Report 4
Problem R4.1 Expanding Particular Solutions and the Coefficient Matrix A
edit
4.1 from lecture notes R4.1 Lect. 7c pgs. 19-22
Given the general form of polynomial excitation.
y
″
+
a
y
′
+
b
y
=
∑
j
=
0
n
d
j
x
j
{\displaystyle \displaystyle y''+ay'+by=\sum _{j=0}^{n}d_{j}x^{j}}
(1.1)
The particular solution that satisfies:
y
p
(
x
)
=
∑
j
=
0
n
c
j
x
j
{\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x^{j}}
(1.2)
The first and second derivative of the particular solution that solves the original polynomial excitation equation.
y
p
′
(
x
)
=
∑
j
=
0
n
−
1
c
j
+
1
(
j
+
1
)
x
j
{\displaystyle \displaystyle y_{p}'(x)=\sum _{j=0}^{n-1}c_{j+1}(j+1)x^{j}}
(1.3)
y
p
″
(
x
)
=
∑
j
=
0
n
−
2
c
j
+
2
(
j
+
2
)
(
j
+
1
)
x
j
{\displaystyle \displaystyle y_{p}''(x)=\sum _{j=0}^{n-2}c_{j+2}(j+2)(j+1)x^{j}}
(1.4)
The particular solutions are put into the polynomial excitation equation to give the general summation form:
∑
j
=
0
n
−
2
[
c
j
+
2
(
j
+
2
)
(
j
+
1
)
+
a
c
j
+
1
(
j
+
1
)
+
b
c
j
]
x
j
+
a
c
n
n
x
n
−
1
+
b
[
c
n
−
1
x
n
−
1
+
c
n
x
n
]
=
∑
j
=
0
n
d
j
x
j
{\displaystyle \displaystyle \sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum _{j=0}^{n}d_{j}x^{j}}
(1.5)
Obtain the equations associated with
d
1
{\displaystyle d_{1}\!}
x
{\displaystyle x\!}
d
2
{\displaystyle d_{2}\!}
x
2
{\displaystyle x^{2}\!}
d
n
−
2
{\displaystyle d_{n-2}\!}
x
n
−
2
{\displaystyle x^{n-2}\!}
d
n
−
1
{\displaystyle d_{n-1}\!}
x
n
−
1
{\displaystyle x^{n-1}\!}
d
n
{\displaystyle d_{n}\!}
x
n
{\displaystyle x^{n}\!}
Also set up the matrix
A
{\displaystyle \mathbf {A} \!}
A
c
=
d
{\displaystyle \mathbf {A} \mathbf {c} =\mathbf {d} \!}
The given equation associated with
d
0
{\displaystyle d_{0}\!}
j
=
0
{\displaystyle j=0\!}
2
c
2
+
a
c
1
+
b
c
0
=
d
0
{\displaystyle \displaystyle 2c_{2}+ac_{1}+bc_{0}=d_{0}}
(1.6)
Taking
j
=
1
{\displaystyle j=1\!}
[
c
1
+
2
(
1
+
2
)
(
1
+
1
)
+
a
c
1
+
1
(
1
+
1
)
+
b
c
1
]
x
1
=
d
1
x
1
{\displaystyle \displaystyle [c_{1+2}(1+2)(1+1)+ac_{1+1}(1+1)+bc_{1}]x^{1}=d_{1}x^{^{1}}}
(1.7)
[
6
c
3
+
2
a
c
2
+
b
c
1
]
x
=
d
1
x
{\displaystyle \displaystyle [6c_{3}+2ac_{2}+bc_{1}]x=d_{1}x}
(1.8)
The equation associated with
d
1
{\displaystyle d_{1}\!}
x
{\displaystyle x\!}
6
c
3
+
2
a
c
2
+
b
c
1
=
d
1
{\displaystyle \displaystyle 6c_{3}+2ac_{2}+bc_{1}=d_{1}}
(1.9)
Taking
j
=
2
{\displaystyle j=2\!}
[
c
2
+
2
(
2
+
2
)
(
2
+
1
)
+
a
c
2
+
1
(
2
+
1
)
+
b
c
2
]
x
2
=
d
2
x
2
{\displaystyle \displaystyle [c_{2+2}(2+2)(2+1)+ac_{2+1}(2+1)+bc_{2}]x^{2}=d_{2}x^{^{2}}}
(1.10)
[
12
c
4
+
3
a
c
3
+
b
c
2
]
x
2
=
d
2
x
2
{\displaystyle \displaystyle [12c_{4}+3ac_{3}+bc_{2}]x^{2}=d_{2}x^{2}}
(1.11)
The equation associated with
d
2
{\displaystyle d_{2}\!}
x
2
{\displaystyle x^{2}\!}
12
c
4
+
3
a
c
3
+
b
c
2
=
d
2
{\displaystyle \displaystyle 12c_{4}+3ac_{3}+bc_{2}=d_{2}}
(1.12)
Taking
j
=
n
−
2
{\displaystyle j=n-2\!}
[
c
(
n
−
2
)
+
2
(
n
−
2
+
2
)
(
n
−
2
+
1
)
+
a
c
(
n
−
2
+
1
)
(
n
−
2
+
1
)
+
b
c
n
−
2
]
x
n
−
2
=
d
n
−
2
x
n
−
2
{\displaystyle \displaystyle [c_{(n-2)+2}(n-2+2)(n-2+1)+ac_{(n-2+1)}(n-2+1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2}}
(1.13)
[
c
n
n
(
n
−
1
)
+
a
c
(
n
−
1
)
(
n
−
1
)
+
b
c
n
−
2
]
x
n
−
2
=
d
n
−
2
x
n
−
2
{\displaystyle \displaystyle [c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2}}
(1.14)
The equation associated with
d
n
−
2
{\displaystyle d_{n-2}\!}
x
n
−
2
{\displaystyle x^{n-2}\!}
c
n
n
(
n
−
1
)
+
a
c
(
n
−
1
)
(
n
−
1
)
+
b
c
n
−
2
=
d
n
−
2
{\displaystyle \displaystyle c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}=d_{n-2}}
(1.15)
For
n
−
1
{\displaystyle n-1\!}
j
=
n
−
2
{\displaystyle j=n-2\!}
a
c
n
n
x
n
−
1
+
b
c
n
−
1
x
n
−
1
=
d
n
−
1
x
n
−
1
{\displaystyle \displaystyle ac_{n}nx^{n-1}+bc_{n-1}x^{n-1}=d_{n-1}x^{n-1}}
(1.16)
a
c
n
n
+
b
c
(
n
−
1
)
]
x
n
−
1
=
d
n
−
1
x
n
−
1
{\displaystyle \displaystyle ac_{n}n+bc_{(n-1)}]x^{n-1}=d_{n-1}x^{n-1}}
(1.17)
The equation associated with
d
n
−
1
{\displaystyle d_{n-1}\!}
x
n
−
1
{\displaystyle x^{n-1}\!}
a
c
n
n
+
b
c
(
n
−
1
)
=
d
n
−
1
{\displaystyle \displaystyle ac_{n}n+bc_{(n-1)}=d_{n-1}}
(1.18)
For
n
{\displaystyle n\!}
b
c
n
x
n
=
d
n
x
n
{\displaystyle \displaystyle bc_{n}x^{n}=d_{n}x^{^{n}}}
(1.19)
d
n
{\displaystyle d_{n}\!}
x
n
{\displaystyle x^{n}\!}
b
c
n
=
d
n
{\displaystyle \displaystyle bc_{n}=d_{n}}
(1.20)
Now set up the equation:
A
c
=
d
{\displaystyle \displaystyle \mathbf {A} \mathbf {c} =\mathbf {d} }
(1.21)
[
E
q
u
a
t
i
o
n
f
o
r
d
0
E
q
u
a
t
i
o
n
f
o
r
d
1
E
q
u
a
t
i
o
n
f
o
r
d
2
.
.
.
E
q
u
a
t
i
o
n
f
o
r
d
n
−
2
E
q
u
a
t
i
o
n
f
o
r
d
n
−
1
E
q
u
a
t
i
o
n
f
o
r
d
n
]
[
c
0
c
1
c
2
.
.
.
c
n
−
2
c
n
−
1
c
n
]
=
[
d
0
d
1
d
2
.
.
.
d
n
−
2
d
n
−
1
d
n
]
{\displaystyle \displaystyle {\begin{bmatrix}Equationford_{0}\\Equationford_{1}\\Equationford_{2}\\...\\Equationford_{n-2}\\Equationford_{n-1}\\Equationford_{n}\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\...\\c_{n-2}\\c_{n-1}\\c_{n}\end{bmatrix}}={\begin{bmatrix}d_{0}\\d_{1}\\d_{2}\\...\\d_{n-2}\\d_{n-1}\\d_{n}\end{bmatrix}}}
(1.22)
Therefore the matrix
A
{\displaystyle \mathbf {A} \!}
A
=
[
b
a
2
b
2
a
6
b
3
a
12
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
b
a
(
n
−
1
)
n
(
n
−
1
)
b
a
n
b
]
{\displaystyle \displaystyle \mathbf {A} ={\begin{bmatrix}b&a&2&&&&\\&b&2a&6&&&\\&&b&3a&12&&\\...&...&...&...&...&...&...\\&&&&b&a(n-1)&n(n-1)\\&&&&&b&an\\&&&&&&b\end{bmatrix}}}
(1.23)
Solved and Typed By - Chris Stewart Egm4313.s12.team1.stewart (talk ) -- 21:12, 11 March 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.durrance (talk ) 23:14, 11 March 2012 (UTC)
edit
Consider the L2-ODE-CC with
sin
x
{\displaystyle \sin x\!}
(see R4.2 Lect. 7c pgs. 26-27) :
y
″
−
3
y
′
+
2
y
=
r
(
x
)
{\displaystyle y''-3y'+2y=r(x)\!}
r
(
x
)
=
sin
x
{\displaystyle r(x)=\sin x\!}
and the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0\!}
1) Use the Taylor series for
sin
x
{\displaystyle \sin x\!}
2) Let
y
p
,
n
(
x
)
{\displaystyle y_{p,n}(x)\!}
r
n
(
x
)
{\displaystyle r_{n}(x)\!}
y
p
,
n
″
+
a
y
p
,
n
′
+
b
y
p
,
n
=
r
n
(
x
)
{\displaystyle y''_{p,n}+ay_{p,n}'+by_{p,n}=r_{n}(x)\!}
Let
r
n
(
x
)
{\displaystyle r_{n}(x)\!}
sin
x
{\displaystyle \sin x\!}
r
n
(
x
)
:=
∑
k
=
0
n
(
−
1
)
k
t
2
k
+
1
(
2
k
+
1
)
!
=
t
−
t
3
3
!
+
.
.
.
+
(
−
1
)
n
t
2
n
+
1
(
2
n
+
1
)
!
{\displaystyle r_{n}(x):=\sum _{k=0}^{n}{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}=t-{\frac {t^{3}}{3!}}+...+{\frac {(-1)^{n}t^{2n+1}}{(2n+1)!}}\!}
Let
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)\!}
Find
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
3)Find the exact overall soln
y
(
x
)
{\displaystyle y(x)\!}
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
To approximate the value of the excitation, the Taylor expansion must be found:
f
(
x
)
=
∑
n
=
0
∞
f
(
n
)
(
x
^
)
n
!
(
x
−
x
^
)
n
{\displaystyle \displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}({\hat {x}})}{n!}}(x-{\hat {x}})^{n}}
(2.0)
For the sine function, the Taylor series approximated to 13 places is:
sin
x
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
x
9
9
!
−
x
11
11
!
+
x
13
13
!
{\displaystyle \displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}+{\frac {x^{13}}{13!}}}
(2.1)
Any lower approximation would include all of the terms above without any terms with a higher order than the desired order:
Plotting every order approximation of the Taylor series up to n=13 with the actual sine function produces the following:
Figure 4.2-1
In order to find the overall solution for the L2-ODE-CC corresponding to the Taylor series expansion of the sine function, both the homogenous and particular solutions must be found. The homogenous equation can be found through this method:
y
″
−
3
y
′
+
2
y
=
0
{\displaystyle \displaystyle y''-3y'+2y=0}
(2.2)
λ
2
−
3
λ
+
2
=
0
→
λ
=
1
,
2
{\displaystyle \displaystyle \lambda ^{2}-3\lambda +2=0\rightarrow \lambda =1,2}
(2.3)
y
h
,
n
(
x
)
=
C
1
e
x
+
C
2
e
2
x
{\displaystyle \displaystyle y_{h,n}(x)=C_{1}e^{x}+C_{2}e^{2x}}
(2.4)
Next, the excitation must be expanded to the desired n. The following shows the excitation expanded to n=3, 5, and 9:
r
3
(
x
)
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
{\displaystyle \displaystyle r_{3}(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}}
(2.5)
r
5
(
x
)
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
x
9
9
!
−
x
11
11
!
{\displaystyle \displaystyle r_{5}(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}}
(2.6)
r
9
(
x
)
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
x
9
9
!
−
x
11
11
!
+
x
13
13
!
−
x
15
15
!
+
x
17
17
!
−
x
19
19
!
{\displaystyle \displaystyle r_{9}(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}+{\frac {x^{13}}{13!}}-{\frac {x^{15}}{15!}}+{\frac {x^{17}}{17!}}-{\frac {x^{19}}{19!}}}
(2.7)
Next, the particular solution must be found. The particular solution will be of the form:
y
p
,
n
=
∑
i
=
0
n
k
i
x
i
{\displaystyle \displaystyle y_{p,n}=\sum _{i=0}^{n}k_{i}x^{i}}
(2.8)
Using the derivation discussed in R4.1, a matrix equation in the form Ak = d can be found, where k is the matrix containing the coefficients of the particular solution and d is the matrix containing the coefficients of each power of x in the expansion of the excitation. The general formula for A is:
[
b
a
2
b
2
a
6
b
3
a
12
.
.
.
.
.
.
.
.
.
b
a
(
n
−
1
)
n
(
n
−
1
)
b
a
n
b
]
{\displaystyle \displaystyle {\begin{bmatrix}b&a&2&&&&\\&b&2a&6&&&\\&&b&3a&12&&\\&&&...&...&...&\\&&&&b&a(n-1)&n(n-1)\\&&&&&b&an\\&&&&&&b\end{bmatrix}}}
(2.9)
n=3
A
=
[
2
−
3
2
2
−
6
6
2
−
9
12
2
−
12
20
2
−
15
30
2
−
18
42
2
−
21
2
]
{\displaystyle \displaystyle A={\begin{bmatrix}2&-3&2&&&&&\\&2&-6&6&&&&\\&&2&-9&12&&&\\&&&2&-12&20&&\\&&&&2&-15&30&\\&&&&&2&-18&42\\&&&&&&2&-21\\&&&&&&&2\end{bmatrix}}}
(2.10)
Therefore, the matrix equation is:
[
2
−
3
2
2
−
6
6
2
−
9
12
2
−
12
20
2
−
15
30
2
−
18
42
2
−
21
2
]
[
k
0
k
1
k
2
k
3
k
4
k
5
k
6
k
7
]
=
[
0
1
0
−
1
6
0
1
120
0
−
1
5040
]
{\displaystyle \displaystyle {\begin{bmatrix}2&-3&2&&&&&\\&2&-6&6&&&&\\&&2&-9&12&&&\\&&&2&-12&20&&\\&&&&2&-15&30&\\&&&&&2&-18&42\\&&&&&&2&-21\\&&&&&&&2\end{bmatrix}}{\begin{bmatrix}k_{0}\\k_{1}\\k_{2}\\k_{3}\\k_{4}\\k_{5}\\k_{6}\\k_{7}\end{bmatrix}}={\begin{bmatrix}0\\1\\0\\-{\frac {1}{6}}\\0\\{\frac {1}{120}}\\0\\-{\frac {1}{5040}}\end{bmatrix}}}
(2.11)
[
k
0
k
1
k
2
k
3
k
4
k
5
k
6
k
7
]
=
[
−
0.1992
−
0.3984
−
0.3984
−
0.0990
−
0.0078
−
0.0031
−
0.0010
−
9.9206
×
10
−
5
]
{\displaystyle \displaystyle {\begin{bmatrix}k_{0}\\k_{1}\\k_{2}\\k_{3}\\k_{4}\\k_{5}\\k_{6}\\k_{7}\end{bmatrix}}={\begin{bmatrix}-0.1992\\-0.3984\\-0.3984\\-0.0990\\-0.0078\\-0.0031\\-0.0010\\-9.9206\times 10^{-5}\end{bmatrix}}}
(2.12)
Therefore the particular solution is:
y
p
,
3
(
x
)
=
−
9.9206
×
10
−
5
x
7
−
0.0010
x
6
−
0.0031
x
5
−
0.0078
x
4
−
0.0990
x
3
−
0.3984
x
2
−
0.3984
x
−
0.1992
{\displaystyle \displaystyle y_{p,3}(x)=-9.9206\times 10^{-5}x^{7}-0.0010x^{6}-0.0031x^{5}-0.0078x^{4}-0.0990x^{3}-0.3984x^{2}-0.3984x-0.1992}
(2.13)
And the overall solution for n=3 is:
y
3
(
x
)
=
C
1
e
x
+
C
2
e
2
x
−
9.9206
×
10
−
5
x
7
−
0.0010
x
6
−
0.0031
x
5
−
0.0078
x
4
−
0.0990
x
3
−
0.3984
x
2
−
0.3984
x
−
0.1992
{\displaystyle \displaystyle y_{3}(x)=C_{1}e^{x}+C_{2}e^{2x}-9.9206\times 10^{-5}x^{7}-0.0010x^{6}-0.0031x^{5}-0.0078x^{4}-0.0990x^{3}-0.3984x^{2}-0.3984x-0.1992}
(2.14)
Using the initial conditions:
1
=
C
1
+
C
2
−
0.1992
&
0
=
C
1
+
2
C
2
−
0.3984
{\displaystyle \displaystyle 1=C_{1}+C_{2}-0.1992\&0=C_{1}+2C_{2}-0.3984}
(2.15)
Solving yields
C
1
=
2
&
C
2
=
−
0.8008
{\displaystyle C_{1}=2\&C_{2}=-0.8008\!}
y
3
(
x
)
=
2
e
x
−
0.8008
e
2
x
−
9.9206
×
10
−
5
x
7
−
0.0010
x
6
−
0.0031
x
5
−
0.0078
x
4
−
0.0990
x
3
−
0.3984
x
2
−
0.3984
x
−
0.1992
{\displaystyle \displaystyle y_{3}(x)=2e^{x}-0.8008e^{2x}-9.9206\times 10^{-5}x^{7}-0.0010x^{6}-0.0031x^{5}-0.0078x^{4}-0.0990x^{3}-0.3984x^{2}-0.3984x-0.1992}
(2.16)
n=5
Therefore the particular solution is:
y
p
,
5
(
x
)
=
−
1.2526
×
10
−
8
x
11
−
2.0668
×
10
−
7
x
10
−
1.0334
×
10
−
6
x
9
−
4.6503
×
10
−
6
x
8
−
1.1781
×
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.0999
x
3
−
0.3999
x
2
−
0.3999
x
−
0.2000
{\displaystyle \displaystyle y_{p,5}(x)=-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^{9}-4.6503\times 10^{-6}x^{8}-1.1781\times 10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.3999x-0.2000}
(2.18)
And the overall solution for n=5 is:
y
5
(
x
)
=
C
1
e
x
+
C
2
e
2
x
−
1.2526
×
10
−
8
x
11
−
2.0668
×
10
−
7
x
10
−
1.0334
×
10
−
6
x
9
−
4.6503
×
10
−
6
x
8
−
1.1781
×
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.0999
x
3
−
0.3999
x
2
−
0.3999
x
−
0.2000
{\displaystyle \displaystyle y_{5}(x)=C_{1}e^{x}+C_{2}e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^{9}-4.6503\times 10^{-6}x^{8}-1.1781\times 10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.3999x-0.2000}
(2.19)
Using the initial conditions:
1
=
C
1
+
C
2
−
0.2
&
0
=
C
1
+
2
C
2
−
0.3999
{\displaystyle \displaystyle 1=C_{1}+C_{2}-0.2\&0=C_{1}+2C_{2}-0.3999}
(2.20)
Solving yields
C
1
=
2.0001
&
C
2
=
−
0.8001
{\displaystyle C_{1}=2.0001\&C_{2}=-0.8001\!}
y
5
(
x
)
=
2.0001
e
x
−
0.8001
e
2
x
−
1.2526
×
10
−
8
x
11
−
2.0668
×
10
−
7
x
10
−
1.0334
×
10
−
6
x
9
−
4.6503
×
10
−
6
x
8
−
1.1781
×
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.0999
x
3
−
0.3999
x
2
−
0.3999
x
−
0.2
{\displaystyle \displaystyle y_{5}(x)=2.0001e^{x}-0.8001e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^{9}-4.6503\times 10^{-6}x^{8}-1.1781\times 10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.3999x-0.2}
(2.21)
n=9
Therefore the particular solution is:
y
p
,
9
(
x
)
=
−
4.1103
×
10
−
18
x
19
−
1.1714
×
10
−
16
x
18
−
1.0543
×
10
−
15
x
17
−
8.9615
×
10
−
15
x
16
−
4.5405
×
10
−
13
x
15
−
9.1408
×
10
−
12
x
14
−
6.3985
×
10
−
11
x
13
−
4.1590
×
10
−
10
x
12
−
1.5021
×
10
−
8
x
11
−
2.2040
×
10
−
7
x
10
−
1.1020
×
10
−
6
x
9
−
4.9591
×
10
−
6
x
8
−
1.1904
×
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.1000
x
3
−
0.4000
x
2
−
0.4000
x
−
0.2000
{\displaystyle \displaystyle y_{p,9}(x)=-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^{9}-4.9591\times 10^{-6}x^{8}-1.1904\times 10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000}
(2.23)
And the overall solution for n=9 is:
y
9
(
x
)
=
C
1
e
x
+
C
2
e
2
x
−
4.1103
×
10
−
18
x
19
−
1.1714
×
10
−
16
x
18
−
1.0543
×
10
−
15
x
17
−
8.9615
×
10
−
15
x
16
−
4.5405
×
10
−
13
x
15
−
9.1408
×
10
−
12
x
14
−
6.3985
×
10
−
11
x
13
−
4.1590
×
10
−
10
x
12
−
1.5021
×
10
−
8
x
11
−
2.2040
×
10
−
7
x
10
−
1.1020
×
10
−
6
x
9
−
4.9591
×
10
−
6
x
8
−
1.1904
×
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.1000
x
3
−
0.4000
x
2
−
0.4000
x
−
0.2000
{\displaystyle \displaystyle y_{9}(x)=C_{1}e^{x}+C_{2}e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^{9}-4.9591\times 10^{-6}x^{8}-1.1904\times 10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000}
(2.24)
Using the initial conditions:
1
=
C
1
+
C
2
−
0.2
&
0
=
C
1
+
2
C
2
−
0.4
{\displaystyle \displaystyle 1=C_{1}+C_{2}-0.2\&0=C_{1}+2C_{2}-0.4}
(2.25)
Solving yields
C
1
=
2
&
C
2
=
−
0.8
{\displaystyle C_{1}=2\&C_{2}=-0.8\!}
y
9
(
x
)
=
2
e
x
−
0.8
e
2
x
−
4.1103
×
10
−
18
x
19
−
1.1714
×
10
−
16
x
18
−
1.0543
×
10
−
15
x
17
−
8.9615
×
10
−
15
x
16
−
4.5405
×
10
−
13
x
15
−
9.1408
×
10
−
12
x
14
−
6.3985
×
10
−
11
x
13
−
4.1590
×
10
−
10
x
12
−
1.5021
×
10
−
8
x
11
−
2.2040
×
10
−
7
x
10
−
1.1020
×
10
−
6
x
9
−
4.9591
×
10
−
6
x
8
−
1.1904
×
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.1000
x
3
−
0.4000
x
2
−
0.4000
x
−
0.2000
{\displaystyle \displaystyle y_{9}(x)=2e^{x}-0.8e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^{9}-4.9591\times 10^{-6}x^{8}-1.1904\times 10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000}
(2.26)
Figure 4.2-2
The near perfect overlap between all three graphs shows that the approximations converge quickly for very low values of n.
Next, the exact ODE will be calculated to find the accuracy of the above approximations. The homogeneous solution is the same as the homogeneous solution above, only with different coefficients. The particular solution will be in the form:
y
p
(
x
)
=
K
cos
x
+
M
sin
x
{\displaystyle \displaystyle y_{p}(x)=K\cos x+M\sin x}
(2.27)
y
p
′
(
x
)
=
−
K
sin
x
+
M
cos
x
{\displaystyle \displaystyle y_{p}'(x)=-K\sin x+M\cos x}
(2.28)
y
p
″
(
x
)
=
−
K
cos
x
−
M
sin
x
{\displaystyle \displaystyle y_{p}''(x)=-K\cos x-M\sin x}
(2.29)
Plugging these values into the ODE:
−
K
cos
x
−
M
sin
x
+
3
K
sin
x
−
3
M
cos
x
+
2
K
cos
x
+
2
M
sin
x
=
sin
x
{\displaystyle \displaystyle -K\cos x-M\sin x+3K\sin x-3M\cos x+2K\cos x+2M\sin x=\sin x}
(2.30)
Separating sine and cosine terms yield two linear equations which can be used to solve for the unknown coefficients:
K
−
3
M
=
0
&
3
K
+
M
=
1
{\displaystyle \displaystyle K-3M=0\&3K+M=1}
(2.31)
Solving these equations yield K = 0.3 and M = 0.1.
y
p
(
x
)
=
0.3
cos
x
+
0.1
sin
x
{\displaystyle \displaystyle y_{p}(x)=0.3\cos x+0.1\sin x}
(2.32)
Therefore the exact overall solution is:
y
(
x
)
=
C
1
e
x
+
C
2
e
2
x
+
0.3
cos
x
+
0.1
sin
x
{\displaystyle \displaystyle y(x)=C_{1}e^{x}+C_{2}e^{2x}+0.3\cos x+0.1\sin x}
(2.33)
y
′
(
x
)
=
C
1
e
x
+
2
C
2
e
2
x
−
0.3
sin
x
+
0.1
cos
x
{\displaystyle \displaystyle y'(x)=C_{1}e^{x}+2C_{2}e^{2x}-0.3\sin x+0.1\cos x}
(2.34)
Using the initial conditions yield the equations:
1
=
C
1
+
C
2
+
0.3
&
0
=
C
1
+
2
C
2
+
0.1
{\displaystyle \displaystyle 1=C_{1}+C_{2}+0.3\&0=C_{1}+2C_{2}+0.1}
(2.35)
Solving these two equations yield
C
1
=
1.5
&
C
2
=
−
0.8
{\displaystyle C_{1}=1.5\&C_{2}=-0.8\!}
y
(
x
)
=
1.5
e
x
−
0.8
e
2
x
+
0.3
cos
x
+
0.1
sin
x
{\displaystyle \displaystyle y(x)=1.5e^{x}-0.8e^{2x}+0.3\cos x+0.1\sin x}
(2.36)
The following figure shows the plot of the exact solution over the plot in Fig. 4.2-2:
Figure 4.2-3
The overlap of the plots shows that the Taylor series approximation approach to the ODE is actually accurate to a very large degree with respect to the exact solution of the ODE.
Solved and Typed By - Egm4313.s12.team1.armanious (talk ) 04:02, 14 March 2012 (UTC)
Reviewed By - Chris Stewart Egm4313.s12.team1.stewart (talk ) 04:52, 14 March 2012 (UTC)
Consider the L2-ODE-CC (5) p7b-7 with log(1+x) as excitation:
y
″
−
3
y
′
+
2
y
=
r
(
x
)
{\displaystyle \displaystyle y''-3y'+2y=r(x)}
(3.0)
r
(
x
)
=
log
(
1
+
x
)
{\displaystyle \displaystyle r(x)=\log(1+x)}
(3.1)
y
(
−
3
4
)
=
1
,
y
′
(
−
3
4
)
=
0
{\displaystyle \displaystyle y(-{\frac {3}{4}})=1,y'(-{\frac {3}{4}})=0}
(3.2)
1. Develop log(1+x) in Taylor series, about
x
^
=
0
{\displaystyle \displaystyle {\hat {x}}=0}
2. Let
r
n
(
x
)
{\displaystyle \displaystyle r_{n}(x)}
Find
y
n
(
x
)
{\displaystyle \displaystyle y_{n}(x)}
with the same initial conditions (2).
Plot
y
n
(
x
)
{\displaystyle \displaystyle y_{n}(x)}
[
−
3
4
,
3
]
{\displaystyle \displaystyle [-{\frac {3}{4}},3]}
y
n
(
x
)
{\displaystyle \displaystyle y_{n}(x)}
1. Developing the Taylor series
f
(
x
)
=
∑
n
=
0
∞
f
(
n
)
(
x
^
)
n
!
(
x
−
x
^
)
n
{\displaystyle \displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}({\hat {x}})}{n!}}(x-{\hat {x}})^{n}}
(3.3)
l
o
g
(
1
+
x
)
=
1
1
!
(
x
)
1
+
−
1
2
!
(
x
)
2
+
1
3
!
(
x
)
3
+
−
1
4
!
(
x
)
4
.
.
.
{\displaystyle \displaystyle log(1+x)={\frac {1}{1!}}(x)^{1}+{\frac {-1}{2!}}(x)^{2}+{\frac {1}{3!}}(x)^{3}+{\frac {-1}{4!}}(x)^{4}...}
(3.4)
2. Yn(x) at different n values
First we create the characteristic equation in standard form:
λ
2
−
3
λ
+
2
=
0
{\displaystyle \displaystyle {\lambda ^{2}-3\lambda +2=0}}
(3.5)
Then, by setting it equal to zero, we can find what
λ
{\displaystyle \lambda \!}
(
λ
−
2
)
(
λ
−
1
)
=
0
{\displaystyle \displaystyle {(\lambda -2)(\lambda -1)=0}}
(3.6)
λ
=
2
,
λ
=
1
{\displaystyle \displaystyle {\lambda =2,\lambda =1}}
(3.7)
y
h
(
x
)
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle \displaystyle y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}}
(3.8)
The particular solution at n=4 is of the form:
y
p
(
x
)
=
A
4
x
4
+
A
3
x
3
+
A
2
x
2
+
A
1
x
+
A
0
{\displaystyle \displaystyle y_{p}(x)=A_{4}x^{4}+A_{3}x^{3}+A_{2}x^{2}+A_{1}x+A_{0}}
(3.9)
It's derivative would look like this:
y
p
′
(
x
)
=
4
A
4
x
3
+
3
A
3
x
2
+
2
A
2
x
+
A
1
{\displaystyle \displaystyle y_{p}'(x)=4A_{4}x^{3}+3A_{3}x^{2}+2A_{2}x+A_{1}}
(3.10)
And the second derivative to follow would then become:
y
p
″
(
x
)
=
12
A
4
x
2
+
6
A
3
x
+
2
A
2
{\displaystyle \displaystyle y_{p}''(x)=12A_{4}x^{2}+6A_{3}x+2A_{2}}
(3.11)
Based on the coefficients, the following system of equations exists:
2
A
4
=
−
1
24
{\displaystyle \displaystyle 2A_{4}={\frac {-1}{24}}}
(3.12)
−
12
A
4
+
2
A
3
=
1
6
{\displaystyle \displaystyle -12A_{4}+2A_{3}={\frac {1}{6}}}
(3.13)
12
A
4
−
9
A
3
+
2
A
2
=
−
1
2
{\displaystyle \displaystyle 12A_{4}-9A_{3}+2A_{2}={\frac {-1}{2}}}
(3.14)
6
A
3
−
6
A
2
+
2
A
1
=
1
{\displaystyle \displaystyle 6A_{3}-6A_{2}+2A_{1}=1}
(3.15)
2
A
2
−
3
A
1
+
2
A
0
=
0
{\displaystyle \displaystyle 2A_{2}-3A_{1}+2A_{0}=0}
(3.16)
The results of this set of equations make the coefficients of A's:
A
4
=
−
1
48
{\displaystyle \displaystyle A_{4}={\frac {-1}{48}}}
A
3
=
−
1
24
{\displaystyle \displaystyle A_{3}={\frac {-1}{24}}}
A
2
=
−
3
16
{\displaystyle \displaystyle A_{2}={\frac {-3}{16}}}
A
1
=
1
16
{\displaystyle \displaystyle A_{1}={\frac {1}{16}}}
A
0
=
9
32
{\displaystyle \displaystyle A_{0}={\frac {9}{32}}}
The resulting particular equation looks like this:
y
p
(
x
)
=
−
1
48
x
4
+
−
1
24
x
3
+
−
3
16
x
2
+
1
16
x
+
9
32
{\displaystyle \displaystyle y_{p}(x)={\frac {-1}{48}}x^{4}+{\frac {-1}{24}}x^{3}+{\frac {-3}{16}}x^{2}+{\frac {1}{16}}x+{\frac {9}{32}}}
(3.17)
By adding the particular and homogeneous solutions, we get the complete solution:
−
1
48
x
4
+
−
1
24
x
3
+
−
3
16
x
2
+
1
16
x
+
9
32
+
C
1
e
2
x
+
C
2
e
x
=
y
n
=
4
(
x
)
{\displaystyle \displaystyle {\frac {-1}{48}}x^{4}+{\frac {-1}{24}}x^{3}+{\frac {-3}{16}}x^{2}+{\frac {1}{16}}x+{\frac {9}{32}}+C_{1}e^{2x}+C_{2}e^{x}=y_{n=4}(x)}
(3.18)
We consider the initial conditions by taking the first derivative of the complete solution:
4
−
1
48
x
3
+
3
−
1
24
x
2
+
2
−
3
16
x
+
1
16
+
2
C
1
e
2
x
+
C
2
e
x
=
y
′
{\displaystyle \displaystyle 4{\frac {-1}{48}}x^{3}+3{\frac {-1}{24}}x^{2}+2{\frac {-3}{16}}x+{\frac {1}{16}}+2C_{1}e^{2x}+C_{2}e^{x}=y'}
(3.19)
By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants
C
1
,
C
2
{\displaystyle C_{1},C_{2}\!}
1
=
−
1
48
(
−
.75
)
4
+
−
1
24
(
−
.75
)
3
+
−
3
16
(
−
.75
)
2
+
1
16
(
−
.75
)
+
9
32
+
C
1
e
2
(
−
.75
)
+
C
2
e
−
.75
=
y
n
=
4
(
−
.75
)
{\displaystyle \displaystyle 1={\frac {-1}{48}}(-.75)^{4}+{\frac {-1}{24}}(-.75)^{3}+{\frac {-3}{16}}(-.75)^{2}+{\frac {1}{16}}(-.75)+{\frac {9}{32}}+C_{1}e^{2(-.75)}+C_{2}e^{-.75}=y_{n=4}(-.75)}
(3.20)
0
=
4
−
1
48
(
−
.75
)
3
+
3
−
1
24
(
−
.75
)
2
+
2
−
3
16
(
−
.75
)
+
1
16
+
2
C
1
e
2
(
−
.75
)
+
C
2
e
(
−
.75
)
=
y
n
=
4
′
(
−
.75
)
{\displaystyle \displaystyle 0=4{\frac {-1}{48}}(-.75)^{3}+3{\frac {-1}{24}}(-.75)^{2}+2{\frac {-3}{16}}(-.75)+{\frac {1}{16}}+2C_{1}e^{2(-.75)}+C_{2}e^{(-.75)}=y'_{n=4}(-.75)}
(3.21)
Solving the equations proves that
C
1
=
−
4.8322
,
C
2
=
4.0745
{\displaystyle C_{1}=-4.8322,C_{2}=4.0745\!}
−
1
48
x
4
+
−
1
24
x
3
+
−
3
16
x
2
+
1
16
x
+
9
32
+
−
4.8322
e
2
x
+
4.0745
e
x
=
y
n
=
4
(
x
)
{\displaystyle \displaystyle {\frac {-1}{48}}x^{4}+{\frac {-1}{24}}x^{3}+{\frac {-3}{16}}x^{2}+{\frac {1}{16}}x+{\frac {9}{32}}+-4.8322e^{2x}+4.0745e^{x}=y_{n=4}(x)}
(3.22)
The particular solution at n=7 is found using this matrix equation:
[
2
−
3
2
0
0
0
0
0
2
−
6
6
0
0
0
0
0
2
−
9
12
0
0
0
0
0
2
−
12
20
0
0
0
0
0
2
−
15
30
0
0
0
0
0
2
−
18
0
0
0
0
0
0
2
]
⋅
(
A
0
A
1
A
2
A
3
A
4
A
5
A
6
)
=
{
0
1
−
1
/
2
1
/
6
−
1
/
24
1
/
120
−
1
/
720
}
{\displaystyle \displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0\\0&2&-6&6&0&0&0\\0&0&2&-9&12&0&0\\0&0&0&2&-12&20&0\\0&0&0&0&2&-15&30\\0&0&0&0&0&2&-18\\0&0&0&0&0&0&2\\\end{bmatrix}}\cdot {\begin{pmatrix}{A}_{0}\\{A}_{1}\\{A}_{2}\\{A}_{3}\\{A}_{4}\\{A}_{5}\\{A}_{6}\end{pmatrix}}={\begin{Bmatrix}0\\1\\-1/2\\1/6\\-1/24\\1/120\\-1/720\end{Bmatrix}}}
(3.23)
The values for A then look like this:
A
0
=
−
21
128
{\displaystyle \displaystyle A_{0}={\frac {-21}{128}}}
A
1
=
−
21
64
{\displaystyle \displaystyle A_{1}={\frac {-21}{64}}}
A
2
=
−
21
64
{\displaystyle \displaystyle A_{2}={\frac {-21}{64}}}
A
3
=
−
5
96
{\displaystyle \displaystyle A_{3}={\frac {-5}{96}}}
A
4
=
−
5
192
{\displaystyle \displaystyle A_{4}={\frac {-5}{192}}}
A
5
=
−
1
480
{\displaystyle \displaystyle A_{5}={\frac {-1}{480}}}
A
6
=
−
6.944
E
−
4
{\displaystyle \displaystyle A_{6}=-6.944E-4}
The resulting particular equation looks like this:
y
p
(
x
)
=
−
6.944
E
−
4
x
7
+
−
1
480
x
6
+
−
5
192
x
5
+
−
5
96
x
4
+
−
21
64
x
3
+
−
21
64
x
2
+
−
21
128
x
{\displaystyle \displaystyle y_{p}(x)=-6.944E-4x^{7}+{\frac {-1}{480}}x^{6}+{\frac {-5}{192}}x^{5}+{\frac {-5}{96}}x^{4}+{\frac {-21}{64}}x^{3}+{\frac {-21}{64}}x^{2}+{\frac {-21}{128}}x}
(3.24)
By adding the particular and homogeneous solutions, we get the complete solution:
−
6.944
E
−
4
x
7
+
−
1
480
x
6
+
−
5
192
x
5
+
−
5
96
x
4
+
−
21
64
x
3
+
−
21
64
x
2
+
−
21
128
x
+
C
1
e
2
x
+
C
2
e
x
=
y
n
=
7
(
x
)
{\displaystyle \displaystyle -6.944E-4x^{7}+{\frac {-1}{480}}x^{6}+{\frac {-5}{192}}x^{5}+{\frac {-5}{96}}x^{4}+{\frac {-21}{64}}x^{3}+{\frac {-21}{64}}x^{2}+{\frac {-21}{128}}x+C_{1}e^{2x}+C_{2}e^{x}=y_{n=7}(x)}
(3.25)
We consider the initial conditions by taking the first derivative of the complete solution:
(
7
)
−
6.944
E
−
4
x
6
+
(
6
)
−
1
480
x
5
+
(
5
)
−
5
192
x
4
+
(
4
)
−
5
96
x
3
+
(
3
)
−
21
64
x
2
+
(
2
)
−
21
64
x
+
−
21
128
+
2
C
1
e
2
x
+
C
2
e
x
=
y
n
=
7
′
(
x
)
{\displaystyle \displaystyle (7)-6.944E-4x^{6}+(6){\frac {-1}{480}}x^{5}+(5){\frac {-5}{192}}x^{4}+(4){\frac {-5}{96}}x^{3}+(3){\frac {-21}{64}}x^{2}+(2){\frac {-21}{64}}x+{\frac {-21}{128}}+2C_{1}e^{2x}+C_{2}e^{x}=y'_{n=7}(x)}
(3.26)
By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants
C
1
,
C
2
{\displaystyle C_{1},C_{2}\!}
1
=
−
6.944
E
−
4
(
−
.75
)
7
+
−
1
480
(
−
.75
)
6
+
−
5
192
(
−
.75
)
5
+
−
5
96
(
−
.75
)
4
+
−
21
64
(
−
.75
)
3
+
−
21
64
(
−
.75
)
2
+
−
21
128
(
−
.75
)
+
C
1
e
2
(
−
.75
)
+
C
2
e
(
−
.75
)
=
y
n
=
7
(
−
.75
)
{\displaystyle \displaystyle 1=-6.944E-4(-.75)^{7}+{\frac {-1}{480}}(-.75)^{6}+{\frac {-5}{192}}(-.75)^{5}+{\frac {-5}{96}}(-.75)^{4}+{\frac {-21}{64}}(-.75)^{3}+{\frac {-21}{64}}(-.75)^{2}+{\frac {-21}{128}}(-.75)+C_{1}e^{2(-.75)}+C_{2}e^{(-.75)}=y_{n=7}(-.75)}
(3.27)
0
=
(
7
)
−
6.944
E
−
4
(
−
.75
)
6
+
(
6
)
−
1
480
(
−
.75
)
5
+
(
5
)
−
5
192
(
−
.75
)
4
+
(
4
)
−
5
96
(
−
.75
)
3
+
(
3
)
−
21
64
(
−
.75
)
2
+
(
2
)
−
21
64
(
−
.75
)
+
−
21
128
+
2
C
1
e
2
(
−
.75
)
+
C
2
e
(
−
.75
)
=
y
n
=
7
′
(
(
−
.75
)
)
{\displaystyle \displaystyle 0=(7)-6.944E-4(-.75)^{6}+(6){\frac {-1}{480}}(-.75)^{5}+(5){\frac {-5}{192}}(-.75)^{4}+(4){\frac {-5}{96}}(-.75)^{3}+(3){\frac {-21}{64}}(-.75)^{2}+(2){\frac {-21}{64}}(-.75)+{\frac {-21}{128}}+2C_{1}e^{2(-.75)}+C_{2}e^{(-.75)}=y'_{n=7}((-.75))}
(3.28)
Solving the equations proves that
C
1
=
−
3.3921
,
C
2
=
3.5789
{\displaystyle C_{1}=-3.3921,C_{2}=3.5789\!}
−
6.944
E
−
4
x
7
+
−
1
480
x
6
+
−
5
192
x
5
+
−
5
96
x
4
+
−
21
64
x
3
+
−
21
64
x
2
+
−
21
128
x
+
(
−
3.3921
)
e
2
x
+
(
3.5789
)
e
x
=
y
n
=
7
(
x
)
{\displaystyle \displaystyle -6.944E-4x^{7}+{\frac {-1}{480}}x^{6}+{\frac {-5}{192}}x^{5}+{\frac {-5}{96}}x^{4}+{\frac {-21}{64}}x^{3}+{\frac {-21}{64}}x^{2}+{\frac {-21}{128}}x+(-3.3921)e^{2x}+(3.5789)e^{x}=y_{n=7}(x)}
(3.29)
The particular solution at n=11 is found using this matrix equation:
(3.30)
The values for A then look like this:
A
0
=
2481
2048
{\displaystyle \displaystyle A_{0}={\frac {2481}{2048}}}
A
1
=
36635
3072
{\displaystyle \displaystyle A_{1}={\frac {36635}{3072}}}
A
2
=
5897
1024
{\displaystyle \displaystyle A_{2}={\frac {5897}{1024}}}
A
3
=
8987
4608
{\displaystyle \displaystyle A_{3}={\frac {8987}{4608}}}
A
4
=
1417
3072
{\displaystyle \displaystyle A_{4}={\frac {1417}{3072}}}
A
5
=
415
4608
{\displaystyle \displaystyle A_{5}={\frac {415}{4608}}}
A
6
=
33
2560
{\displaystyle \displaystyle A_{6}={\frac {33}{2560}}}
A
7
=
.001434
{\displaystyle \displaystyle A_{7}=.001434}
A
8
=
5.27
E
−
5
{\displaystyle \displaystyle A_{8}=5.27E-5}
A
9
=
3.44
E
−
6
{\displaystyle \displaystyle A_{9}=3.44E-6}
A
10
=
1.37786
E
−
7
{\displaystyle \displaystyle A_{10}=1.37786E-7}
The resulting particular equation looks like this:
y
p
(
x
)
=
1.37786
E
−
7
x
11
+
3.44
E
−
6
x
10
+
5.27
E
−
5
x
9
+
.001434
x
8
+
33
2560
x
7
+
415
4608
x
6
+
1417
3072
x
5
+
8987
4608
x
4
+
5897
1024
x
3
+
36635
3072
x
2
+
2481
2048
x
{\displaystyle \displaystyle y_{p}(x)=1.37786E-7x^{11}+3.44E-6x^{10}+5.27E-5x^{9}+.001434x^{8}+{\frac {33}{2560}}x^{7}+{\frac {415}{4608}}x^{6}+{\frac {1417}{3072}}x^{5}+{\frac {8987}{4608}}x^{4}+{\frac {5897}{1024}}x^{3}+{\frac {36635}{3072}}x^{2}+{\frac {2481}{2048}}x}
(3.31)
By adding the particular and homogeneous solutions, we get the complete solution:
1.37786
E
−
7
x
11
+
3.44
E
−
6
x
10
+
5.27
E
−
5
x
9
+
.001434
x
8
+
33
2560
x
7
+
415
4608
x
6
+
1417
3072
x
5
+
8987
4608
x
4
+
5897
1024
x
3
+
36635
3072
x
2
+
2481
2048
x
+
C
1
e
2
x
+
C
2
e
x
=
y
n
=
11
(
x
)
{\displaystyle \displaystyle 1.37786E-7x^{11}+3.44E-6x^{10}+5.27E-5x^{9}+.001434x^{8}+{\frac {33}{2560}}x^{7}+{\frac {415}{4608}}x^{6}+{\frac {1417}{3072}}x^{5}+{\frac {8987}{4608}}x^{4}+{\frac {5897}{1024}}x^{3}+{\frac {36635}{3072}}x^{2}+{\frac {2481}{2048}}x+C_{1}e^{2x}+C_{2}e^{x}=y_{n=11}(x)}
(3.32)
We consider the initial conditions by taking the first derivative of the complete solution:
(
11
)
1.37786
E
−
7
x
10
+
(
10
)
3.44
E
−
6
x
9
+
(
9
)
5.27
E
−
5
x
8
+
(
8
)
.001434
x
7
+
(
7
)
33
2560
x
6
+
(
6
)
415
4608
x
5
+
(
5
)
1417
3072
x
4
+
(
4
)
8987
4608
x
3
+
(
3
)
5897
1024
x
2
+
(
2
)
36635
3072
x
+
2481
2048
+
C
1
e
2
x
+
C
2
e
x
=
y
n
=
11
′
(
x
)
{\displaystyle \displaystyle (11)1.37786E-7x^{10}+(10)3.44E-6x^{9}+(9)5.27E-5x^{8}+(8).001434x^{7}+(7){\frac {33}{2560}}x^{6}+(6){\frac {415}{4608}}x^{5}+(5){\frac {1417}{3072}}x^{4}+(4){\frac {8987}{4608}}x^{3}+(3){\frac {5897}{1024}}x^{2}+(2){\frac {36635}{3072}}x+{\frac {2481}{2048}}+C_{1}e^{2x}+C_{2}e^{x}=y'_{n=11}(x)}
(3.33)
By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants
C
1
,
C
2
{\displaystyle C_{1},C_{2}\!}
1
=
1.37786
E
−
7
(
−
.75
)
11
+
3.44
E
−
6
(
−
.75
)
10
+
5.27
E
−
5
(
−
.75
)
9
+
.001434
(
−
.75
)
8
+
33
2560
(
−
.75
)
7
+
415
4608
(
−
.75
)
6
+
1417
3072
(
−
.75
)
5
+
8987
4608
(
−
.75
)
4
+
5897
1024
(
−
.75
)
3
+
36635
3072
(
−
.75
)
2
+
2481
2048
(
−
.75
)
+
C
1
e
2
(
−
.75
)
+
C
2
e
(
−
.75
)
=
y
n
=
11
(
−
.75
)
{\displaystyle \displaystyle 1=1.37786E-7(-.75)^{11}+3.44E-6(-.75)^{10}+5.27E-5(-.75)^{9}+.001434(-.75)^{8}+{\frac {33}{2560}}(-.75)^{7}+{\frac {415}{4608}}(-.75)^{6}+{\frac {1417}{3072}}(-.75)^{5}+{\frac {8987}{4608}}(-.75)^{4}+{\frac {5897}{1024}}(-.75)^{3}+{\frac {36635}{3072}}(-.75)^{2}+{\frac {2481}{2048}}(-.75)+C_{1}e^{2(-.75)}+C_{2}e^{(-.75)}=y_{n=11}(-.75)}
(3.34)
0
=
(
11
)
1.37786
E
−
7
(
−
.75
)
10
+
(
10
)
3.44
E
−
6
(
−
.75
)
9
+
(
9
)
5.27
E
−
5
(
−
.75
)
8
+
(
8
)
.001434
(
−
.75
)
7
+
(
7
)
33
2560
(
−
.75
)
6
+
(
6
)
415
4608
(
−
.75
)
5
+
(
5
)
1417
3072
(
−
.75
)
4
+
(
4
)
8987
4608
(
−
.75
)
3
+
(
3
)
5897
1024
(
−
.75
)
2
+
(
2
)
36635
3072
(
−
.75
)
+
2481
2048
+
C
1
e
2
(
−
.75
)
+
C
2
e
(
−
.75
)
=
y
n
=
11
′
(
−
.75
)
{\displaystyle \displaystyle 0=(11)1.37786E-7(-.75)^{10}+(10)3.44E-6(-.75)^{9}+(9)5.27E-5(-.75)^{8}+(8).001434(-.75)^{7}+(7){\frac {33}{2560}}(-.75)^{6}+(6){\frac {415}{4608}}(-.75)^{5}+(5){\frac {1417}{3072}}(-.75)^{4}+(4){\frac {8987}{4608}}(-.75)^{3}+(3){\frac {5897}{1024}}(-.75)^{2}+(2){\frac {36635}{3072}}(-.75)+{\frac {2481}{2048}}+C_{1}e^{2(-.75)}+C_{2}e^{(-.75)}=y'_{n=11}(-.75)}
(3.35)
Solving the equations proves that
C
1
=
56.1374
,
C
2
=
−
32.64
{\displaystyle C_{1}=56.1374,C_{2}=-32.64\!}
1.37786
E
−
7
x
11
+
3.44
E
−
6
x
10
+
5.27
E
−
5
x
9
+
.001434
x
8
+
33
2560
x
7
+
415
4608
x
6
+
1417
3072
x
5
+
8987
4608
x
4
+
5897
1024
x
3
+
36635
3072
x
2
+
2481
2048
x
+
56.1374
e
2
x
+
−
32.64
e
x
=
y
n
=
11
(
x
)
{\displaystyle \displaystyle 1.37786E-7x^{11}+3.44E-6x^{10}+5.27E-5x^{9}+.001434x^{8}+{\frac {33}{2560}}x^{7}+{\frac {415}{4608}}x^{6}+{\frac {1417}{3072}}x^{5}+{\frac {8987}{4608}}x^{4}+{\frac {5897}{1024}}x^{3}+{\frac {36635}{3072}}x^{2}+{\frac {2481}{2048}}x+56.1374e^{2x}+-32.64e^{x}=y_{n=11}(x)}
(3.36)
Plot of all n=4 (red), n=7 (blue), n=11 (green).
3. Plotting the actual y(x) against approximations
Solved and Typed By - Egm4313.s12.team1.silvestri (talk ) 23:17, 12 March 2012 (UTC)
Egm4313.s12.team1.rosenberg (talk ) 03:10, 13 March 2012 (UTC)
4.4 from lecture notes R4.1 Lect. 7c pgs. 29-30
Extend the accuracy of the solution beyond
x
^
=
1
{\displaystyle {\widehat {x}}=1\!}
1. Back up away a little from the brink of non-convergence at
x
=
1
{\displaystyle x=1\!}
l
o
g
(
1
+
x
)
{\displaystyle log(1+x)\!}
x
^
=
0
{\displaystyle {\widehat {x}}=0\!}
x
1
=
0.9
{\displaystyle x_{1}=0.9\!}
Find the value of
y
n
(
x
1
)
,
y
n
′
(
x
1
)
{\displaystyle y_{n}(x_{1}),y'_{n}(x_{1})\!}
n
{\displaystyle n\!}
y
n
(
x
1
)
,
y
n
′
(
x
1
)
{\displaystyle y_{n}(x_{1}),y'_{n}(x_{1})\!}
10
−
5
{\displaystyle 10^{-5}\!}
2. Develop
l
o
g
(
1
+
x
)
{\displaystyle log(1+x)\!}
x
^
=
1
{\displaystyle {\widehat {x}}=1\!}
n
=
4
,
7
,
11
{\displaystyle n=4,7,11\!}
What is now the domain of convergence? (by observation of your results.)
First, we have the general form of a Taylor series:
f
(
x
)
=
∑
n
=
0
∞
f
(
n
)
(
x
^
)
n
!
(
x
−
x
^
)
n
{\displaystyle \displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}({\hat {x}})}{n!}}(x-{\hat {x}})^{n}}
(4.2.0)
Next, we set
f
(
x
)
=
r
(
x
)
=
l
o
g
(
1
+
x
)
{\displaystyle f(x)=r(x)=log(1+x)\!}
f
(
x
)
{\displaystyle f(x)\!}
f
(
x
)
=
l
o
g
(
1
+
x
)
→
f
′
(
x
)
=
1
(
x
+
1
)
,
f
″
(
x
)
=
−
1
(
x
+
1
)
2
,
f
‴
(
x
)
=
2
(
x
+
1
)
3
.
.
.
{\displaystyle \displaystyle f(x)=log(1+x)\rightarrow f'(x)={\frac {1}{(x+1)}},f''(x)={\frac {-1}{(x+1)^{2}}},f'''(x)={\frac {2}{(x+1)^{3}}}...}
Using these derivatives, we can now develop
l
o
g
(
1
+
x
)
{\displaystyle log(1+x)\!}
x
^
=
1
{\displaystyle {\hat {x}}=1\!}
r
(
x
)
=
l
o
g
(
1
+
x
)
=
f
(
x
)
=
∑
n
=
0
∞
f
(
n
)
(
1
)
n
!
(
x
−
1
)
n
=
1
0
!
(
x
−
1
)
0
+
1
1
!
(
x
−
1
)
1
−
1
2
!
(
x
−
1
)
2
+
2
3
!
(
x
−
1
)
3
−
6
4
!
(
x
−
1
)
4
+
.
.
.
{\displaystyle \displaystyle r(x)=log(1+x)=f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(1)}{n!}}(x-1)^{n}={\frac {1}{0!}}(x-1)^{0}+{\frac {1}{1!}}(x-1)^{1}-{\frac {1}{2!}}(x-1)^{2}+{\frac {2}{3!}}(x-1)^{3}-{\frac {6}{4!}}(x-1)^{4}+...}
(4.2.1)
r
(
x
)
=
l
o
g
(
1
+
x
)
=
f
(
x
)
=
∑
n
=
0
∞
(
−
1
)
n
−
1
f
(
n
)
(
x
−
1
)
n
n
!
{\displaystyle \displaystyle r(x)=log(1+x)=f(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n-1}f^{(n)}(x-1)^{n}}{n!}}}
(4.2.2)
Now the Taylor series for
n
=
4
,
7
,
11
{\displaystyle n=4,7,11\!}
For
n
=
4
{\displaystyle n=4\!}
f
(
x
)
=
∑
n
=
0
∞
(
−
1
)
3
f
(
4
)
(
x
−
1
)
4
4
!
=
(
x
−
1
)
−
1
2
(
x
−
1
)
2
+
1
6
(
x
−
1
)
3
−
1
24
(
x
−
1
)
4
{\displaystyle \displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{3}f^{(4)}(x-1)^{4}}{4!}}=(x-1)-{\frac {1}{2}}(x-1)^{2}+{\frac {1}{6}}(x-1)^{3}-{\frac {1}{24}}(x-1)^{4}}
(4.2.3)
For
n
=
7
{\displaystyle n=7\!}
f
(
x
)
=
∑
n
=
0
∞
(
−
1
)
6
f
(
7
)
(
x
−
1
)
7
7
!
=
(
x
−
1
)
−
1
2
(
x
−
1
)
2
+
1
6
(
x
−
1
)
3
−
1
24
(
x
−
1
)
4
+
1
120
(
x
−
1
)
5
−
1
720
(
x
−
1
)
6
+
1
5040
(
x
−
1
)
7
{\displaystyle \displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{6}f^{(7)}(x-1)^{7}}{7!}}=(x-1)-{\frac {1}{2}}(x-1)^{2}+{\frac {1}{6}}(x-1)^{3}-{\frac {1}{24}}(x-1)^{4}+{\frac {1}{120}}(x-1)^{5}-{\frac {1}{720}}(x-1)^{6}+{\frac {1}{5040}}(x-1)^{7}}
(4.2.4)
For
n
=
11
{\displaystyle n=11\!}
f
(
x
)
=
∑
n
=
0
∞
(
−
1
)
10
f
(
11
)
(
x
−
1
)
11
11
!
=
(
x
−
1
)
−
1
2
(
x
−
1
)
2
+
1
6
(
x
−
1
)
3
−
1
24
(
x
−
1
)
4
+
1
120
(
x
−
1
)
5
−
1
720
(
x
−
1
)
6
+
1
5040
(
x
−
1
)
7
−
1
40320
(
x
−
1
)
8
+
1
362880
(
x
−
1
)
9
−
1
3628800
(
x
−
1
)
10
+
1
39916800
(
x
−
1
)
11
{\displaystyle \displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{10}f^{(11)}(x-1)^{11}}{11!}}=(x-1)-{\frac {1}{2}}(x-1)^{2}+{\frac {1}{6}}(x-1)^{3}-{\frac {1}{24}}(x-1)^{4}+{\frac {1}{120}}(x-1)^{5}-{\frac {1}{720}}(x-1)^{6}+{\frac {1}{5040}}(x-1)^{7}-{\frac {1}{40320}}(x-1)^{8}+{\frac {1}{362880}}(x-1)^{9}-{\frac {1}{3628800}}(x-1)^{10}+{\frac {1}{39916800}}(x-1)^{11}}
(4.2.5)
3. Find
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
n
=
4
,
7
,
11
{\displaystyle n=4,7,11\!}
y
n
″
+
a
y
n
′
+
b
y
n
=
r
n
(
x
)
{\displaystyle y''_{n}+ay'_{n}+by_{n}=r_{n}(x)\!}
x
{\displaystyle x\!}
[
0.9
,
3
]
{\displaystyle [0.9,3]\!}
y
n
(
x
1
)
,
y
n
′
(
x
1
)
{\displaystyle y_{n}(x_{1}),y'_{n}(x_{1})\!}
Plot
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
n
=
4
,
7
,
11
{\displaystyle n=4,7,11\!}
x
{\displaystyle x\!}
[
0.9
,
3
]
{\displaystyle [0.9,3]\!}
Using the same characteristic in problem 4.3 (3.5), we see that the homogenous solution is the same (3.8):
y
h
(
x
)
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle \displaystyle y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}}
(4.3.0)
The particular solution at n=4 is of the form:
y
p
(
x
)
=
A
4
(
x
−
1
)
4
+
A
3
(
x
−
1
)
3
+
A
2
(
x
−
1
)
2
+
A
1
(
x
−
1
)
+
A
0
{\displaystyle \displaystyle y_{p}(x)=A_{4}(x-1)^{4}+A_{3}(x-1)^{3}+A_{2}(x-1)^{2}+A_{1}(x-1)+A_{0}}
(4.3.1)
It's derivative would look like this:
y
p
′
(
x
)
=
4
A
4
(
x
−
1
)
3
+
3
A
3
(
x
−
1
)
2
+
2
A
2
(
x
−
1
)
+
A
1
{\displaystyle \displaystyle y_{p}'(x)=4A_{4}(x-1)^{3}+3A_{3}(x-1)^{2}+2A_{2}(x-1)+A_{1}}
(4.3.2)
And the second derivative to follow would then become:
y
p
″
(
x
)
=
12
A
4
(
x
−
1
)
2
+
6
A
3
(
x
−
1
)
+
2
A
2
{\displaystyle \displaystyle y_{p}''(x)=12A_{4}(x-1)^{2}+6A_{3}(x-1)+2A_{2}}
(4.3.3)
Based on the coefficients, the following system of equations exists:
2
A
4
=
−
1
4
{\displaystyle \displaystyle 2A_{4}={\frac {-1}{4}}}
(4.3.4)
−
12
A
4
+
2
A
3
=
1
3
{\displaystyle \displaystyle -12A_{4}+2A_{3}={\frac {1}{3}}}
(4.3.5)
12
A
4
−
9
A
3
+
2
A
2
=
−
1
2
{\displaystyle \displaystyle 12A_{4}-9A_{3}+2A_{2}={\frac {-1}{2}}}
(4.3.6)
6
A
3
−
6
A
2
+
2
A
1
=
1
{\displaystyle \displaystyle 6A_{3}-6A_{2}+2A_{1}=1}
(4.3.7)
2
A
2
−
3
A
1
+
2
A
0
=
0
{\displaystyle \displaystyle 2A_{2}-3A_{1}+2A_{0}=0}
(4.3.8)
The results of this set of equations make the coefficients of A's:
A
4
=
−
1
8
{\displaystyle \displaystyle A_{4}={\frac {-1}{8}}}
A
3
=
−
7
12
{\displaystyle \displaystyle A_{3}={\frac {-7}{12}}}
A
2
=
−
17
8
{\displaystyle \displaystyle A_{2}={\frac {-17}{8}}}
A
1
=
−
33
8
{\displaystyle \displaystyle A_{1}={\frac {-33}{8}}}
A
0
=
−
65
16
{\displaystyle \displaystyle A_{0}={\frac {-65}{16}}}
The resulting particular equation looks like this:
y
p
(
x
)
=
−
1
8
(
x
−
1
)
4
+
−
7
12
(
x
−
1
)
3
+
−
17
8
(
x
−
1
)
2
+
−
33
8
(
x
−
1
)
+
−
65
16
{\displaystyle \displaystyle y_{p}(x)={\frac {-1}{8}}(x-1)^{4}+{\frac {-7}{12}}(x-1)^{3}+{\frac {-17}{8}}(x-1)^{2}+{\frac {-33}{8}}(x-1)+{\frac {-65}{16}}}
(4.3.9)
By adding the particular and homogeneous solutions, we get the complete solution:
y
n
=
4
(
x
)
=
C
1
e
2
x
+
C
2
e
x
+
−
1
8
(
x
−
1
)
4
+
−
7
12
(
x
−
1
)
3
+
−
17
8
(
x
−
1
)
2
+
−
33
8
(
x
−
1
)
+
−
65
16
{\displaystyle \displaystyle y_{n=4}(x)=C_{1}e^{2x}+C_{2}e^{x}+{\frac {-1}{8}}(x-1)^{4}+{\frac {-7}{12}}(x-1)^{3}+{\frac {-17}{8}}(x-1)^{2}+{\frac {-33}{8}}(x-1)+{\frac {-65}{16}}}
(4.3.10)
The particular solution at n=7 is found using this matrix equation:
[
2
−
3
2
0
0
0
0
0
2
−
6
6
0
0
0
0
0
2
−
9
12
0
0
0
0
0
2
−
12
20
0
0
0
0
0
2
−
15
30
0
0
0
0
0
2
−
18
0
0
0
0
0
0
2
]
⋅
(
A
0
A
1
A
2
A
3
A
4
A
5
A
6
)
=
{
0
1
−
1
/
2
1
/
3
−
1
/
4
1
/
5
−
1
/
6
}
{\displaystyle \displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0\\0&2&-6&6&0&0&0\\0&0&2&-9&12&0&0\\0&0&0&2&-12&20&0\\0&0&0&0&2&-15&30\\0&0&0&0&0&2&-18\\0&0&0&0&0&0&2\\\end{bmatrix}}\cdot {\begin{pmatrix}{A}_{0}\\{A}_{1}\\{A}_{2}\\{A}_{3}\\{A}_{4}\\{A}_{5}\\{A}_{6}\end{pmatrix}}={\begin{Bmatrix}0\\1\\-1/2\\1/3\\-1/4\\1/5\\-1/6\end{Bmatrix}}}
(4.3.11)
The values for A then look like this:
A
0
=
−
99
2
{\displaystyle \displaystyle A_{0}={\frac {-99}{2}}}
A
1
=
−
99
{\displaystyle \displaystyle A_{1}=-99}
A
2
=
−
49
{\displaystyle \displaystyle A_{2}=-49}
A
3
=
−
95
6
{\displaystyle \displaystyle A_{3}={\frac {-95}{6}}}
A
4
=
−
15
4
{\displaystyle \displaystyle A_{4}={\frac {-15}{4}}}
A
5
=
−
13
20
{\displaystyle \displaystyle A_{5}={\frac {-13}{20}}}
A
6
=
−
1
12
{\displaystyle \displaystyle A_{6}={\frac {-1}{12}}}
The resulting particular equation looks like this:
y
p
(
x
)
=
−
1
12
(
x
−
1
)
7
+
−
13
20
(
x
−
1
)
6
+
−
15
4
(
x
−
1
)
5
+
−
95
6
(
x
−
1
)
4
+
−
49
(
x
−
1
)
3
+
−
99
(
x
−
1
)
2
+
−
199
2
(
x
−
1
)
{\displaystyle \displaystyle y_{p}(x)={\frac {-1}{12}}(x-1)^{7}+{\frac {-13}{20}}(x-1)^{6}+{\frac {-15}{4}}(x-1)^{5}+{\frac {-95}{6}}(x-1)^{4}+-49(x-1)^{3}+-99(x-1)^{2}+{\frac {-199}{2}}(x-1)}
(4.3.12)
By adding the particular and homogeneous solutions, we get the complete solution:
y
n
=
7
(
x
)
=
C
1
e
2
x
+
C
2
e
x
+
−
1
12
(
x
−
1
)
7
+
−
13
20
(
x
−
1
)
6
+
−
15
4
(
x
−
1
)
5
+
−
95
6
(
x
−
1
)
4
+
−
49
(
x
−
1
)
3
+
−
99
(
x
−
1
)
2
+
−
199
2
(
x
−
1
)
{\displaystyle \displaystyle y_{n=7}(x)=C_{1}e^{2x}+C_{2}e^{x}+{\frac {-1}{12}}(x-1)^{7}+{\frac {-13}{20}}(x-1)^{6}+{\frac {-15}{4}}(x-1)^{5}+{\frac {-95}{6}}(x-1)^{4}+-49(x-1)^{3}+-99(x-1)^{2}+{\frac {-199}{2}}(x-1)}
(4.3.13)
The particular solution at n=11 is found using this matrix equation:
4. Use the matlab command ode45 to integrate numerically
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)\!}
y
n
(
x
1
)
,
y
n
′
(
x
1
)
{\displaystyle y_{n}(x_{1}),y'_{n}(x_{1})\!}
y
(
x
)
{\displaystyle y(x)\!}
Plot
y
(
x
)
{\displaystyle y(x)\!}
y
n
(
x
)
{\displaystyle y_{n}(x)\!}
Solved and Typed By - --Egm4313.s12.team1.wyattling (talk ) 19:44, 14 March 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein (talk ) 19:52, 14 March 2012 (UTC)
Contributing Members
edit
Team Contribution Table
Problem Number Lecture Assigned To Solved By Typed By Proofread By
4.1
R4.1 Lect. 7c pgs. 19-22 Chris Stewart
Chris Stewart
Chris Stewart
Jesse Durrance
4.2
R4.2 Lect. 7c pgs. 26-27 George Armanious
George Armanious
George Armanious
Chris Stewart
4.3
[Lecture link]
Emotion Silvestri
Emotion Silvestri
Emotion Silvestri
Steven Rosenberg
4.4
R4.1 Lect. 7c pgs. 29-30 Wyatt Ling
Wyatt Ling
Wyatt Ling
Eric Essenwein
![{\displaystyle \displaystyle \sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum _{j=0}^{n}d_{j}x^{j}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26abde87ec508415f9dc5c6291152d54fa4f487c)
![{\displaystyle \displaystyle [c_{1+2}(1+2)(1+1)+ac_{1+1}(1+1)+bc_{1}]x^{1}=d_{1}x^{^{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/772467a65d68a2c203222a343881e5d2b618eb85)
![{\displaystyle \displaystyle [6c_{3}+2ac_{2}+bc_{1}]x=d_{1}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd211d80fe182396e5c0b511006c390b75f3bfab)
![{\displaystyle \displaystyle [c_{2+2}(2+2)(2+1)+ac_{2+1}(2+1)+bc_{2}]x^{2}=d_{2}x^{^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b64f91fa46469fcb590ec054f5e512080538b0bb)
![{\displaystyle \displaystyle [12c_{4}+3ac_{3}+bc_{2}]x^{2}=d_{2}x^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53fe6260c48db7ccb70a9bae2fc5ecafae9ea0ea)
![{\displaystyle \displaystyle [c_{(n-2)+2}(n-2+2)(n-2+1)+ac_{(n-2+1)}(n-2+1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/774eba2ccf1c2ee6523852766ccb20c608df236a)
![{\displaystyle \displaystyle [c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cee5e5ffcd7eb236931108f0dd8ce3d867610e1c)
![{\displaystyle \displaystyle ac_{n}n+bc_{(n-1)}]x^{n-1}=d_{n-1}x^{n-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30bd9813c6417259456eb9a9582c9b18549febfe)
![{\displaystyle \displaystyle [-{\frac {3}{4}},3]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51eafeeb79de2e1c3d77b6df66e9817a5c555879)
![{\displaystyle [0.9,3]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb2c515c8c74080e5a7916ac7bd889dc9164916e)
