ページ3:

EX. 1
·Y" + y = 0.0017²
y (0) = 0, y' (0) - 1.5
=
·r.(x).
Step 1 find yo
=
set y₁₁ e^r y₁ = ne" y "
Yn
,
(x² + 1) e^x = 0, λ = ti
Y₁ = A cos x + B sin x
Step 2 set Yp
2.1 Basic rule r(x) → Exh
ki
Set yp k²x² + k₁x + ko
=
Yp² = 2k2x + ki
Yp" = 2 k z
H D t t =
2
ak² + k²x² + k₁x + ko
242+k
·x k₁ = 0
= 1²
= 0.001 x²
= 0 0 = -0.002
k₂ = 0.001

ページ4:

Step 3 y = Yn + Yp
=
yAcosx Bsinx + 0,00-0002
Step. 4 use
J =
boundary conditions
y 10) = A- 0.002 = 0
y' (0) = B = 1.5
·Y'(0)
,
A = 0.002
= 0.002 cos x +1.5 sinx +0.00 | x² - 0.002

ページ5:

EX. 2
y" + 3y+2.25y=-
+ 3y / +2.25y = -10 e 1
Step 1 find yn
Set yh = e^t yn' = re^t,
,
(^² +377 +2.25) e^t o
7112
士
2
19-9
0
y = C₁ e²² x + C₂ x e²±² x
Step 2 set y
α =
2.1 Basic rule
Yh" =^e^t
2
P
=
-15X
Yp= c e
2.2 modification rule
-5X
Yp = x²ce"
=
2x Ce
Yp" = 2ce1sx
-
5X
-1·5x² Ce
4,5X
3xce
"-3xce is x +2.25x" ce
7.58

ページ6:

代回原式
4.5x
7.5x
5X
2
2 Ce 15 x 6x6 +2.25 x Ce + 3 (2 x Ce - 1.sx² Ce 15 x )
-1.5X
= -
+2.25 X² C è'
15
2C-bxc+ 2. 25x²C + 6xc - 9.5x²C +2.25 x²C = -10
C=-5
Step 3 y = yn + Yp
=
J. c. ecx ex's ex
-5X
-

ページ7:

=
EX. 3 y" + zy' +0.75y 2 cosx-0.25 sinx + 0.09x
(Y2(X)
·r₁(x)
Step 1 find yn
=
Y (0) - 278, y'10) = -0.43
2
Set yh = e^t Yh² = 2 e^z, y "" = x² e^t
( 1² +27 +0.75) e^ ² = 0
71,2 = 2 + √4-3
Jn. C. et ₂ ex
=
Step 2 set Yp
+
Yp. = k cos x + M sin x.
=
gp -ksinx + M cosx
Yp" = - koosx - Msinx
Sub into
(Kosx - Msinx)+2-sinx + M cosx)
tong lk cosx + m sin x) 2 COSX - 0.25 sinx
=

ページ8:

Cosx
-16+214 +0.75 = 2
Sinx -M-2k +0.75M = -0.25
M=1, K=0
Yp₂ = k₁x+ ko
Y pź = ki
,
Yp₁ = sinx
Yp₁ = 0
Sub into
X':
0 + 2k₁ +0.75 (k.x+ko) = 0.09x
1=0.12
x² = 2k, +0.75 k₁ = 0, ko = -0.32
Yp₂ = 0.12x -0.32
Step. 3. y = yn + Yp
y = C₁ e ± x + Cz e ³ x + sin x + 0.12X -0.32
y=ce

ページ9:

Step 4 use
boundary conditions
910) = 2.78 = C₁ + C₂ -0.32
y' lo) = -0.43 = = C₁ = C₂ +1 +0.12
C₁ + C₂ = 3, 2
C₁ + 3C2 = 3.)
-
2 C 2 = - O. 1, C₂ = -0.05
C₁ = 3.25

ページ10:

Mass - spring system
my" + cy' +ky = rit) = Fo cos (wt)
k.
Ek spring coefficient
mass = m
damping coefficient c
F = Fo cos (wt)
Step 1 find yn y. C₁ e^it + C₂ et
Step 2 set Yp
Yp
=
= a cos ct +
b
Sin wt
Yp' = - aw sin wt + bw coswt
yp" = -aw² coscut - bw² sinwt.

ページ11:

Sub into
m(-aw² coscot - bw³sinwt) + C (-aw sin wt + bw coswt) + k (a as wt + b sin wt).
- Focos wit
=
coswt (k-mw³) a Cub-Fo
T
Sinwt
-cwa + (k-mw³) b=
Yp (t) Fo (k-mw²)
(k - mw³² )² + w² c²
0
cos wt +.
F. wc
(k - mw³²)² + w²c²
Sinwt
·√a² + b²
Yp = acos wt + b sin wt
b
a
a
=
cos wt +
b
√α² + b²
sin (wt))
=
√a² + b²
(
cos O coscut + Sin O sin wt)
√a²+b² cos (wt-0)
=
C* (w) cos (wt-0)
最大振辐

ページ12:

C = 0 -
→ Wo
=
m
Resonance * *, Wo= W
共振
y" + W³y = Fo cos wt
m
set Yp = t (a cos wt + b sin wt)
zw (-a sinwt +b coscut) - w³t (a cos ct + b sin wt)
Yp"
=
Yp (+) =
Fo
2mWo
t sin (Wot)
Beats 沒有共振,Wo≠W
y" + Way = Fo
y (t) =
= 2F.
m (wo²-w³)
Cos wt
Sin (wo+wt) sin (wo-wt)

ページ13:

System
of ODE as models.
= a.. Y₁ + a.. Yo
y
All aiz
A21 Azz
Aiz
12
A22
= A₂, y + A²² Y=
eigenvalues & eigenvectors
J' = A ÿ
入特徵值
100加侖
我特徵向量
2 gal/min
2 gal/min
150磅(26) 肥料 = yz
T₂
100 (gal)
2
ih
y₁ = 200 y =
2
System of tanks
out
2
100
Y₁ = 2 y₁ - 2 Yz
100
·set ÿ = x ext
y = λx e²²
-0.02 0.02
y' = [0.022-002 ] J
0.02 -0.02
J' = A ÿ
①

ページ14:

1 = ②
A ÿ = λx e₁₂
A(x ext) = λx ent
A x - λ x = 0
eigen vector
(A-^ 1 ) x = 0
eigen value
[bi]
if x+0, A-^1=0
det (A-11) = 0
-0.02-7
0.02
=
0.02
-0.02-2
0
z
0.02² + 0.0 4 7 + x² -0.02²
入(入 +0.04) = 0
入
=
,
For λ = 0
=
0,
-0.04
有無限多組解,代一組就好
[ 0.02 -002 ] [ X ] = 0 5 X 1 3 10 l; 10 -30 30 42
0.02-0.02
X₁ = X 2
=
|
“”
= []
e^
gọi tet lie
=

ページ15:

X
x = [ 4 ]
=
x" e^zt (1) e 42
-0.04t
For ₂ = -0.04
[ 0.02 002 ] [ X ] = 0
0.02
X₁ = 1, X 2 = -|
Cz
y= C₁ j " + C₂ j "
=
• C₁ [1] +
C₂
[1]
"y"
=
-0.04t

ページ16:

降階
2nd order 1st order t
:
2nd order y" + y² + y = 0
set y = y y' = Y₂y" = "Y"
Ist order
-
2
=
→
Y₁' = - Y₂- Y₁
j = [01]J

ページ17:

EX. y " + y + Q.
y" 0.25 y
=
0
set y = y. y' = Yzy" y=""
-Y2-0.25 Y
Y₁₁ = Y₂
Y₂ = -y=-0.25y
det Ax-71-0
→>>
J' = [ 00351] J
0.25
一入
|
x² +7 +0.25 = 0
= 0
入===
2
-0.25 -1-入
For -0.5
=
Xz
]
[02-05]
[-0.5 -05 ] [ X
0.5X, +X2=0 x"
= 0
=
=-0.5
[3], ju= [ 3 ] eas
-0.5t
but you, you are linear dependent

ページ18:

set y₁ = e^¹² + tx" ext
^₁ x = AX"
yos = ū^ie^it + x e^t + nit x ent
=
y'. Ay Ag". A (üe^" + " te"") //
=
(A-1₁2) ū = X"
0.5
1
-0.250.5
u = [is]
-0.5t
9" [is] est [i]ease
=
-0.51
-0.51
-0.51
J = C₁ ( [ 1 ] e 0 st ) + C₁ ( [is] e ³ ³ 4+ t[i]e-ost)
least).

ページ19:

phase plane method
Constant coefficient
system
ODE
det
011-λ
012
=
021
det | Añ-11 |= 0
=
- (all -λ) (a..-λ) - a₁₂ a+ -x²- anλ-a-λ) + a. azz - a₁₂ a.
λ² - ( a + a² ) λ
+ det | A] =λ² - p7+9=0
71,2
1
=
z
£, √1-49, p=49
0
Po (critical point) s-p²-49 7. Dz=det|A| -q P=^it^2 stability
tare
node
670
7.77270
Same Sign
900
Po
λ, and a <o attractive stable
Pro
^ and Az 70
unstable
Y₂
Saddle
point
real
ay,
7.77240.900
opposite sign
unstable
P=0
71.772
Center
29.
+bi
(a+b) (a-bi)
•p=0·
Stable
imaginary
=a+b² 70
1-0
Pto
spiral
970
atbi
P70 unstable
pco
→ attractive stable

ページ20:

q
spiral
unstable
spiral
Stable
Stable
(attractive)
node
→Center
>=0'
unstable
node
940
saddle point
unstable
Y₂
node
Po (critical point) = p² 49 ^.^2 = Jet|A| -q
= a₁, + are
Pco
stability
λ, and a co
∙y.
670
7.77270,970
Same Sign
→ attractive Stable
Po
^ and Az 70
unstable
Saddle
point
real
7177240.900
unstable
opposite sign
P=0
7.772
Center
Ibi
=(a+bi) (a-bi)
pro
Stable
imaginary
·=a² + b² 70
150
P+o
spiral
970
P70
→ unstable
atbi
pco
→ attractive stable

ページ21:

EX. Improper node]
J = [ "$ {} ] j
P = a + Azz = 11772 = -3+(-3)=-6
₁₁
=
q= det/Al = ^i^2-9-1 = 8 70
7172
√p=4q
-6
71,2 = + + √1:49
For λ = -4
+4
[3.4
4] [X]
+X₂ = 0
X₁ + x 2 = 0
For A₂ = -2
+ 2
****
-3+2
-X + X₂ = 0
X₁ - X₂ = 0
36-32
2
=
-3 + 1
= -
-4,-2
= 0
x = [4]
= 0
* = []

ページ22:

-4+
j⋅ α. [4] € * + G² [1] e "
=
graph: Step 1. It Y, & Y₂ 67 13.
Step
set Co
Y₁ = C₂ e²² t
Set C2
=
=
Y₂ = Cz ez z
Czé
-27
-2t
Yz = - Cze²e
2. It critical point
Y₁ = Y₂
Y₁ = - Yz
題目
-34, +42
→(y,,y.) (0,0)
=
0
找方向
Step 3. I4 15 17
-* 1 1, 2 < 0 1 A
9₁-3y=
:
* 2 When (y,, J.) = (0,1)
Y₁ = -3y₁ + Yz
Yo' = Y₁ - 392
20
<0

ページ23:

When too, y₁
Yz
When t→∞
Yz
attractive stable
= y z
y
-zt
Cie + Cze²²+
-*t
Cl + Cze
4t
-2t
0.2
-4t
Y₂ - c, e + c₂ et.
=1

ページ24:

Ex. proper node
A1 + Azz
·P = ai +1
q= det/Al
71,27
=
For = 1
=
=
2
|
P₁ √p=49
乙
J₁ = [bi] y
=
士
4-4
[:]]。
t
Y = C₁ [ 6 ] e ² + c ₁₂ [ i ] e t
:
X = [:]
x
= [ i ]
graph Step 1. Ft Y, & Yz 67 B 13.
Y₁ = C₁et
Yz = C ₂ e t
Step 2. It critical point
題目
Y Y
=
Y₂' Jz
=
C₁
Cz
:dy,
=
=
Constant
1
→ (y,, y) = (0, 0)

ページ25:

Step 3. 28 15 19
1270
向外
unstable
Yz

ページ26:

EX Saddle point
P = A₁ + Azz
q= det/Al
71,2
=
£1.
For ₁ = 1
H
0
0
-2X2
For 112=-1
1+1 0
=
=
y' - [14] ÿ
0
- |
1p² 49.
= 0
X2
[H] [
0 -1.71
2X,
=
0
= 0
<0
=
0 ± 2
隨便代個數字
x = [b]
x^
=
[1]
t
-t
ÿ = a [ i ] e² + c [ i ] e t

ページ27:

y = c₁ e² y = c₂ e²
Y
y₁ y₂ = C. C₂ = constant
graph: Step 1. It y. & y. 67 B 14.
set C₁ = 0
=
0
Y ₂ = ē
Y₁ =
= 0
Set C 2 = 0
J₁ = e²
Yz = 0
→ Y ₂ = 0
2
Step
2. I critical point
dy,
26 A
0
y₁
-Y₂
→ (y₁, y₁) = (0, 0)
0
找方向
Step 3. It' 517
when (Y₁, J.) = (0,1)
"
-Y= <0
↓

ページ28:

Yz
unstable
Yo

ページ29:

EX Center
Azz
P = a₁ + A²² = 0
q = det/A1 = 4
√√√ 1,2 =
[-4 0 ] ÿ
P₁ √p=49
士
For a = zi
-21
:
0 + zi
]]
-21
- 2 ; X + X 2
-4X,
-
For 入z=-zi
= 0
21 X 2 = 0
D
[24][]=
-4
2ī x+ X2 = D
-4 x,
+ 21×2 = 0
X"
:
X (2)
]

ページ30:

zit
-zit
J = C [ 1 ] e² ² + C₁₂ [ 1 ] e π t
graph: Step 1. It Y, & Yz 57 B 13.
ix
tulais formula ex cosx ti sinx
Eular's é
=
Y₁ = c, coszt + Co isinat
= C₂ coszt - C₁ sinzt
y₁² + (c) = constant
Step 2. It critical point
dy,
題目
0
Yz
→
-49,
· (y₁, y₁) = (0, 0)
。
Step 3. 28 519
when (Y₁, Y₂) = (0,1)
dy₁ Y z > 0
→
(順時針)
题目

ページ31:

Yz
Stable
.Y,

ページ32:

EX. Spiral] J' [+] ÿ
螺旋
2 = A₁₁ + Azz
=
44 J
= - 2 ¥ 0
q= det/Al
= 2
^\₁₁₂ = £ + √p=49
71,2
For λ = -1 +5
[4]Q
- 1 x + x 2 = 0
-TX,
+X
- X₁ + x 2 = 0
For ^2 = -1-1
√4-8.
士
=-127
如:[]
=
= 0
1 x + x 2 = 0
X₁ + 1 X 2 = 0
x-(4)

ページ33:

J = C₁ [1]
e
(-1+i)t
(-1-1)t
+ e
C [;)] C
graph: Step 1. FE Yi & Yz by B 13.
9 = e² (c, [ ] e
+
C. [ ;]*)
y₁ = e-t [ C, cost + C₂ sint]
Y
·Y₁ = et [ C. cost - ci sint ]
-zt
+
Constant
七个半径2
Step
2. It critical point.
dy.
dy,
0
Step3.找方向
題目
-x 1 211, 2 <0 1
-Y₁+yz
-Y₁-y
* 2 when (Y₁, Y₂) = (0,1)
dy₁ = Y₂ ?
70
• (y₁, y.) - (0,0)

ページ34:

Yz
attractive stable

ページ35:

Lotka-Volterra population model.
non-linear system.
= ay₁- by₁yz = f₁
·Y ₂ = k Y₁ Yz -l Yz = fz
Step 1. It critical point
•Aayi
ayi- by y
.ky Yo -lyz
=
dy,
0
y₁ (a-by)
=0
Yz ( ky₁ - 1) = 0
→
depending variables
Y₁ number of rabbits.
:
Yo number of faxes
:
a growth rate of
b: killing rate
(relative)
k:
rate of
k growth rate
(relative)
y₂
I death rate of Yz
(y,, y₁) = (0, 0), ( ± 1 )
Step 2 把方程式變成線性化
For Po(0, 0)
+41(0,0)
ry,
=
a-by= =
= a
f. (0,0) =
dy,
· ky z = 0
-
= 0
8f210,0) 0-by₁ = 0
ryz
ky₁ -1 - -1
Jyz

ページ36:

Taylor expansion
:
f(x.) + f'(x) (x-xo) + f (x) (x-x₁)² ...
2 variables
Y₁ = dfil 0,0)
2!
dy,
Y; +
+f, 10,02
Y₂ = ay
y' =
fa(0, 0)
dy
y₁ + + f₂ (0,0) You - Yz
aya
J = [ a i ] ]
P = A1 + Azz
=
=
a-l
q = det|A| al <0
-
Saddle point
=
0-p²-49
70
Y₂

ページ37:

For P. 1 1/2, 2² ²)
set J₁ = y. - £
J₁ ya
=
-
sub into
ay₁- by₁y = f₁
y₁ = ay₁.
yz
Y₁ = k J₁ yz - Yz = fz
J₁a ( Ĵ₁ + — — ) - b ( Ĵ₁ + — —²) (g₁ + 1 )
N
十
- ( √₁ + 1 ) ( a − b (J₁ + 1 )
)
=
N
十
-
by
+
J.
Yz
=
=
-
+
-) (- by ₁)
- by₁ j₁-eb y₁
l
(J) (+) (+)
+
a
-
(1) (k-1)
(9+号)(kg)
kỹ, ta k ỹ,

ページ38:

:
New P. ( 1/1, 1 1 )
a
→
ŷ.' - bj, J. eb J.
= -b
=
-
Po' (0,0)
=
fi'
ỹ kỹ, tak ỹ, tỉ
+
=
fi'10,0)
dyz
- by, - eb
Yz
afi' (0,0)
-bŷ
ay
f₂10,0)
=
ay,
ky + ak
f.100)
诩
2
=
y' = df, 10.0) √ + of 10,0) J. = lb y
dy,
Yi
Jyz
J₁ = f(0.0) √ + of 10,0) J = ok J
dy,
J' = [
P = a₁ + Azz
=
0
ak
0
-
0
k
q = det/Al -- ( - eb ok) = la> 0
0
=
-p²-49<0
(Center)

ページ39:

找方向
when (J₁, J₁) = (0,1)
dy, = y.
Y₂
2
<0 (逆時針)
> Y₁

ページ40:

damped pendulum equation
8
T
mg sin e
mg
(a) Pendulum
m
"0" + c o² + k s i n o = 0
set 0" = y."
=
Y₂ = -k siny, cyz
J' = [ i
Step 2. It critical point.
0
dy,
For Po (TV, 0)
題目
Yz
l ] J
= Ay
-C
-ksiny₁-cy=
(±π; 0), (±3π; 0).
C: damping coef.
k = 2
Yi is small
I
-ky-cy₂
=
→ (Y₁, y₁) (nTv, o).
→
J₁ = Y₁-π, J.-Y--O, J₁ = y, J₁ = y'
=
-
=
J₁ = -ksin sin
3- [14]
=
k-c
=
Sin
-

ページ41:

P = A₁ + Azz
= -C
q = det | Al = -c -k < o
0
=
p²-49
20
1. TA damping (C>O):
11. 2 damping (C-0).
For Po (0, 0), (± 2 TV, 0), (± 41, 0)
=
Ali + Azz
22 = -C
q = det 1 Al = -C+k
(Saddle point)
undamped cck, 920, oco Spiral
✓
overdamped cak, qco, 070 Saddle point
1. A damping (CO):
i
IV. 2 damping (C-O) =

ページ42:

power series
00
Σ am (x − x₁) m = a + a₁ (x − xo) + a₂(x − xo)² + ...
m=0
-
Set y
00
·for x=0.
= Σ Am X" = ao + A₁ X + A₂ X² + ...
m=0
Taylor expansion
f(x) = f(0) + + f (0) x + z = f ( 9 ) x² + ..
1
1-x
= x" =1+
n-0
R = 1
ex =
00 x"
n= n!
x
= 1+
+
sin x = 2(-1)"
+
1! 2! 3!
+2+1
R = 00
x³
x7
=x-
-0
(2n+1)!
+
3! 5!
-
+
R = 00
7!
COS X =
00
2(-1)"
n=0
(2n)!
x2n
= 1
-
00
x2n+1
tan 'x=(-1)"
n-0
00
In(1 + x) = ∑(-1)^-1 =x-
2n+1
x"
n
+
2!
4!
6!
r3
x³
x7
+
5
7
x² x
x 3
x4
+
-
+...
2
3
4
00
k(k-1)
k(k-1)(k-2)
=
x"=1+ kx+
+
R = 1
I I
-0 n
2!
3!
R = 00
R = 1
R = 1

ページ43:

radius of
convergence
00
Σ Am X
m= 0
m
R.
=
R
lim amti
moos
am
|
lim
m->ex
mam

ページ44:

linear ODE with x coefficients.
y" + plx y² + q(x)y = 0
Step 1. check
1. check convergence
Step 2.
00
m
2. set y = Σ am X, X.-0
M= 0
8
y' = Em Am X
y"
==
m = |
00
m-1
Σ m (m-1) am X
m=2
Step 3. sub y, y', y" into ODE
1x
Step 4. find am, x 11 xp -k
Step 5 y G₁y + Gry
t
z
m-2

ページ45:

Ex. 1 y" +y=0
Step 2
2.
m
X"
set y = Σ am X
M= 0.
8
'y' = Em Am X'
J₁
=
m = 1.
00
m-I
Σm (m-1) am X
m=2
Step 3. sub y, y', y" into ODE
0
m-2
00
m
m-2
I m (n-1) a = X" ² + ₤ an X " -0
m=2
Step 4. find am
Am =
m=0
,
讓X的次戶一樣
Set
5=3-2
S=m
(m=5+2)
0
Σ X.
(S+2) (S+1) Astz x² + Σ A≤ X² = 0
(S+2) (S+1) as+z+ As = 0
+2
[
05-0
-1
recurrence relation
05+2=
(S+3) (S+1)
as
Lot

ページ46:

|
S=0
5=2
A ₂ = = a.
2x1
-1
A4 = 4x3 A₂ =
a3 =
S=1
5=3
a 5
=
-1
a₁
3x2
5x4
03
=
a
Step 5. y = C₁y₁ + Cayz
00
Set As=m, y = Σ amx"
M=0
y = a. ( x + 1 ) + a₁ (1 x x T x²₁₁)
2!
ao COSXA ai sinh

ページ47:

Ex. 2 (1-x³) y" -2xy + 2y = 0
F&B" y " - x²y" - 2xy + zy = 0
00
= Σ
Step 2 set y- amx"
m= 0.
J'- Zmax ""
=
m = 1.
00
I m (m·1) am X
m=2
Step 3. sub y, y', y" into ODE
m-2
m
00
m-2
m
0
m
+25 am X =
I m (m-1) am X" - 2 Σ m Am X"
-----
I m (m-1) am X"
= 0.
m=2
m=0
1
m = 0
Step 4. find am, x 11 xp -k
Set
4-3-2
(m = 4+2)
Sem
·S=m
0
0
5=0
∞
+
(-x-xx-
5=0
5=0

ページ48:

00
Σ (sta) (S+1) a st₂ x²- = S (S-1) asx³-2 = sas x² + 2 as X² = 0
5=0
5=0
5=0
5=0
2
Za₂ + 2 a 0 = 0
A₂ = - A₁
"
S=1
3:203
-
|
2a + 2a₁ = 0
03=0
,
5=2
A4
5=0
4·3 a 4-2 Az - 4a₂ +2α₂ = 0
(Sto) (S+1) a st₂+ (-5 (5-1)-25 + 2) as=0
a5+2
5+2
(5+2)(5-1)
as
(S+2) (S+1)
5-2
4-1
A 4.
02
=
=
4-3
5=3
as
0
6.3
5=4
ab
a4
=
6.5
=
t
Step 5. y = C₁y₁ + Czyz
y = a x + a₁ (1 - x² - — x²- —≤ x²· ·)

ページ49:

Legendre's equation
n
(1-x') y" -axy² + (n+1)=0
00
Step 2 set y = Σ Am X
M= 0
00
y' = Em Am X'
"y"
m = 1
1-11
= I m (m-1) am Xm-2
m = 2
Step 3. sub y, y', y" into ODE
m-2
0
m-1
m
(1-x) m (m-1) am X - 2x Σ m Am X + h (n+1) Σ am X = 0
m=2
m-200
m = 1
m
m=0
00
m
I m (m-1) am X "² I m (m-1) am X" - z 2 m Am X" + k z Am X" = 0
m = 2
m=2
m = 1
Step 4. find am, X 1 XP - k
·Set S-m-2
(m = 5+²)
00
S= m
m= 0
S=m
00
= (S₁=) (SH) a₁ x² - 2 S (S-1) a₁ x² - 2 ², sa, x² + k = as X ²= 0
5=0
= 1
S=0

ページ50:

— (STO) (SHI) a₁ x² - 2 ≤ (S-1) a, X³ - 2 = sa, x² + k = a, X² = 0
5=2
S=1
·S=0
5=0
5=0
a
2α₂ + kao = 0
5=1
5-2
>
az
=
·n (nti)
a.
2
0 A3 = ~(n-1) (n+2)
3⋅ 2 a₂-za, - ka₁ = 0,
3
+
31
-a₁
·4.3 A4 -2a24a
+
ka₂ = 0;
A4 =
(n-2)(n+1)(n+3)
4:
+ α, (x
y = a₁ ( 1- nin+) x² +
0
3
(n-2) n (n+1) (n+3)
4!
4
x"....)
(n+2)(n-1)
(n-3) (n-1) (n+2) (n+4)
-
X +
3!
5!
(x³...)
=aoy + a₁yz
Legendre functions (Y, & Y=)
Y₁ = 1 + 2 | (-1)"
m=1
00
Y₂ = x + Σ
(2m)!
m
(-1)m
•·m=||(2m+1)!
2m
rm
1/17 (n+ (-1)* (k-1)) x * )
TV
k =
The ( n + (-1) k) x²+1]
k=1

ページ51:

lim
·Am+1.
Am
= lim
=
m700
lim
m78
2 (m+1),
TV
1=1
((1-1), (1) +U) 12 (11)
·⋅(-1)
(2m+2)!
(一)m
2m
(1-UP) 1" (n+ (-1) (1-1))
=
| =
(1+12+1) (12-11)
(2m+²) (2m+1)
|
Legendre polynomials
Yn (x)
=
M
(1-)
m=0
m
(24-2m)!
2"m! (n-m)! (n-am)
12-11
For
even
n-> M = 0
» M = n
2
= ²d 1 = 42=U
(1-4) = | <= |=|12-u
odd
n = 3; M =) P3 =
(xε - x5) F