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