[ edit | edit source ] Mtg 21:
P.20-4 (continued)
⇒
a
x
−
1
{\displaystyle \Rightarrow ax^{-1}\ }
∀
a
{\displaystyle \forall \!\,a\ }
u
1
(
x
)
=
c
1
x
−
1
{\displaystyle u_{1}(x)=c_{1}x^{-1}\ }
2)
b
=
2
⇒
a
x
2
{\displaystyle b=2\Rightarrow ax^{2}\ }
b
2
−
b
−
2
=
0
{\displaystyle b^{2}-b-2=0\ }
u
2
(
x
)
=
c
2
x
2
{\displaystyle u_{2}(x)=c_{2}x^{2}\ }
u
1
{\displaystyle u_{1}\ }
u
2
{\displaystyle u_{2}\ }
W
{\displaystyle W\ }
3)
b
=
6
,
a
=
1
4
⇒
{\displaystyle b=6,a={\frac {1}{4}}\Rightarrow \ }
=
7
x
4
{\displaystyle =7x^{4}\ }
=
7
x
4
{\displaystyle =7x^{4}\ }
b
=
6
,
a
=
1
4
⇒
1
4
x
6
{\displaystyle b=6,a={\frac {1}{4}}\Rightarrow {\frac {1}{4}}x^{6}\ }
4)
b
=
5
,
a
=
1
6
⇒
{\displaystyle b=5,a={\frac {1}{6}}\Rightarrow \ }
=
3
x
3
{\displaystyle =3x^{3}\ }
=
3
x
3
{\displaystyle =3x^{3}\ }
b
=
5
,
a
=
1
6
⇒
1
6
x
5
{\displaystyle b=5,a={\frac {1}{6}}\Rightarrow {\frac {1}{6}}x^{5}\ }
Llinearity of ordinary differential equation
⇒
{\displaystyle \Rightarrow \ }
y
P
(
x
)
=
1
4
x
6
+
1
6
x
5
{\displaystyle y_{P}(x)={\frac {1}{4}}x^{6}+{\frac {1}{6}}x^{5}\ }
y
(
x
)
=
c
1
x
−
1
+
c
2
x
2
+
y
P
(
x
)
{\displaystyle y(x)=c_{1}x^{-1}+c_{2}x^{2}+y_{P}(x)\ }
c
1
x
−
1
+
c
2
x
2
=
y
H
(
x
)
{\displaystyle c_{1}x^{-1}+c_{2}x^{2}=y_{H}(x)\ }
u
2
{\displaystyle u_{2}\ }
Eq.(1) P.3-1 =
y
″
+
a
1
(
x
)
y
′
+
a
0
(
x
)
y
=
f
(
x
)
{\displaystyle y''+a_{1}(x)y'+a_{0}(x)y=f(x)\ }
u
1
(
x
)
{\displaystyle u_{1}(x)\ }
u
1
″
+
a
1
(
x
)
u
1
′
+
a
0
(
x
)
u
1
=
0
{\displaystyle u_{1}''+a_{1}(x)u_{1}'+a_{0}(x)u_{1}=0\ }
y
(
x
)
=
U
(
x
)
u
1
(
x
)
{\displaystyle y(x)=U(x)u_{1}(x)\ }
U
(
x
)
{\displaystyle U(x)\ }
Follow the same argument as on P.17-2 to obtain:
f
(
x
)
=
U
′
(
a
1
u
1
+
2
u
1
′
)
+
U
″
u
1
{\displaystyle \displaystyle {\begin{aligned}f(x)=U'(a_{1}u_{1}+2u_{1}')+U''u_{1}\end{aligned}}}
(1)
NOTE: this equation is missing the dependant variable
U
{\displaystyle U\ }
U
′
{\displaystyle U'\ }
ϕ
{\displaystyle \phi \ \ }
Z
(
x
)
:=
U
′
(
x
)
{\displaystyle \displaystyle {\begin{aligned}Z(x):=U'(x)\end{aligned}}}
(2)
⇒
u
1
(
x
)
Z
′
+
[
a
1
(
x
)
u
1
(
x
)
+
2
u
1
′
(
x
)
]
Z
=
f
(
x
)
{\displaystyle \displaystyle {\begin{aligned}\Rightarrow u_{1}(x)Z'+\left[a_{1}(x)u_{1}(x)+2u_{1}'(x)\right]Z=f(x)\end{aligned}}}
(3)
where
u
1
(
x
)
{\displaystyle u_{1}(x)\ }
[
a
1
(
x
)
u
1
(
x
)
+
2
u
1
′
(
x
)
]
{\displaystyle \left[a_{1}(x)u_{1}(x)+2u_{1}'(x)\right]\ }
Z
(
x
)
{\displaystyle Z(x)\ }
Eq.(4) P.8-2
U
(
x
)
=
∫
x
Z
(
s
)
d
s
{\displaystyle \displaystyle {\begin{aligned}U(x)=\int _{}^{x}Z(s)\,ds\end{aligned}}}
(4)
y
(
x
)
=
U
(
x
)
u
1
(
x
)
{\displaystyle \displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)\end{aligned}}}
(5)
K p.28, problem 1.1ab
a)
u
1
(
x
)
=
e
x
{\displaystyle u_{1}(x)=e^{x}\ }
(
x
−
1
)
y
″
−
x
y
′
+
y
=
0
{\displaystyle (x-1)y''-xy'+y=0\ }
y
(
x
)
=
e
r
x
{\displaystyle y(x)=e^{rx}\ }
r
=
{\displaystyle r=\ }
Find
r
1
,
r
2
{\displaystyle r_{1},r_{2}\ }
u
1
=
e
r
1
x
{\displaystyle u_{1}=e^{r_{1}x}\ }
u
2
{\displaystyle u_{2}\ }
![{\displaystyle \displaystyle {\begin{aligned}\Rightarrow u_{1}(x)Z'+\left[a_{1}(x)u_{1}(x)+2u_{1}'(x)\right]Z=f(x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ead73dc619d2938abc6fae193d0028c165527b2)
![{\displaystyle \left[a_{1}(x)u_{1}(x)+2u_{1}'(x)\right]\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3253add6f44741f03d5795fe69c1cc40ba4778b4)
