233
4-6 Variation of Parameters
The method can solve the particular solution for any linear DE
( ) ( 1)1 1 0( )n nn na x y x a x y x a x y x a x y g x
(1) May not have constant coefficients (2) g(x) may not be of the special forms
4-6-1 方法的限制
2344-6-2 Case of the 2nd order linear DE
2 1 0( ) ( )a x y x a x y x a x y g x
Suppose that the solution of the associated homogeneous equation is
1 1 2 2( ) ( )c y x c y x
Then the particular solution is assumed as:
1 0( ) ( ) 0na x y x a x y x a x y associated homogeneous equation:
1 1 2 2( ) ( ) ( ) ( )py u x y x u x y x
(方法的基本精神)
2351 1 2 2( ) ( ) ( ) ( )py u x y x u x y x
1 1 1 1 2 2 2 2py u y u y u y u y
1 1 1 1 1 1 2 2 2 2 2 22 2py u y u y u y u y u y u y
( ) ( )y x P x y x Q x y f x 代入
012 2 2
( )( ) ( ), ,( ) ( ) ( )
a xa x g xP x Q x f xa x a x a x
1 2 1 1
1 1 2 2 2 2 1 1
1 1 1 2 2
2 2
2
2 2p p p y Py Qy y Py Qyy P x y Q x y u u y u
u y y u u y P y u y u
zero zero
代入原式後,總是可以簡化
236 ,p p py P x y Q x y f x
1 1 2 2 1 1 2 2 1 1 2 2d y u y u P y u y u y u y u f xdx
簡化
1 1 2 2py u y u y
進一步簡化:
假設 1 1 2 2 0y u y u
1 1 2 2y u y u f x 聯立方程式
1 1 2 2
1 1 2 2
0y u y uy u y u f x
1 1 1 1 2 2 2 2 1 1 2 22 2p p py P x y Q x y y u u y y u u y P y u y u
代入
237
1 1 2 2
1 1 2 2
0y u y uy u y u f x
211
y f xWuW W
122
y f xWuW W
1 1u u x dx
2 2u u x dx
where
| |: determinant
1 2
1 2
y yW
y y
2
12
0( )
yW
f x y
1
21
0( )
yW
y f x
1 1 2 2py x u x y x u x y x
可以和 1st order case (page 62) 相比較
238Determinant:
a bad bc
c d
a b cd e f aei bfg cdh afh bdi cegg h i
Matrix Inverse
If 1 12 2
x ya bx yc d
then
11 1
2 2
x ya bx yc d
1
2
1 yd bab bc yc a
1 2
1 2
1 dy byab bc cy ay
11
2
a bc dxy by d
21
2
a bc dxa yb y
2394-6-3 Process for the 2nd Order CaseStep 2-1 變成 standard form
( ) ( )y x P x y x Q x y f x
Step 2-2 1 21 2
y yW
y y
2
12
0( )
yW
f x y
1
21
0( )
yW
y f x
11
WuW
22WuW
Step 2-3
Step 2-4 1 1u u x dx 2 2u u x dx
1 1 2 2py x u x y x u x y x Step 2-5
240
Example 1 (text page 162)24 4 ( 1) xy y y x e
4 4 0 :y y y Step 1: solution of 2 2
1 2x x
cy c e c xe
Step 2-2:
2 24
2 2 22 2
x xx
x x x
e xeW e
e xe e
1 1 22 2
1 22 , ,x x
p y e yy u y xu ey
24
1 2 2 2
0( 1)
( 1) 2
xx
x x x
xeW x xe
x e xe e
2
42 2 2
0( 1)
2 ( 1)
xx
x x
eW x e
e x e
211
Wu x xW
22 1Wu xW
Step 2-3:
4-6-4 Examples
241Step 2-4: 2 3 21 1 11 1( ) 3 2u u dx x x dx x x c
2
2 2 21( 1) 2u u dx x dx x x c
3 2 2 32 2 221 1 1 1 1( ) ( ) ( )3 2 2 6 2p
x x xy e xex x x x x ex
3 222 21 2
1 1( )6 2x x xy c e c xe ex x Step 3:
Step 2-5:
242Example 2 (text page 163) 4 36 csc3y y x
( ) csc3 / 4f x x
4 36 0 :y y Step 1: solution of 1 2cos3 sin3cy c x c x
Step 2-1: standard form: 9 csc3 / 4y y x
cos3 sin33
3sin3 3cos3x x
Wx x
2
cos3 0cos31
1 4 sin3sin3 csc34
xxW xx x
1
0 sin31/ 41 csc3 3cos34
xW
x x
11
112
WuW
22cos31
12 sin3W xu xW
1 cos312 sin 3
xdxx
1 12xu
21 ln sin336u x
注意 算法
Step 2-2:
Step 2-3:
Step 2-4:
(未完待續)
243
Note: 課本 Interval (0, /6) 應該改為(0, /3)
Step 2-5: 1cos3 sin3 ln sin312 36pxy x x x
Step 3: 1 2 1cos3 sin3 cos3 sin3 ln sin312 36c pxy y y c x c x x x x
244Example 3 (text page 164)
( ) 1/f x x
Note: 沒有 analytic 的解
所以直接表示成 (複習 page 48)
xe dxx
0
tx
x
e dtt
1/y y x
1 2x x
cy c e c e
2x x
x x
e eW
e e
2454-6-5 Case of the Higher Order Linear DE
( ) ( 1)1 1 0( )n nn na x y x a x y x a x y x a x y g x
Solution of the associated homogeneous equation:
1 1 2 2 3 3( ) ( ) ( ) ( )c n ny c y x c y x c y x c y x
The particular solution is assumed as:
1 1 2 2 3 3( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )p n ny u x y x u x y x u x y x u x y x
( ) kkWu xW
( ) ( )k ku x u x dx
246
( ) kkWu xW
1 2 3
1 2 3
1 2 3
( 1) ( 1) ( 1) ( 1)1 2 3
n
n
n
n n n nn
y y y yy y y y
W y y y y
y y y y
1 2 1 1
1 2 1 1
( 2) ( 2) ( 2) ( 2) ( 2)1 2 1 1( 1) ( 1) ( 1) ( 1) ( 1)1 2 1 1
00
0( )
k k n
k k n
kn n n n n
k k nn n n n n
k k n
y y y y yy y y y y
Wy y y y yy y y f x y y
/ nf x g x a x
24700
0( )f x
Wk: replace the kth column of W by
For example, when n = 3,
2 3
1 2 3
2 3
00( )
y yW y y
f x y y
1 3
2 1 3
1 3
00( )
y yW y y
y f x y
1 2
3 1 2
1 2
00( )
y yW y y
y y f x
n
g xf x
a x
2484-6-6 Process of the Higher Order Case
Step 2-1 變成 standard form
Step 2-2 Calculate W, W1, W2, …., Wn (see page 247)
11
WuW
22WuW
Step 2-3 ………
Step 2-4 ……. 1 1u u x dx 2 2u u x dx
1 1 2 2p n ny x u x y x u x y x u x y x Step 2-5
( ) ( 1)1 1 0( )n nnn n n n
a x a x a x g xy x y x y x y
a x a x a x a x
nn
WuW
n nu u x dx
249Exercise 26 4 sec2y y x
Complementary function: 1 2 3cos 2 sin 2cy c c x c x
1 cos2 sin 20 2sin 2 2cos2 80 4cos2 4sin 2
x xW x x
x x
1
0 cos2 sin 20 2sin 2 2cos2 2sec2
sec2 4cos 2 4sin 2
x xW x x x
x x x
2
1 0 sin 20 0 2cos 2 20 sec2 4sin 2
xW x
x x
3
1 cos2 00 2sin 2 0 2 tan 20 4cos 2 sec2
xW x x
x x
2501
1sec2
4W xuW
221
4WuW
33tan 2
4W xuW
11 ln sec2 tan 28
u x x 2 4xu 3
1 ln cos 28
u x
1 2 3cos2 sin 2
1 1ln sec2 tan 2 cos 2 ln cos 2 sin 28 4 8
y x c c x c xxx x x x x
for -/4 < x < /4
Note: -/4 , /4 are singular points
251
(1)養成先解 associated homogeneous equation 的習慣
(2) 記熟幾個重要公式
(3) 這裡 | | 指的是 determinant
(4) 算出 u1(x) 和 u2(x) 後別忘了作積分
(5) f(x) = g(x)/an(x) (和 1st order 的情形一樣,使用 standard form)
(6) 計算 u1'(x) 和 u2'(x) 的積分時,+ c 可忽略
因為我們的目的是算particular solution ypyp 是任何一個能滿足原式的解
(7) 這方法解的範圍,不包含 an(x) = 0 的地方
4-6-7 本節需注意的地方
特別要小心
252
4-7 Cauchy-Euler Equation
( ) 1 ( 1)1 1 0( )n n n nn na x y x a x y x a xy x a y g x
not constant coefficients
but the coefficients of y(k)(x) have the form of
ak is some constant
kka x
associated homogeneous equation
particular solution
( ) 1 ( 1)11 0( ) 0
n n n nn na x y x a x y x
a xy x a y
k
4-7-1 解法限制條件
253
Guess the solution as y(x) = xm , then
Associated homogeneous equation of the Cauchy-Euler equation
( ) 1 ( 1)1 1 0( ) 0n n n nn na x y x a x y x a xy x a y
1 11
2 22
11
0
( 1)( 2) 1
( 1)( 2) 2
( 1)( 2) 3
0
n m nn
n m nn
n m nn
m
m
a x m m m m n x
a x m m m m n x
a x m m m m n x
a xmx
a x
4-7-2 解法
254
1
2
1
0
( 1)( 2) 1( 1)( 2) 2( 1)( 2) 3
0
n
n
n
a m m m m na m m m m na m m m m n
a ma
auxiliary function
比較: 和 constant coefficient 時有何不同?
kkk
dxdx
規則:把 變成!
( )!m
m k
2554-7-3 For the 2nd Order Case
22 1 0( ) 0a x y x a xy x a y
auxiliary function: 2 1 01 0a m m a m a 22 1 2 0 0a m a a m a
22 1 1 2 2 01
2
42
a a a a a am
a
2
2 1 1 2 2 02
2
42
a a a a a am
a
roots
[Case 1]: m1 m2 and m1, m2 are real
two independent solution of the homogeneous part:1 2 and m mx x
1 21 2
m mcy c x c x
256[Case 2]: m1 = m2
Use the method of reduction of order1
1my x
1
21
1
( )
2 1 2 21
P x dxa dx
a xm
mey x y x dxy
ex dxx x
01 22 2
( ) 0,aay x y x ya x a x
12
aP xa x
Note 1: 原式
Note 2: 此時 2 11 222
a am ma
257
1 11
2 22
1 1 1
1 1 1
1 1 21
1 2 12 12
2
ln
2 2 2
1 n1 l
a a adx xa x a am m mm m m
a a aaaa mm a m
xe ex dx x dx x dxx x x
x x x
y x
dx xx xx dx
If y2(x) is a solution of a homogeneous DE
then c y2(x) is also a solution of the homogeneous DE
If we constrain that x > 0, then 12 lnmy x x
1 11 2 ln
m mcy c x c x x
258[Case 3]: m1 m2 and m1, m2 are the form of
1m j 2m j
two independent solution of the homogeneous part:
and j jx x
1 2j j
cy C x C x
( ) ln ln ln
cos( ln
( )
) sin( ln )
j j j x x j xlnx e e e
x x j x
x e
cos( ln ) sin( ln )jx x x j x 同理 1 2 1 2[( )cos ln ( )sin ln ]cy x C C x j C C x
1 2[ cos ln sin ln ]cy x c x c x
259Example 1 (text page 167)
2 2 ( ) 4 0x y x xy x y
Example 2 (text page 168) 24 8 ( ) 0x y x xy x y
260
Example 3 (text page 169)
24 17 0x y x y 1 1y 11 2y
2614-7-4 For the Higher Order Case
auxiliary function
roots
solution of the nth order associated homogeneous equation
Step 1-1
n independent solutions Step 1-2
Step 1-3
Process:
262
(1) 若 auxiliary function 在 m0 的地方只有一個根
是 associated homogeneous equation 的其中一個解0mx
(2) 若 auxiliary function 在 m0 的地方有 k 個重根
皆為 associated homogeneous equation 的解
0 0 0 02 1ln (ln ) (ln, , , ),m m m m kx x x x x x x
263
(3) 若 auxiliary function 在 + j和 − j的地方各有一個根(未出現重根)
是 associated homogeneous equation 的其中二個解
,cos ln sin lnx x x x
(4) 若 auxiliary function 在 + j和 − j的地方皆有 k 個重根
是 associated homogeneous equation 的其中2k 個解
2
1
2
1
, , , ,cos ln cos ln ln cos ln (ln )
cos ln (ln )
sin ln sin ln ln sin ln (ln )
sin ln (ln
, , , ,
)
k
k
x x x x x x x x
x x x
x x x x x x x x
x x x
264Example 4 (text page 169)
3 25 7 ( ) 8 0x y x x y x xy x y
22 4 0m m
1 2 5 1 7 8 0m m m m m m auxiliary function
3 2 23 2 5 5 7 8 0m m m m m m 3 22 4 8 0m m m
2654-7-5 Nonhomogeneous Case
To solve the nonhomogeneous Cauchy-Euler equation:
Method 1: (See Example 5)
(1) Find the complementary function (general solutions of the associated homogeneous equation) from the rules on pages 255-258, 262-263.
(2) Use the method in Sec.4-6 (Variation of Parameters) to find the particular solution.
(3) Solution = complementary function + particular solution
Method 2: See Example 6,很重要
Set x = et, t = ln x
266Example 5 (text page 169, illustration for method 1)
2 43 ( ) 3 2 xx y x xy x y x e
auxiliary function 1 3 3 0m m m
31 2cy c x c x
2 4 3 0m m
2 3m 1 1m
Step 1 solution of the associated homogeneous equation
Step 2-2 Particular solution 3
32
21 3x x
W xx
35
1 22
02
2 3xxxW
xe
xex
211
xWu x eW
23
2
02
1 2x
xx ex
W x e
22
xWu eW
Step 2-3
267
2 2xu u dx e
21 1 2 2
x x xu u dx x e xe e Step 2-4
21 1 2 2 2 2
x xpy u y u y x e xe
Step 3 3 21 2 2 2
x xy c x c x x e xe
Step 2-5
268
1dy dt dy dydx dx dt x dt
2
2
2 2
2 2 2 2
1 1
1 1 1 1
d y d dy dt d dy d dydx dx dx dx dt dx x dt x dt
d y d dy d y dyx dt x dt x dt x dt dt
Example 6 (text page 170, illustration for method 2)
2 ( ) lnx y x xy x y x
Set x = et, t = ln x
(chain rule)
Therefore, the original equation is changed into
2
2 2 ( ) ( )d dy t y t y t tdtdt
269
2
2 2 ( ) ( )d dy t y t y t tdtdt
1 2( ) 2t ty t c e c te t
1 2( ) ln ln 2y x c x c x x x (別忘了 t = ln x 要代回來)
Note 2: 簡化計算的小技巧: 使用Cauchy-Euler equation 的 auxiliary function
1 1k
kt t tk
d yx D k D D ydx
Note 1: 以此類推
means tdDdt
270
(1) 本節公式記憶的方法:把 Section 4-3 的 ex 改成 x,x 改成 ln(x)把 auxiliary function 的 mn 改成
(2) 如何解 particular solution? Variation of Parameters 的方法
(3) 解的範圍將不包括 x = 0 的地方 (Why?)
4-7-6 本節要注意的地方
( 1)( 2) 1m m m m n
271還有很多 linear DE 沒有辦法解,怎麼辦
(1) numerical approach (Section 4-9-3)
(2) using special function (Chap. 6)
(3) Laplace transform and Fourier transform (Chaps. 7, 11, 14)
(4) 查表 (table lookup)
272
(1) 即使用了 Section 4-7 的方法,大部分的 DE還是沒有辦法解
(2) 所幸,自然界真的有不少的例子是 linear DE
甚至是 constant coefficient linear DE
273
Chapter 5 Modeling with Higher OrderDifferential Equations
自然界,有不少的系統可以用 linear DE 來表示
其中有不少的系統可以進一步簡化成 linear DE with constant coefficients
Chapter 4 的應用題
274
5-1 Linear Models: Initial Value Problem
位置:x, 速度: 加速度d xdt2
2d xdt
F ma2
2d xF mdt
F v ma
v: 速度, v: 磨擦力
2
2dx d xF mdt dt
275
彈力 F
Figure 5.1.1 Spring/mass system
5-1-1 ~ 5-1-3 Spring / Mass Systems
2762
2d xm Fdt
2
2d xm kxdt
2
2 0d xm kxdt
Solution: 1 2cos sinx t c t c t km
Figure 5.1.4
277
彈力 F摩擦力
dxdt
Figure 5.1.5
278
2
2d x dxm kxdtdt
2
2d x dxm Fdtdt
2
2 0d x dxm kxdtdt
解有成三種情形
(1)
(2)
(3)
2 4 0mk 2 4 0mk 2 4 0mk
1 21 2
n t n tx c e c e 2
1 24, 2
mkn n m
1 2nt ntx c e c te / 2n m
1 2cos sinatx e c bt c bt
/ 2a m 24 / 2b mk m
279需要注意的概念
(1) 名詞一
( ) ( 1)1 1 0n nn na t y t a t y t a t y t a t y t g t
g t 被稱作 input 或 deriving function 或 forcing function y t 被稱作 output 或 response
280(2) 名詞二
對一個 2nd order linear DE with constant coefficients 2 1 0 0a y x a y x a y x
auxiliary function 22 1 0 0a m a m a
當 時,稱作 overdamped21 2 04 0a a a
當 時,稱作 critical damped21 2 04 0a a a
當 時,稱作 underdamped21 2 04 0a a a
當 , a1 = 0 時,稱作 undamped2 04 0a a
281(3) 當中
a1 的值將影藉衰減速度
當 a2 , a1 , a0 的值皆為正, a1/ a2 的值越大,衰減的進度越快
2 1 0a y x a y x a y x g x
1 2cos sinxcy e c x c x 21 2 04 0a a a When1 2/ 2 ,a a
When 21 2 04 0a a a 1 21 2m x m x
cy c e c e 2
1 1 2 01
2
42
a a a am
a
2
1 1 2 02
2
42
a a a am
a
282
2
2q dq d qR L E tC dt dt
5-1-4 RLC circuit
inductance 的電壓 diLdt
capacitor 的電壓 qC
resistor 的電壓 Ri
q diRi L E tC dt using dqi
dt
一定可以解
283
2
2q dq d qR L E tC dt dt
2 1/ 0Lm Rm C auxiliary function
roots: 2
14 /
2R R L Cm
L
2
24 /
2R R L Cm
L
Case 1: R2 4L/C > 0
(m1 m2, m1, m2 are real)
(overdamped)
1 21 2m t m tcq t c e c e
註:由於 R, L, C 的值都是正的, 必定可以滿足
所以 m1, m2 都是負的
2 4 /R L C R
0cq t when t
Complementary function:
284Particular solution (1) E(t) 有的時候可用"guess” 的方法來解
(2) E(t) 用 variation of parameters 的方法一定解得出來(但較耗時)
1 2
1 2
1 2
( )2 1
1 2
m t m tm m t
m t m t
e eW m m e
m e m e
2
212
??0 m t
m t
eW
m e
1
121
?0?
m t
m t
eW
m e
2
1 2
11 ( )
2 1
( ) /( )
m t
m m tW e E t LuW m m e
1
12 1
( )( )
m tE t e dtu
L m m
1
1 2
22 ( )
2 1
( ) /( )
m t
m m tW e E t LuW m m e
2
22 1
( )( )
m te E t dtu
L m m
285
1 21 2
2 1 2 1
( ) ( )( ) ( )
m t m tm t m t
p
e E t e dt e E t e dtq t
L m m L m m
1 1 2 2
1 21 2
2 1 2 1
( ) ( )( ) ( )
m t m t m t m tm t m t e E t e dt e E t e dtq t c e c e
L m m L m m
Specially, when E(t) = E0 where E0 is some constant
1 21 2
0 0 0
2 1 2 1 2 1 1 2
0
10
2
1 1( ) ( ) ( )
m t m tm m t
p
tE e e dt E e e dt EL m m L m m L m m m m
E
q t
E CLm m
(m1m2 = 1/LC)
1 21 2 0m t m tq t c e c e E C
286Case 2: R2 4L/C = 0
(m1 = m2 = R/2L)
/ 2 / 21 2Rt L Rt Lcq t c e c t e
(critically damped)
0cq t when t
Particular solution
/ 2
/ 2 / 2( ) ( )Rt L
Rt L Rt Lp
eq t t E t e dt E t te dtL
When E(t) = E0 ,
0pq t E C
/ 2
/ 2 / 2 / 2 / 21 2 ( ) ( )
Rt LRt L R t L Rt L Rt Leq t c e c t e t E t e dt E t te dt
L
287Case 3: R2 4L/C < 0
1m j
24 /,2 2
R L C RL L
2m j
(underdamped)
1 2cos sintcq t e c t c t
Particular solution
0cq t
sin ( ) cos cos ( ) sint
t tp
eq t t E t e tdt t E t e tdtL
1 2cos sin
sin ( ) cos cos ( ) sin
t
tt t
q t e c t c t
e t E t e tdt t E t e tdtL
General solution
when t
288When E(t) = E0 where E0 is some constant
1 2 0cos sintq t e c t c t E C
When R = 0 , then = 0
1 21cos sin sin ( )cos cos ( )sinq t c t c t t E t tdt t E t tdt
L
When R = 0 , E(t) = E0 1 2 0cos sinq t c t c t E C
1 2( ) cos sindi t q t d t d tdt 1 2d c 2 1d c
289
以 DE 的觀點來解釋 RLC 電路的問題
(1) R2 < 4L/C 產生弦波(2) R 越小,弦波衰減得越慢
290
2
2d q dq qL R E tdt dt C
E(t) = 1, L = 0.25, C = 0.01
2 21 2 04 100a a a R
例子
291
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.005
0.01
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.005
0.01
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.005
0.01
R = 100
R = 25
R = 10
292
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.005
0.01
0.015
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.005
0.01
0.015
0.02
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.005
0.01
0.015
0.02
R = 5
R = 1.5
R = 0.2
2935-1-5 Express the Solutions by Other Forms
(1) Express the Solution by the Form of Amplitude and Phase
2 1 0 0a y x a y x a y x
當 時,solution 為21 2 04 0a a a 1 2cos sinxy e c x c x
sinxy Ae x
Solution 可改寫成1 2/ 2 ,a a
22 0 1 24 / 2a a a a
2 21 2A c c
1 11 2sin / cos /c A c A
294
sinxy Ae x
:xAe damped amplitude
:2
damped frequency
: phase angle
295
(2) Express the Solution by Hyperbolic Functions
2 1 0 0a y x a y x a y x 當 a1 = 0
且 a2 > 0, a0 < 0 (或 a0 > 0, a2 < 0)
1 11 2
m x m xy c e c e
3 1 4 1cosh sinhy c m x c m x
1 11cosh 2m x m xe em x
1 11sinh 2m x m xe em x
3 1 2c c c
4 1 2c c c
01
2
am a
2965-1-6 本節要注意的地方
(1) 將力學現象寫成 DE 時,要注意正負號 (根據力的方向)
(2) 注意 page 280 的四個專有名詞
(3) 注意 linear DE with constant coefficients 的解,有其他的寫法
(see pages 293 and 295)
297
(A) Linear DE Complementary Function 3 大解法
(3) Cauchy-Euler Equation (Section 4-7)
( ) 1 ( 1)1 1 0( ) 0n n n nn na x y x a x y x a xy x a y
1 1 0! ! ! 0( )! ( 1)! ( 1)!n n
m m ma a a x am n m n m
適用情形:
附錄八(續):Reviews for Higher Order DE(接續 page 230)
298(B) Linear DE Particular solution 3 大解法
(3) Variation of parameters (Section 4-6) 1 1 2 2p n ny u y u y u y
( ) kkWu xW
1 2 3
1 2 3
1 2 3
( 1) ( 1) ( 1) ( 1)1 2 3
n
n
n
n n n nn
y y y yy y y yy y y y
y y
W
y y
Wk : replace the kth column of W by00
0( )f x
n
g xf x
a x
適用情形:
299(4) For Cauchy-Euler Equation (Section 4-7)
可採用二種方法
(1) 先用
1 1 0! ! ! 0( )! ( 1)! ( 1)!n n
m m ma a a x am n m n m
再用 Variation of parameters 解 particular solution
解 complementary function
(2) Use the method on pages 268, 269
Set x = et, t = ln x
300Exercises for practicing
Section 4-6 4, 5, 8, 13, 14, 17, 18, 21, 28, 29, 34
Section 4-7 11, 17, 18, 20, 21, 24, 32, 34, 36, 37, 40, 42
Review 4 27, 28, 29, 30, 32, 42
Section 5-1 1, 11, 29, 43, 44, 49, 52, 56, 60
Review 5 12, 21, 22