245

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

246

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

247

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

: constant coefficient

kkk

dxdx

!

( )!m

m k

2484-7-3 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

249[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:

Note 2: 2 11 222

a am ma

250

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

251[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 1 2 1 2[( )cos ln ( )sin ln ]cy x C C x C C x

1 2[ cos ln sin ln ]cy x c x c x

252Example 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

253

Example 3 (text page 169)

24 17 0x y x y 1 1y 11 2y

2544-7-4 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:

255

(1) auxiliary function m0

associated homogeneous equation 0mx

(2) auxiliary function m0 k

associated homogeneous equation

0 0 0 02 1ln (ln ) (ln, , , ),m m m m kx x x x x x x

256

(3) auxiliary function + j j()

associated homogeneous equation

,cos ln sin lnx x x x

(4) auxiliary function + j j k

associated homogeneous equation 2k

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

257Example 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

2584-7-5 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 248-251, 255-256.

(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

Set x = et, t = ln x

259Example 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

260

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

261

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

262

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 ( t = ln x )

Note 2: nonhomogeneous Cauchy-Euler equation

1 1 1k

t t tk kd y D k D D ydx x

Note 1:

means tdDdt

263

(1) Section 4-3 ex xx ln(x) auxiliary function mn

(2) particular solution? Variation of Parameters

(3) x = 0 (Why?)

4-7-6

( 1)( 2) 1m m m m n

264 linear DE

(1) numerical approach (Section 4-9-3)

(2) using special function (Chap. 6)

(3) Laplace transform and Fourier transform (Chaps. 7, 11, 14)

(4) (table lookup)

265

(1) Section 4-7 DE

(2) linear DE

constant coefficient linear DE

266

Exercise for practice

Section 4-4 5, 6, 14, 17, 18, 24, 26, 32, 33, 39, 42

Section 4-5 2, 7, 8, 13, 18, 31, 45, 60, 62, 69, 70

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 2, 21, 22, 25, 27, 28, 29, 30, 32, 33, 34, 37, 42

267

Chapter 5 Modeling with Higher OrderDifferential Equations

linear DE

linear DE with constant coefficients

Chapter 4

268

5-1 Linear Models: Initial Value Problem

x, d xdt2

2d xdt

F ma2

2d xF mdt

F v ma

v: v:

2

2dx d xF mdt dt

2695-1-1 ~ 5-1-3 Spring / Mass Systems

F

2702

2d xm Fdt

2

2d xm kxdt

2

2 0d xm kxdt

Solution: 1 2cos sinx t c t c t km

271

dxdt

F

272

2

2d x dxm kxdtdt

2

2d x dxm Fdtdt

2

2 0d x dxm kxdtdt

2 4 0mk 2 4 0mk 2 4 0mk

273

(1)

( ) ( 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 input deriving function forcing function y t output response

274(2)

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

overdamped21 2 04 0a a a

critical damped21 2 04 0a a a

underdamped21 2 04 0a a a

, a1 = 0 undamped2 04 0a a

275(3)

a1

a2 , a1 , a0 a1/ a2

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

276

2

2q dq d qR L E tC dt dt

5-1-4 RLC circuit

inductance diLdt

capacitor qC

resistor Ri

q diRi L E tC dt using dqi

dt

277

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

R, L, C ,

m1, m2

2 4 /R L C R

0cq t when t

Complementary function:

278Particular solution (1) E(t) guess

(2) E(t) variation of parameters ()

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

279

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

280Case 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

281Case 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

282When 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

283

DE RLC

(1) R2 < 4L/C (2) R

284

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

285

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

286

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

2875-1-5 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

solution 21 2 04 0a a a 1 2cos sinxy e c x c x

sinxy Ae x

Solution 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

288

sinxy Ae x

:xAe damped amplitude

:2

damped frequency

: phase angle

289

(2) Express the Solution by Hyperbolic Functions

2 1 0 0a y x a y x a y x a1 = 0

a2 > 0, a0 < 0 ( 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

2905-1-6

(1) DE ()

(2) page 274

(3) linear DE with constant coefficients

(see pages 287 and 289)

291

5-2 Linear Models: Boundary-Value Problem

Section 5-2 Section 5-1

( Linear DE)

initial value problems boundary value problems

IVP boundary value problems solution

()

292Section 5-1

(1)

(2) (subsection 5-1-1~5-1-3)

(3) RLC Circuit (subsection 5-1-4)

Section 5-2 (1) (a) (subsection 5-2-1)

(b) (subsection 5-2-2)

(2) (subsection 5-2-3)

293

4-9 Solving Systems of Linear Equations by Elimination

2 dependent variables

Section 3-3

linear and constant coefficients

4-9-1

294

(Step 1) D n

(Step 2) ()

dependent variable DE

(Step 3) Sections 4-3, 4-4

dependent variables

(Step 4) unknowns

( page 298)

n

nddt

4-9-2

295

21 1 2 2 1 3( )di t

L R R i t R i t E tdt

31 2 2 1 3di t

R i t L R i t E tdt

L1 = L2 = 1, R1 = 4, R2 = 6, E(t) = 10

2 2 310 4 10di t

i t i tdt

32 34 4 10di t

i t i tdt

Figure 3.3.4 (See Pages 96, 97)

4-9-3

296

2 3( 10) 4 10D i i

2 34 ( 4) 10i D i .. ( 1).. ( 2)

(D + 4) ( 1) 4 ( 2)

2 2( 4)( 10) 16 ( 4)10 4 10D D i i D Note: ( 4)10 10 4 10 40dD dt

22( 14 24) 0D D i

2 14 24 0m m auxiliary function:

1 2m 2 12m roots

2 122 1 2t ti t c e c e

Step 1

Step 2-1

Step 3-1

2974 ( 1) (D+10) ( 2)

3 316 ( 4)( 10) 4 10 ( 10)10i D D i D 2

3( 14 24) 60D D i 2

3( 14 24) 60D D i

auxiliary function: 2 14 24 0m m

roots 1 2m 2 12m complementary function for i3,c(t) = 3 4

2 12t tc ce e

Particular solution: i3, p(t) = A24 60 , 5 / 2A A

2 123 3 4 5 / 2t ti t c e c e

Step 2-2

Step 3-2

298Step 4

2 122 1 2t ti t c e c e 2 123 3 4 5 / 2t ti t c e c e

( 1)

2 121 3 2 4(8 4 ) ( 2 4 ) 10 10

t tc c e c c e

1 3

2 4

8 4 02 4 0c cc c

3 1

4 2

2/ 2

c cc c

2 123 1 22 / 2 5/ 2t ti t c e c e

2 122 1 2t ti t c e c e

i2(t) i3(t)

299

2 122 1 2t ti t c e c e 2 123 1 22 / 2 5/ 2t ti t c e c e

( 2)

2 121 2(4 4 8) (4 6 2) 10 10

t tc e c e

c1 c2

300

Step 2-2 i2(t)

i2(t) (1)2 12

1 2 3( 10)( ) 4 10t tD c e c e i

2 2 12 21 1 2 32 10 12 10 4 10

t t t tc e c e c e e i 2 12

3 1 24 8 2 10t ti c e c e

2 123 1 22 / 2 5/ 2

t ti c e c e

( dependent variable i3(t) )

301

Example 1 (text page 185)

( 2) 0( 3) 2 0Dx D yD x y

Example on text page 184 3 02 0Dx y

x Dy

1 1 2

2 1 2

2 125 50

2 225 25

d x x xdtd x x xdt

Example 3 (text page 186)

302Example 2 (text page 185)

240

x x y tx x y

2 2( 4)( 1) 0D x D y tD x Dy

Step 1

Step 2-1 (2) D (1)

(1)(2)

2 2( 4)D x t

complementary function: 1 2cos 2 sin 2cx c t c t particular solution: 2px At Bt C

2 24 4 4 2At Bt C A t 1/ 4, 0, 1/8A B C 21 1

4 8px t

21 2 1 1cos 2 sin 2 4 8x c t c t t

Step 3-1

303Step 2-2 (1) (D+1) (2) (D4)

3 2 2( 4 ) ( 1) 2D D y D t t t

3 4 5cos2 sin 2cy c c t c t Step 3-2 complementary function:

particular solution 3 2

p At Bt Cy t

2py At Bt C

2 2 212 8 6 4 2At Bt A C t t 1/12, 1/ 4, 1/8A B C

3 21 1 112 4 8py t t t

3 23 4 5

1 1 1cos2 sin 2 12 4 8y c c t c t t t t

304Step 4 (2) (21)

1 2 5 2 1 4( 2 2 )cos 2 ( 2 2 )sin 2 0c c c t c c c t

5 2 1 4 2 11 1,2 2c c c c c c

3 23 2 1 2 1

1 1 1 1 1( )cos2 ( )sin 22 2 12 4 8y c c c t c c t t t t

21 2 1 1cos 2 sin 2 4 8x c t c t t

3054-9-4 Dependent Variables

Exercise 19 6d x ydtd y x zdtd z x ydt

6 000

Dx yx Dy zx y Dz

Steps 2, 3 x, y, z DE

(2) D + (3) 2( 1) (1 ) 0D x D y

(4) 6 + (1) (1D2)

(1(2 (3

(4

3( 7 6) 0D D x 3 7 6 0m m m = 1, 2, 3

2 31 2 3

t t tx c e c e c e

306(4) D (1) (1+D)

3( 7 6) 0D D y 2 34 5 6t t ty c e c e c e

(3) 6 + (1) ( 6) 6D x Dz (5

(1) D (2) 6 2( 6) 6 0D x z (6

(5) (D2 6) + (6) (D+6)3(6 42 36) 0D D z 2 37 8 9

t t tz c e c e c e

Step 4 c4, c5, c6, c7, c8, c9 c1, c2, c3

(1) 2 31 2 3t t tx c x c x c x

2 34 5 6

t t ty c x c x c x

4 1 5 2 6 36 , 6 2 , 6 3c c c c c c

2 31 2 3

1 1 16 3 2

t t ty c e c e c e

307

y z

(2) 2 3

1 2 3t t tx c e c e c e

2 31 2 3

1 1 16 3 2

t t ty c e c e c e 2 3

7 8 9t t tz c e c e c e

7 1 8 2 9 35 1 1, ,6 3 2c c c c c c

2 31 2 3

5 1 16 3 2

t t tz c e c e c e

308

(1) N dependent variables,

N DE solutions

(2) DE orders k1, k2, k3, .., kN order k1 + k2 + k3 + .. + kN DE

(3) N 1 unknowns

Higher order

309

A circuit that can be modeled by a 2nd order polynomial.

310

(1) Section 4.9 constant coefficients

(2) dependent variable

homogeneous

(3)

()

(4)

(5) Step 4 unknowns

(6)

4-9-5

311

4-10 Nonlinear Differential Equations

Method 1: Reduction of Order

Method 2: Taylor Series

Method 3: Numerical Approach

3124-10-1 Method 1: Reduction of Order

1st order DE

1st order DE

( Section 4-2 linear

)

The DE should have the form of

22, , 0d dF x y ydx dx 2

2, , 0d dF y y ydx dx

(Without the term x)(Without the term y)

or

Case 1, page 313 Case 2, page 315

313Case 1: The 2nd order DE has the form of

(Without the term y) 22, , 0d dF x y ydx dx

(Step 1) Set

DE ( u 1st order DE)

(Step 2) u ( Section 2 )

(Step 3) u y

du ydx

d udxu

, , 0dF x u udx

314Example 1 (text page 189)

22y x y

(Step 1) u y

22d u xudx

21u

x c

u

21 ?y dx

x c

(Step 2)

(Step 3)

315Case 2: The 2nd order DE has the form of

(Without the term x)

22, , 0d dF y y ydx dx

(Step 1) Set

(Chain rule)

DE

( u 1st order DE, independent variable y)

du ydx

2

2dyd du udx dx

d dy u udydx dy

, , 0dF y u u udy

u du udy

316(Step 2) u ( Section 2 )

, u y

(Step 3)

separable variable

1u F y

1dy F ydx 1

dy dxF y

317Example 2 (text page 190)

2yy y du ydx

(Step 1) Set

2dy u u udy

dyduu y 1ln lnu y c 1

cu y e

2u c y 12( )cc e

2dy c ydx 2

dy c dxy 2 3ln y c x c

24

c xy c e

32 cc xy e e

34( )

cc e

(Step 2)

(Step 3)

3184-10-2 Method 2: Taylor Series

(4)

2 3 40 0 0 001! 2! 3! 4!

y y y yy x y x x x x

2 30 0 00 0 0 0

(4)40

0

( ) ( ) ( )1! 2! 3!

( )4!

y x y x y xy x y x x x x x x x

y xx x

Step 1 (4)0 0 0 0 0, , , , ,y x y x y x y x y x

Step 2 Taylor series

319Example 3 (text page 190)2y x y y (0) 1y (0) 1y

2y x y y 20 0 ( 1) 1 2y 2( ) 1 2dy x y y y y ydx

0 4y

(4) 2(1 2 ) 2 2( )dy y y y y y y ydx (4) 0 8y

(5) 2( 2 2( ) ) 2 6dy y y y y y y y y ydx (5) 0 24y

:

2 3 4 52 1 113 3 5

y x x x x x x

Taylor series

320

(1) y(x) x0 analytic,

(x = x0 singular point)

(2) nth order DE y(x0) y'(x0) y''(x0) ..

y(n1)(x0)

(3) x0

(1) Taylor series

(2) |x x0|

321

x0 =0

3224-10-3 Method 3: Numerical Method

2

2 , ,d y f x y ydx

0 0y x y 0 0y x u

, ,y u

u f x y u

subject to 0 0 0 0,y x y u x u

Section 2-6 Eulers Method

1 1( )n n n n ny x y x x x y x

1 1( )n n n n ny x y x x x u x

11

1( ) , ( ), (( )

)n

n n n n n n

n n n nu xu x x x f x y x u xu x x x u x

323

Recursive

, ,y u

u f x y u

0 0 0 0,y x y u x u

Initial: n = 0 0 0 0 0,y x y u x u

1 1( )n n n n ny x y x x x u x

n = n + 1

1 1( ) , ( ), ( )n n n n n n nu x u x x x f x y x u x

324

( 1), , , , ,k

kk

d y f x y y y ydx

0 0,0y x y 0 0,1y x y 0 0,2y x y ( 1) 0 0, 1k ky x y

( 1)

1

1 2

2 3

2 1

1 1 2 1, , , , ,k k

k

k

k

y uu uu u

u uu f x y u

y

uy

u

y

0 0,0

1 0 0,1

2 0 0,2

1 0 0, 1k k

y x yu x yu x y

u x y

subject to

325Recursive

0 0,0 1 0 0,1 2 0 0,2

1 0 0, 1

Initial : , ,

0

,

, ,k k

y x y u x y u x y

u x ny

1 1 1 1 1 2 1, ( ), ( ), ( ), , ( )k n k n n n n n n n k nu x u x x x f x y x u x u x u x

1 1 1( )n n n n ny x y x x x u x

1 1 1 1 2( )n n n n nu x u x x x u x

2 1 2 1 3( )n n n n nu x u x x x u x

2 1 2 1 1( )k n k n n n k nu x u x x x u x

n = n + 1

326

( 1), , , , ,k

kk

d y f x y y y ydx

(1) ( singular point)

(2) k initial conditions

( 1), , , , , kf x y y y y ( 1), , , , , kf x y y y y

327

(1) Section 4.10

(2) Section 4.1 ( exercises 4.10 1, 2 )

(3) Method 1

(4) Method 1 u dy/dx

4-10-4

328

5-3 Nonlinear Models (text pages 222, 223)

(text pages 223, 224, 225)

(text pages 225, 226)

(text page 226)

(text pages 227, 228)

329

F = ma2

2( )d y tF m

dt

F y ()2

2 2( )

( )d y tmMk m

y t dt

M: m:

5-3-1

2

0 2 2( )

( )d y tmMF k m

y t dt

3305-3-2 dF mvdt

m: , v: , mv:

m x (), m = kx(t)20 N

(weight) = x(t)

In Example 4, chain weight = 1 N/m

(mass) = x(t)/9.8

331

0 0

2

22

02

0 2

, ( )

,

d dF mg v m m v F F mgdt dt

d d dF kgx t x t kx t kx t x tdt dt

d dkx t x t k x t kgx t Fdt dt

dt

F0: , k: , x(t) ()

g = 9.8 metres per s2 g = 32 feet per s2

332

2 22 ( ) 9.8 196

d x dxx xdtdt

Example 4 (text page 227)k = 1/9.8g = 9.8F0 = 20

333

(1) Cauchy-Euler equation

(2) Numerical Method

334

(1) DE

DE

(2) (

(3) tan =

(4)

5-3-3

335Reviews for Higher Order DE:(A) Linear DE Complementary Function 3

(1) Reduction of Order (Section 4-2)

(2) Auxiliary Function (Section 4-3)

2 1 21

P x dxey x y x dxy x

( ) ( 1)1 1 0( ) 0n nn na y x a y x a y x a y

11 1 0 0

n nn na m a m a m a

4 Cases (See pages 182, 183)

336(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 am n m n m

337(B) Linear DE Particular solution 3

(1) Guess (Section 4-4) ( page 194 ) yp should be a linear combination of g(x), g'(x),

g'' (x), g'''(x), g(4)(x), g(5)(x), .

(2) Annihilator (Section 4-5)

DE L[y(x)] = g(x)

x lnx

Annihilator: L1[ g(x)] = 0 Particular solution L1{L[y(x)]} = 0

( L[y(x)] = 0 ) c py y y

Annihilator Pages 213-215

338(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

339(4) For Cauchy-Euler Equation (Section 4-7)

(1)

1 1 0! ! ! 0( )! ( 1)! ( 1)!n n

m m ma a a am n m n m

Variation of parameters particular solution

complementary function

(2) Use the method on pages 261, 262

Set x = et, t = ln x

1 1k

kt t tk

d yx D k D D ydx

then

340(C) Combination of Linear DEs (Section 4-9)

Step 1: Dn

Steps 2, 3: dependent variable

DE dependent variable

Step 4: dependent variable ck

n

nddt

341(D) Nonlinear DE 3 (Section 4-10)

(1) Reduction of Order

(1-1)

Set

(1-2)

Set

22, , 0d dF x y ydx dx du ydx

, , 0dF x u udx

22, , 0d dF y y ydx dx du ydx

, , 0dF y u u udy

342(2) Taylor Series

(3) Numerical Method

( ) ( 1), , , , ,n ny F x y y y y

( ) ( 1), , , , ,n ny F x y y y y

( 1), , , , , nF x y y y y

( 1), , , , , nF x y y y y

2 30 0 00 0 0 0

(4)40

0

( ) ( ) ( )1! 2! 3!

( )4!

y x y x y xy x y x x x x x x x

y xx x

343Exercises for practicing

Section 4-9 5, 8, 10, 14, 17, 18, 20, 22, 23

Section 4-10 1, 4, 5, 8, 10, 12, 15, 16, 19, 21, 22, 23

Review 4 43, 44, 48, 50

Section 5-1 1, 11, 29, 43, 44, 49, 52, 56, 60

Section 5-3 14, 15, 16

Review 5 12, 21, 22

20122016, 2017