(
x
)
s
i
n
(
n
π
x
L
)
d
x
{\displaystyle b_{n}={\frac {1}{L}}\int _{0.25}^{2.25}f(x)sin({\frac {n\pi x}{L}})\ dx\!}
b
n
=
1
2
∫
0.25
2.25
A
s
i
n
(
n
π
x
2
)
d
x
{\displaystyle b_{n}={\frac {1}{2}}\int _{0.25}^{2.25}Asin({\frac {n\pi x}{2}})\ dx\!}
b
n
=
2
A
2
n
π
(
c
o
s
(
2.25
n
π
2
)
−
c
o
s
(
0.25
n
π
2
)
{\displaystyle b_{n}={\frac {2A}{2n\pi }}(cos({\frac {2.25n\pi }{2}})-cos({\frac {0.25n\pi }{2}})\!}
b
n
=
A
n
π
(
A
c
o
s
(
n
π
8
)
−
A
c
o
s
(
9
n
π
8
)
{\displaystyle b_{n}={\frac {A}{n\pi }}(Acos({\frac {n\pi }{8}})-Acos({\frac {9n\pi }{8}})\!}
The coefficients give the Fourier series:
f
(
x
)
=
a
0
+
∑
n
=
1
∞
[
a
n
c
o
s
(
n
π
x
L
)
+
b
n
s
i
n
(
n
π
x
L
)
]
{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos({\frac {n\pi x}{L}})+b_{n}sin({\frac {n\pi x}{L}})]\!}
f
(
x
)
=
A
2
+
∑
n
=
1
∞
[
A
n
π
(
s
i
n
(
9
n
π
8
)
−
s
i
n
(
n
π
8
)
)
c
o
s
(
n
π
x
2
)
{\displaystyle f(x)={\frac {A}{2}}+\sum _{n=1}^{\infty }[{\frac {A}{n\pi }}(sin({\frac {9n\pi }{8}})-sin({\frac {n\pi }{8}}))cos({\frac {n\pi x}{2}})\!}
+
A
n
π
(
c
o
s
(
n
π
8
)
−
c
o
s
(
9
n
π
8
)
)
s
i
n
(
n
π
x
2
)
]
{\displaystyle +{\frac {A}{n\pi }}(cos({\frac {n\pi }{8}})-cos({\frac {9n\pi }{8}}))sin({\frac {n\pi x}{2}})]\!}
Homogeneous Solution
edit
Considering the homogeneous case of our ODE:
y
″
−
3
y
′
+
2
y
=
0
{\displaystyle y''-3y'+2y=0\!}
The characteristic equation is
λ
2
−
3
λ
+
2
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=0\!}
(
λ
−
1
)
(
λ
−
2
)
=
0
{\displaystyle (\lambda -1)(\lambda -2)=0\!}
λ
1
=
1
,
λ
1
=
2
{\displaystyle \lambda _{1}=1,\lambda _{1}=2\!}
Therefore our homogeneous solution is of the form
y
h
=
c
1
e
x
+
c
2
e
2
x
{\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!}
Particular Solution
edit
Considering the case with f(x) as excitation
y
″
−
3
y
′
+
2
y
=
A
2
+
∑
k
=
1
n
A
k
π
[
(
s
i
n
(
9
k
π
8
)
−
s
i
n
(
k
π
8
)
)
c
o
s
(
k
π
x
2
)
{\displaystyle y''-3y'+2y={\frac {A}{2}}+\sum _{k=1}^{n}{\frac {A}{k\pi }}[(sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}}))cos({\frac {k\pi x}{2}})\!}
+
(
c
o
s
(
k
π
8
)
−
c
o
s
(
9
k
π
8
)
)
s
i
n
(
k
π
x
2
)
]
{\displaystyle +(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}
The solution will be of the form
y
n
=
A
0
+
∑
k
=
1
n
A
k
c
o
s
(
k
π
x
2
)
+
∑
k
=
1
n
B
k
s
i
n
(
k
π
x
2
)
{\displaystyle y_{n}=A_{0}+\sum _{k=1}^{n}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{n}B_{k}sin({\frac {k\pi x}{2}})\!}
Taking the derivatives
y
n
′
=
−
A
n
π
2
∑
k
=
2
n
k
s
i
n
(
k
π
x
2
)
+
B
n
π
2
∑
k
=
2
n
k
c
o
s
(
k
π
x
2
)
{\displaystyle y_{n}'=-A_{n}{\frac {\pi }{2}}\sum _{k=2}^{n}ksin({\frac {k\pi x}{2}})+B_{n}{\frac {\pi }{2}}\sum _{k=2}^{n}kcos({\frac {k\pi x}{2}})\!}
y
n
″
=
−
A
n
π
2
4
∑
k
=
3
n
k
2
c
o
s
(
k
π
x
2
)
−
B
n
π
2
4
∑
k
=
3
n
k
2
s
i
n
(
k
π
x
2
)
{\displaystyle y_{n}''=-A_{n}{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}k^{2}cos({\frac {k\pi x}{2}})-B_{n}{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}k^{2}sin({\frac {k\pi x}{2}})\!}
Plugging these back into the ODE:
−
π
2
4
∑
k
=
3
n
A
k
k
2
c
o
s
(
k
π
x
2
)
−
π
2
4
∑
k
=
3
n
B
k
k
2
s
i
n
(
k
π
x
2
)
−
3
[
−
π
2
∑
k
=
2
n
A
k
k
s
i
n
(
k
π
x
2
)
{\displaystyle -{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}A_{k}k^{2}cos({\frac {k\pi x}{2}})-{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}B_{k}k^{2}sin({\frac {k\pi x}{2}})-3[-{\frac {\pi }{2}}\sum _{k=2}^{n}A_{k}ksin({\frac {k\pi x}{2}})\!}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle +\frac{\pi}{2}\sum_{k=2}^{n}B_kkcos(\frac{k\pi x}{2})]+2[A_0+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})]\!}
=
A
2
+
∑
k
=
1
n
A
k
π
[
(
s
i
n
(
9
k
π
8
)
−
s
i
n
(
k
π
8
)
)
c
o
s
(
k
π
x
2
)
+
(
c
o
s
(
k
π
8
)
−
c
o
s
(
9
k
π
8
)
)
s
i
n
(
k
π
x
2
)
]
{\displaystyle ={\frac {A}{2}}+\sum _{k=1}^{n}{\frac {A}{k\pi }}[(sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}}))cos({\frac {k\pi x}{2}})+(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}
Setting the two constants equal
2
A
0
=
A
2
{\displaystyle 2A_{0}={\frac {A}{2}}\!}
A
0
=
A
4
{\displaystyle A_{0}={\frac {A}{4}}\!}
This is valid for all values of n. Since the coefficients of the excitation
s
i
n
(
9
k
π
8
)
−
s
i
n
(
k
π
8
)
{\displaystyle sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}})\!}
c
o
s
(
k
π
8
)
−
c
o
s
(
9
k
π
8
)
{\displaystyle cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}})\!}
A
k
{\displaystyle A_{k}\!}
B
k
{\displaystyle B_{k}\!}
n
=
1
{\displaystyle n=1\!}
[
2
0
0
2
]
∗
[
A
1
B
1
]
=
[
A
π
(
s
i
n
(
9
π
8
)
−
s
i
n
(
π
8
)
)
A
π
(
c
o
s
(
π
8
)
−
c
o
s
(
9
π
8
)
)
]
{\displaystyle {\begin{bmatrix}2&0\\\\0&2\end{bmatrix}}*{\begin{bmatrix}A_{1}\\B_{1}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi }}(sin({\frac {9\pi }{8}})-sin({\frac {\pi }{8}}))\\\ {\frac {A}{\pi }}(cos({\frac {\pi }{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!}
Assuming
A
=
1
{\displaystyle A=1\!}
A
1
=
−
0.1218
,
B
1
=
0.2941
{\displaystyle A_{1}=-0.1218\ ,\ B_{1}=0.2941\!}
For the remaining coefficients to be solved all sums will be used so a more general equation may be written:
[
2
−
(
π
k
)
2
4
−
3
π
k
2
−
3
π
k
2
2
−
(
π
k
)
2
4
]
∗
[
A
k
B
k
]
=
[
A
π
k
(
s
i
n
(
9
π
k
8
)
−
s
i
n
(
π
k
8
)
)
A
π
k
(
c
o
s
(
π
k
8
)
−
c
o
s
(
9
π
8
)
)
]
{\displaystyle {\begin{bmatrix}2-{\frac {(\pi k)^{2}}{4}}&{\frac {-3\pi k}{2}}\\{\frac {-3\pi k}{2}}&2-{\frac {(\pi k)^{2}}{4}}\end{bmatrix}}*{\begin{bmatrix}A_{k}\\B_{k}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi k}}(sin({\frac {9\pi k}{8}})-sin({\frac {\pi k}{8}}))\\\ {\frac {A}{\pi k}}(cos({\frac {\pi k}{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!}
Results of these calculations are shown below:
A
=
[
A
1
A
2
.
.
A
7
A
8
]
=
[
−
0.1218
0
0.0084
0
0.0014
0
0.0001
0
]
,
B
=
[
B
1
B
2
.
.
B
7
B
8
]
=
[
0.2941
0
0.0019
0
0.0014
0
0.0007
0
]
{\displaystyle A={\begin{bmatrix}A_{1}\\A_{2}\\.\\.\\A_{7}\\A_{8}\end{bmatrix}}={\begin{bmatrix}-0.1218\\0\\0.0084\\0\\0.0014\\0\\0.0001\\0\end{bmatrix}},\ B={\begin{bmatrix}B_{1}\\B_{2}\\.\\.\\B_{7}\\B_{8}\end{bmatrix}}={\begin{bmatrix}0.2941\\0\\0.0019\\0\\0.0014\\0\\0.0007\\0\end{bmatrix}}\!}
The solution to the particular case can be written for all n (assuming A=1):
y
n
=
1
4
+
∑
k
=
1
n
A
k
c
o
s
(
k
π
x
2
)
+
∑
k
=
1
n
B
k
s
i
n
(
k
π
x
2
)
{\displaystyle y_{n}={\frac {1}{4}}+\sum _{k=1}^{n}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{n}B_{k}sin({\frac {k\pi x}{2}})\!}
General Solution
edit
The general solution is
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}\!}
where
y
h
=
c
1
e
x
+
c
2
e
2
x
{\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!}
Different coefficients
c
1
,
c
2
{\displaystyle c_{1}\ ,\ c_{2}\!}
y
=
c
1
e
x
+
c
2
e
2
x
+
1
4
+
∑
k
=
1
2
A
k
c
o
s
(
k
π
x
2
)
+
∑
k
=
1
2
B
k
s
i
n
(
k
π
x
2
)
{\displaystyle y=c_{1}e^{x}+c_{2}e^{2x}+{\frac {1}{4}}+\sum _{k=1}^{2}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{2}B_{k}sin({\frac {k\pi x}{2}})\!}
Applying the first initial condition
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
(
0
)
=
c
1
+
c
2
+
A
1
+
A
2
=
1
{\displaystyle y(0)=c_{1}+c_{2}+A_{1}+A_{2}=1\!}
Taking the derivative
y
′
=
c
1
e
x
+
2
c
2
e
2
x
−
∑
k
=
2
2
k
π
2
A
k
s
i
n
(
k
π
x
2
)
+
∑
k
=
2
2
k
π
2
B
k
c
o
s
(
k
π
x
2
)
{\displaystyle y'=c_{1}e^{x}+2c_{2}e^{2x}-\sum _{k=2}^{2}{\frac {k\pi }{2}}A_{k}sin({\frac {k\pi x}{2}})+\sum _{k=2}^{2}{\frac {k\pi }{2}}B_{k}cos({\frac {k\pi x}{2}})\!}
Applying the second initial condition
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
y
′
(
0
)
=
c
1
+
2
c
2
+
2
π
2
B
2
=
0
{\displaystyle y'(0)=c_{1}+2c_{2}+{\frac {2\pi }{2}}B_{2}=0\!}
Solving the two equations for two unknowns yields:
c
1
=
2.2436
,
c
2
=
−
1.1218
{\displaystyle c_{1}=2.2436\ ,\ c_{2}=-1.1218\!}
So the general solution for n=2 is:
y
=
2.2436
e
x
−
1.1218
e
2
x
+
1
4
+
∑
k
=
1
2
A
k
c
o
s
(
k
π
x
2
)
+
∑
k
=
1
2
B
k
s
i
n
(
k
π
x
2
)
{\displaystyle y=2.2436e^{x}-1.1218e^{2x}+{\frac {1}{4}}+\sum _{k=1}^{2}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{2}B_{k}sin({\frac {k\pi x}{2}})\!}
Below is a plot showing the general solutions for n=2,4,8:
Matlab Plots
edit
Using ode45 the following graph was generated for n=0:
and for n=1
Solved by: Gonzalo Perez
Problem Statement
edit
Part R4.2, p.7c-26
edit
For each value of n=3,5,9, re-display the expressions for the 3 functions
y
p
,
n
(
x
)
,
y
h
,
n
(
x
)
,
y
n
(
x
)
{\displaystyle y_{p,n}(x),y_{h,n}(x),y_{n}(x)\!}
[
0
,
20
π
]
{\displaystyle [0,20\pi ]\!}
R4.2 p.7-26: Consider the L2-ODE-CC (5) p.7b-7 with
s
i
n
x
{\displaystyle sinx\!}
y
″
−
3
y
′
+
2
y
=
r
(
x
)
{\displaystyle y''-3y'+2y=r(x)\!}
r
(
x
)
=
s
i
n
x
{\displaystyle r(x)=sinx\!}
and the initial conditions:
y
(
0
)
=
1
,
y
′
(
0
)
=
0.
{\displaystyle y(0)=1,y'(0)=0.\!}
Exact solution:
y
(
x
)
=
y
h
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{h}(x)+y_{p}(x)\!}
Re-display the expressions for
y
p
(
x
)
,
y
h
(
x
)
,
y
(
x
)
{\displaystyle y_{p}(x),y_{h}(x),y(x)\!}
Superpose each of the above plot with that of the exact solution.
Solution
edit
The graphs are to be separately graphed as follows.
For n = 3, the code that will generate the graph looks like this:
For n = 6:
For n = 9:
Part R4.3, p.7c-28
edit
Understand and run the TA's code to produce a similar plot, but over a larger interval
[
0
,
10
]
{\displaystyle [0,10]\!}
x
=
−
0.5
,
0
,
+
0.5
{\displaystyle x=-0.5,0,+0.5\!}
R4.3 p.7-28: Consider the L2-ODE-CC (5) p.7b-7 with
l
o
g
(
1
+
x
)
{\displaystyle log(1+x)\!}
y
″
−
3
y
′
+
2
y
=
r
(
x
)
{\displaystyle y''-3y'+2y=r(x)\!}
r
(
x
)
=
l
o
g
(
1
+
x
)
{\displaystyle r(x)=log(1+x)\!}
and the initial conditions:
y
(
−
3
4
)
=
1
,
y
′
(
−
3
4
)
=
0
{\displaystyle y(-{\frac {3}{4}})=1,y'(-{\frac {3}{4}})=0\!}
Solution
edit
For n=4 (from x = 0 to x = 10):
For n=4 (zoom-in plots around x = -0.5):
For n=4 (zoom-in plots around x = 0):
For n=4 (zoom-in plots around x = +0.5):
For n=7 (from x = 0 to x = 10):
For n=11 (from x = 0 to x = 10):
Where the black line represents n=11 and the blue line represents log(1+x).
The problem asks for n=4, 7, and 11. The TA had up to n=16, but the problem does not ask for this. Manipulating the code and graphs to only account for these n values, n=16 has been disregarded.
Combined plots for n=7 and n=11 (zoom-in plots around x = -0.5):
Combined plots for n=7 and n=11 (zoom-in plots around x = 0):
Combined plots for n=7 and n=11 (zoom-in plots around x = 0.5):
Part R4.4, p.7c-29
edit
Understand and run the TA's code to produce a similar plot, but over a larger interval
[
0.9
,
10
]
{\displaystyle [0.9,10]\!}
x
=
1
,
1.5
,
2
,
2.5
{\displaystyle x=1,1.5,2,2.5\!}
R4.4 p.7-29: Extend the accuracy of the solution beyond
x
^
=
1
{\displaystyle {\hat {x}}=1\!}
Solution
edit
The general code that can be used to graph the plots of n=4,7,11 is:
With the above code, we can add the following code to graph n=4.
For n = 4 (from x = 0.9 to x = 10):
For n = 7 (from x = 0.9 to x = 10):
For n=11 (from x = 0.9 to x = 10):
Repeating the same common part of the code, let's graph the other plots around x = 1, 1.5, 2, 2.5:
For n = 4, 7, 11 at x = 1:
For n = 4, 7, 11 at x = 1.5:
For n = 4, 7, 11 at x = 2:
For n = 4, 7, 11 at x = 2.5:
For the last part, the first (and unchanged TA's code) is the following:
The second part includes the change that had to be made in order to solve this problem:
Problem Statement
edit
Given : For the following differential equation:
y
p
″
+
4
y
p
′
+
3
y
p
=
2
e
2
x
c
o
s
(
3
x
)
{\displaystyle y_{p}''+4y_{p}'+3y_{p}=2e^{2x}cos(3x)\!}
y
p
=
x
e
−
2
x
(
M
c
o
s
(
3
x
)
+
N
s
i
n
(
3
x
)
)
{\displaystyle y_{p}=xe^{-2x}(Mcos(3x)+Nsin(3x))\!}
y
p
=
6
e
−
2
x
(
N
c
o
s
(
3
x
)
−
M
s
i
n
(
3
x
)
)
{\displaystyle y_{p}=6e^{-2x}(Ncos(3x)-Msin(3x))\!}
y
p
″
{\displaystyle y_{p}''\!}
4
y
p
′
{\displaystyle 4y_{p}'\!}
13
y
p
{\displaystyle 13y_{p}\!}
y
p
(
x
)
{\displaystyle y_{p}(x)\!}
Solution
edit
1)
y
p
″
{\displaystyle y_{p}''\!}
[1] Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]]
y
p
″
=
−
6
M
e
−
2
x
s
i
n
(
3
x
)
+
12
M
e
−
2
x
x
s
i
n
(
3
x
)
−
4
M
e
−
2
x
c
o
s
(
3
x
)
−
5
M
e
−
2
x
x
c
o
s
(
3
x
)
{\displaystyle y_{p}''=-6Me^{-2x}sin(3x)+12Me^{-2x}xsin(3x)-4Me^{-2x}cos(3x)-5Me^{-2x}xcos(3x)\!}
−
4
N
e
−
2
x
s
i
n
(
3
x
)
−
5
N
e
−
2
x
x
s
i
n
(
3
x
)
+
6
N
e
−
2
x
c
o
s
(
3
x
)
−
12
N
e
−
2
x
x
c
o
s
(
3
x
)
{\displaystyle -4Ne^{-2x}sin(3x)-5Ne^{-2x}xsin(3x)+6Ne^{-2x}cos(3x)-12Ne^{-2x}xcos(3x)\!}
2)
4
y
p
′
{\displaystyle 4y_{p}'\!}
simplify[4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]]
4
y
p
′
=
−
12
M
e
−
2
x
x
s
i
n
(
3
x
)
+
4
M
e
−
2
x
c
o
s
(
3
x
)
−
8
M
e
−
2
x
x
c
o
s
(
3
x
)
+
4
N
e
−
2
x
s
i
n
(
3
x
)
−
8
N
e
−
2
x
x
s
i
n
(
3
x
)
+
12
N
e
−
2
x
x
c
o
s
(
3
x
)
{\displaystyle 4y_{p}'=-12Me^{-2x}xsin(3x)+4Me^{-2x}cos(3x)-8Me^{-2x}xcos(3x)+4Ne^{-2x}sin(3x)-8Ne^{-2x}xsin(3x)+12Ne^{-2x}xcos(3x)\!}
Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]+ 4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]
y
p
″
+
4
y
p
′
=
−
6
M
e
−
2
x
s
i
n
(
3
x
)
−
13
M
e
−
2
x
x
c
o
s
(
3
x
)
−
13
N
e
−
2
x
x
s
i
n
(
3
x
)
+
6
N
e
−
2
x
c
o
s
(
3
x
)
{\displaystyle y_{p}''+4y_{p}'=-6Me^{-2x}sin(3x)-13Me^{-2x}xcos(3x)-13Ne^{-2x}xsin(3x)+6Ne^{-2x}cos(3x)\!}
3)
13
y
p
{\displaystyle 13y_{p}\!}
y
p
″
+
4
y
p
′
+
13
y
p
=
−
6
M
e
−
2
x
s
i
n
(
3
x
)
−
13
M
e
−
2
x
x
c
o
s
(
3
x
)
−
13
N
e
−
2
x
x
s
i
n
(
3
x
)
+
6
N
e
−
2
x
c
o
s
(
3
x
)
+
13
M
e
−
2
x
x
c
o
s
(
3
x
)
+
13
N
e
−
2
x
x
s
i
n
(
3
x
)
{\displaystyle y_{p}''+4y_{p}'+13y_{p}=-6Me^{-2x}sin(3x)-13Me^{-2x}xcos(3x)-13Ne^{-2x}xsin(3x)+6Ne^{-2x}cos(3x)+13Me^{-2x}xcos(3x)+13Ne^{-2x}xsin(3x)\!}
y
p
″
+
4
y
p
′
+
13
y
p
=
−
6
M
e
−
2
x
s
i
n
(
3
x
)
+
(
13
−
13
)
M
e
−
2
x
x
c
o
s
(
3
x
)
+
(
13
−
13
)
N
e
−
2
x
x
s
i
n
(
3
x
)
+
6
N
e
−
2
x
c
o
s
(
3
x
)
{\displaystyle y_{p}''+4y_{p}'+13y_{p}=-6Me^{-2x}sin(3x)+(13-13)Me^{-2x}xcos(3x)+(13-13)Ne^{-2x}xsin(3x)+6Ne^{-2x}cos(3x)\!}
y
p
″
+
4
y
p
′
+
13
y
p
=
−
6
M
e
−
2
x
s
i
n
(
3
x
)
+
6
N
e
−
2
x
c
o
s
(
3
x
)
{\displaystyle y_{p}''+4y_{p}'+13y_{p}=-6Me^{-2x}sin(3x)+6Ne^{-2x}cos(3x)\!}
y
p
″
+
4
y
p
′
+
13
y
p
=
6
e
−
2
x
(
N
c
o
s
(
3
x
)
−
M
s
i
n
(
3
x
)
)
{\displaystyle y_{p}''+4y_{p}'+13y_{p}=6e^{-2x}(Ncos(3x)-Msin(3x))\!}
4)
y
p
(
x
)
=
6
e
−
2
x
(
N
c
o
s
(
3
x
)
−
M
s
i
n
(
3
x
)
)
{\displaystyle y_{p}(x)=6e^{-2x}(Ncos(3x)-Msin(3x))\!}
↑ www.wolframalpha.com
--Egm4313.s12.team11.sheider (talk ) 22:30, 10 April 2012 (UTC)
Solved by Solved by Daniel Suh
Problem Statement
edit
Find the separated Ordinary Differential Equations for the heat equation.
∂
u
∂
t
=
κ
∂
2
u
∂
x
2
⋯
(
1
)
{\displaystyle {\frac {\partial u}{\partial t}}=\kappa {\frac {\partial ^{2}u}{\partial x^{2}}}\cdots (1)\!}
u
(
x
,
t
)
=
F
(
x
)
⋅
G
(
t
)
{\displaystyle u(x,t)=F(x)\cdot G(t)\!}
Solution
edit
∂
u
∂
t
=
F
(
x
)
⋅
G
˙
(
t
)
⋯
(
2
)
{\displaystyle {\frac {\partial u}{\partial t}}=F(x)\cdot {\dot {G}}(t)\cdots (2)\!}
∂
2
u
∂
x
2
=
F
″
(
x
)
⋅
G
(
t
)
⋯
(
3
)
{\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}=F''(x)\cdot G(t)\cdots (3)\!}
Combine
(
2
)
{\displaystyle (2)\!}
(
3
)
{\displaystyle (3)\!}
(
1
)
{\displaystyle (1)\!}
F
(
x
)
⋅
G
˙
(
t
)
=
κ
F
″
(
x
)
⋅
G
(
t
)
{\displaystyle F(x)\cdot {\dot {G}}(t)=\kappa F''(x)\cdot G(t)\!}
G
˙
(
t
)
κ
G
(
t
)
=
F
″
(
x
)
F
(
x
)
=
c
{\displaystyle {\frac {{\dot {G}}(t)}{\kappa G(t)}}={\frac {F''(x)}{F(x)}}=c\!}
Separate
F
″
(
x
)
−
c
F
(
x
)
=
0
{\displaystyle F''(x)-cF(x)=0\!}
G
′
(
t
)
−
κ
c
G
(
t
)
=
0
{\displaystyle G'(t)-\kappa cG(t)=0\!}
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {sin(n\omega x)}{n\omega }}|_{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/948cfb833c24a8531890b3ac3410eb6e71569a99)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {2sin(n\pi )}{n\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91e23e4538e88251593b0a0c69cd95afe7ce8952)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {cos(n\omega x)}{n\omega }}|_{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2eba41982a5c9a4390f37c5904e426055cbc52b4)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {-cos{n\pi }+cos{n\pi }}{n\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfa42c9d3dba92ee2ccc1ea8a835a5fe220bcfb2)
![{\displaystyle a_{0}={\frac {1}{2}}\int _{-1}^{1}x^{2}dx={\frac {1}{2}}\left[{\frac {1}{3}}x^{3}\right]_{-1}^{1}={\frac {1}{2}}\left[{\frac {1}{3}}+{\frac {1}{3}}\right]={\frac {1}{3}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cdcec4b23b99a873211496d4c7284a7e126a84d)
![{\displaystyle \left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\int _{0}^{1}xsin(\pi nx)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bad33c22c5b16e79133262bd5ec55d4edda8aad9)
![{\displaystyle \left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83b10de0d3425337dcfd8d80e144e393d55c673a)
![{\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a60b2f09abdff1de7c5f4edcb367187a52ca5cb2)
![{\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\left[{\frac {sin(\pi nx)}{\pi n}}\right]_{0}^{1}\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/285797fa8a188f052f64069a7fb8e60d99fabecb)
![{\displaystyle 2\left[\left[0-0\right]+{\frac {2}{\pi ^{2}n^{2}}}\left[cos(\pi n)\right]-{\frac {1}{\pi n}}\left[0-0\right]\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40e36f0954f238cf8f595c4bd95b1cef757d0a54)
![{\displaystyle a_{0}={\frac {1}{4}}\int _{-2}^{2}1-{\frac {x^{2}}{4}}dx={\frac {1}{4}}\left[x-{\frac {1}{12}}x^{3}\right]_{-2}^{2}={\frac {1}{4}}\left[{\frac {24-8}{12}}-{\frac {24+8}{12}}\right]={\frac {2}{3}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35f2149448f2c8078f965e3d71d3e3d957077c8b)
![{\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/b888e20b56e5edf65b6488f76ee37ceb4513bc47)
![{\displaystyle a_{0}=(1/\pi )*[-1\int _{-\pi }^{-\pi /2}(x+\pi )dx+\int _{-\pi /2}^{\pi /2}(x)dx+\int _{\pi /2}^{\pi }(\pi -x)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3515ec49fdd8ef95aedb6e74ee844ba7b39b6ce)
![{\displaystyle a_{0}=1/2[\int _{-1}^{0}(1+x)dx+\int _{0}^{1}(1-x)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/663792d020231d208ed32b73f0c4db7045053632)
![{\displaystyle a_{0}=1/2[1-1/2+1-1/2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57762d639f689e1bcd7d58eeb81c3590c3473f73)
![{\displaystyle a_{n}=1/1[\int _{-1}^{0}(1+x)cos(n\pi *x/1)dx+\int _{0}^{1}(1-x)cos(n\pi *x/1)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecc54c05805473116b69ac5965dd79d368827f34)
![{\displaystyle a_{n}=2[1/(n^{2}\pi ^{2})_{(}-1)^{n}/(n^{2}\pi ^{2})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82a4babc3b94dfce840755f8704669ff666085ce)
![{\displaystyle b_{n}=1/1[\int _{-1}^{0}(1+x)sin(n\pi *x/1)dx+\int _{0}^{1}(1-x)sin(n\pi *x/1)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b3a3c83592e4975230374d189014a6e9669263f)
![{\displaystyle [0,10]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7afee0f848455b66aae9ef3f95de42d3509f2f26)
![{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos({\frac {n\pi x}{L}})+b_{n}sin({\frac {n\pi x}{L}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1882b82559d4f1569411c84a2b6e3d86f161f7fe)
![{\displaystyle +{\frac {A}{n\pi }}(cos({\frac {n\pi }{8}})-cos({\frac {9n\pi }{8}}))sin({\frac {n\pi x}{2}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8893708df79a090f008049edab53aed382f0f66e)
![{\displaystyle +(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cd6ccf84a4564bffcd87fc35ab5fb5eccc20fad)
![{\displaystyle ={\frac {A}{2}}+\sum _{k=1}^{n}{\frac {A}{k\pi }}[(sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}}))cos({\frac {k\pi x}{2}})+(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c55a889240e8411f80b6f6e1b9954e45d32b4c8)
![{\displaystyle [0,20\pi ]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0265ce74a1f75468b07b65508ca66e583cf31727)
![{\displaystyle [0.9,10]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c86a6f6483bf9ba445c11cb83d4e369f736f9cc)
