92 北科

工程數學 詳解

a−δ a a+δ

lxfax

=→

)(lim

陳立、林易、周成 編著

(a,l)

y

x

y=f(x)

δ

ε

l+ε

l−ε

l

92北科-1

92北科土木

The following is a 2-dimensional Laplace Equation problem. (1) Please derive the Laplace equation from the rectangular coordinate system

to the polar coordinate system.(10%) (2) Please solve the following boundary value problem in the polar coordinate

system (i.e., find the steady-state temperature ),( θru in the semicircular plate shown in Fig. 1). (15%)

【92北科土木】

【範圍】§16-1 【詳解】(1)卡氏座標之 Laplace方程式

為 022

2

2

=∂∂

+∂∂

yu

xu

範例 1

92北科-2

令⎩⎨⎧

==

θθ

sincos

ryrx

，即⎪⎩

⎪⎨⎧

=

+=−

xyyxr

1

22

tanθ

則

⎪⎪⎩

⎪⎪⎨

⎧

==+

=∂∂

==+

=∂∂

θθ

θθ

sinsin

coscos

22

22

rr

yxy

yr

rr

yxx

xr

同理

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−==+

=+

=∂∂

−=−

=+−

=+

−=

∂∂

rrr

yxx

xy

xy

rrr

yxy

xy

xy

x

θθθ

θθθ

coscos

)(1

1

sinsin

)(1

2222

2222

2

代入r

uru

xu

xr

ru

xu θ

θθθ

θsincos

∂∂

−∂∂

=∂∂

∂∂

+∂∂

∂∂

=∂∂

則x

xu

xr

rxu

xxu

xu

∂∂

∂∂∂

∂+

∂∂

∂∂∂

∂=

∂∂∂

∂=

∂∂ θ

θ22

= )sincos(sin)sincos(cosr

uru

rru

ru

rθ

θθ

θθθ

θθθ

∂∂

−∂∂

∂∂

−∂∂

−∂∂

∂∂

=θ

θθθ

θθθ∂∂

+∂∂

∂−

∂∂ u

rru

rru

2

2

2

22 cossin2sincos2cos

22

2

22 sinsinθ

θθ∂∂

+∂∂

+u

rru

r

又r

uru

yu

yr

ru

yu θ

θθθ

θcossin

∂∂

+∂∂

=∂∂

∂∂

+∂∂

∂∂

=∂∂

92北科-3

則y

yu

yr

ryu

yyu

yu

∂∂

∂∂∂

∂+

∂∂

∂∂∂

∂=

∂∂∂

∂=

∂∂ θ

θ22

= )cossin(cos)cossin(sinr

uru

rru

ru

rθ

θθ

θθθ

θθθ

∂∂

+∂∂

∂∂

+∂∂

+∂∂

∂∂

=θ

θθθ

θθθ∂∂

−∂∂

∂+

∂∂ u

rru

rru

2

2

2

22 cossin2sincos2sin

22

2

22 coscosθ

θθ∂∂

+∂∂

+u

rru

r

故 2

2

22

2

2

2

2

2 11θ∂∂

+∂∂

+∂∂

=∂∂

+∂∂ u

rru

rru

yu

xu

即極座標之 Laplace方程式為 011 22

22

2

=∂∂

+∂∂

+∂∂

θu

rru

rru

(2) 由分離變數法

令 )()(),( θθ Θ= rRru 代入 P.D.E

得 011 2 =Θ ′′+Θ′+Θ′′ RrR

rR ⇒ λ=

ΘΘ ′′

−=′+′′

Rr

Rr

R

2

1

1

⇒ ⎪⎩

⎪⎨⎧

=Θ=Θ=Θ+Θ′′

=−′+′′

LLLLLLLLL

LLLLLLLLLLLLL

0)()0(；0

01 2

πλ

λ Rr

Rr

R

由○2 2n=λ ( L,3,2,1=n ) θθ nsin)( =Θ

由○1 022 =−′+′′ RnRrRr (Cauchy等維方程式) ⇒ nn BrArrR += −)(

由 B.C. =)0(R 有界 ⇒ 0=A ∴ nBrrR =)(

○1

○2

92北科-4

由重疊原理

令 ∑∞

=

=1

sin),(n

nn nrBru θθ

B.C. ∑∞

=

==1

0 sin),(n

nn ncBucu θθ

⇒ ∫=π

θθπ 0 0

sin2 dnucB nn⎪⎩

⎪⎨⎧

=

==−=L

L

,6,4,20

,5,3,1n4)cos1(20

0

nnu

nnu

πππ

⇒ =nB⎪⎩

⎪⎨⎧

=

=

L

L

,6,4,20

,5,3,14 0

n

ncn

unπ

將 nB 代入 ∑∞

=

=1

sin),(n

nn nrBru θθ ∑

∞

=

=L,5,3,1

0 sin14n

nn nrnc

u θπ

【另解】(1) θeru

re

ruu r

vv

∂∂

+∂∂

=∇1

⎭⎬⎫

⎩⎨⎧

∂∂

∂∂

+∂∂

∂∂

=∇⋅∇=∇ )1()(12θθu

rrur

rruu

= 22

22

2 11θ∂∂

+∂∂

+∂∂ u

rru

rru

故極座標之 Laplace方程式為 011 22

22

2

=∂∂

+∂∂

+∂∂

θu

rru

rru

Consider the spring/mass system of Fig. 2. Let 021 == xx at the equilibrium position, where the weight are at rest. Choose the direction to the right as positive and suppose the weights are at positions )(1 tx and )(2 tx at time t. The equations of motion of the system are as shown as follows.

範例 2

92北科-5

)()()()(

22321222

12212111

tfxkkxkxmtfxkxkkxm

++−=+++−=

&&

&&

These equations assume that damping is negligible but allow for forcing functions acting on each mass. Suppose 121 == mm and 431 == kk while

5.22 =k ; and suppose 0)(2 =tf and )]3(1[2)(1 −−= tHtf . Please use the Laplace transform to solve this system.

(25%)【92 北科土木】

【範圍】§8-3

【詳解】 [ ]⎩⎨⎧

−−+−=+−=

)2()3(125.65.2)1(5.25.6

212

211

LLLL&&

LLLLLLLLLLL&&

tHxxxxxx

取 Laplace變換

⎪⎩

⎪⎨⎧

−=Χ++Χ−

=Χ−Χ+− )1(2)()5.6()(5.2

0)(5.2)()5.6(3

22

1

212

ses

sss

sss

由 Cramer rule

5.6)1(25.20

)(5.65.2

5.25.62312

2

+−

−=Χ

+−−+

− ses

ss

ss

)1(5)()3613( 312 se

ssss −−=Χ++⇒

( )ss

essssss

es 33

1 191

91

41

411

365

)9)(4()1(5)( −

−

−⎟⎠⎞

⎜⎝⎛

++

+−=

++−

=Χ⇒

92北科-6

)3(365

91

41

365)( 941 −−+−=⇒

−− tHeetx tt

)3(91)3(

41 )3(9)3(4 −−−+ −−−− tHetHe tt

由 式 )3(9)3(4945.6)(5.2 )3(9)3(494112 −−−++−=+=

−−−−−− tHetHeeexxtx tttt&&

)3(36

5.3295.6

45.6

365.32 94 −−+−+ −− tHee tt

)3(95.6)3(

45.6 )3(9)3(4 −−−+ −−−− tHetHe tt

)3(36

5.329

5.874

5.2236

5.32 94 −−+−= −− tHee tt

)3(9

5.87)3(4

5.22 )3(9)3(4 −−−+ −−−− tHetHe tt

t9t42 e9

35e49

3613tx −− +−=⇒ )(

)3(935)3(

49)3(

3613 )3(9)3(4 −−−+−− −−−− tHetHetH tt

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎥

⎦

⎤⎢⎣

⎡=+= te

tXtGtAXdt

tdX34

8)(

5133

)()()( , where )(tX is a 2×1 matrix,

and A is 2×2 matrix. (a) Find the solution of the above equation. (15%) (b) Find the matrix A10. (10%) 【92 北科土木】

【範圍】§24-4

範例 3

92北科-7

【詳解】 051

33)det( =

−−

=Ι−λ

λλA 6 2,=⇒ λ

⎥⎦

⎤⎢⎣

⎡−

=⇒=⎥⎦

⎤⎢⎣

⎡=

13

03131

:2 111 kuuλ

⎥⎦

⎤⎢⎣

⎡=⇒=⎥

⎦

⎤⎢⎣

⎡−

−=

11

011

33:6 222 kuuλ

取 ⎥⎦

⎤⎢⎣

⎡=

6002

D , ⎥⎦

⎤⎢⎣

⎡−

=1113

P , ⎥⎦

⎤⎢⎣

⎡ −=−

3111

411P

則可將 A對角化成 DAPP =−1 ，即 1A −= PDP (a) 令座標變換 YPX =

代入 GXAX +=′

得 GPYDY 1−+=′

⇒ ⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎦

⎤

⎢⎢⎣

⎡′

′tey

y

yy

32

1

2

1

48

3111

41

6002

⎪⎩

⎪⎨⎧

++=′−+=′⇒

t

t

eyyeyy

322

311

32622

⎪⎩

⎪⎨⎧

−−=

−−=⇒ tt

tt

eeky

eeky36

22

3211

31

31

⇒ ttt eekekyy

Px

362

21

2

1

2

1

210

323

10

11

13x

⎥⎦

⎤⎢⎣

⎡−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

(b) AQ 可對角化成 1A −= PDP

11010A −=∴ PPD ⎥⎦

⎤⎢⎣

⎡−

=1113

⎥⎦

⎤⎢⎣

⎡10

10

6002

⎥⎦

⎤⎢⎣

⎡ −3111

41

⎥⎦

⎤⎢⎣

⎡

⋅++−⋅+⋅−+⋅

= 1010101010101010

632626323623

41

92北科-8

The system of equations is shown as follow. BAX =

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

432655102

A , ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=41

2B

a. Please use Gauss-Jordan elimination to solve the above equations. (15%) b. Please use Cramer’s rule to solve the above equations. (10%)【92 北科土木】

【範圍】§21-3

【詳解】(a) 由增廣矩陣 ]|[ BA =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−−

41

2

432655102

～

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 366219

100010001

⇒⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

366219

X

(b) 由 Cramer’s rule

=1x 19

432655102434651102

=

−−

−

, =2x 62

432655102442615122

=

−−

−−−

, =3x 36

432655102432155

202

−=

−−

−−−

範例 4

92北科-9

92北科化工

Schrodinger’s Equation is given by

0)(822

2

2

2

2

=Ψ++Ψ

+Ψ

rZeE

hdrd

rdrd µπ , where E＜0

(a) By means of the substitution rx λ2= , where hE /)8( 2µπλ = , the

above equation becomes 0)41(22

2

=Ψ−+Ψ

+Ψ

xp

dxd

xxdd …………………(1)

Determine p. (5%)

(b) By means of the substitution )()( 2 xuexx

−=Ψ , Equation (1) becomes

0)()()2(22

=+−+ xuxgdxdux

dxudx ………………………………………(2)

determine g(x). (5%) (c) By differentiation of the following Laguerre’s differential Equation,

0)1( =+′−+′′ pyyxyx equation (2) can be obtained with the identification of )()( xyxu ′= . Find a solution of Laguerre’s differential Equation. (10%) 【92 北科化工】

【範圍】ch1 【詳解】(a)令 rx λ2=

則dxd

drdx

dxd

drd ψλψψ 2==

範例 1

92北科-10

22

22

2

422dxd

drdx

dxdxdd

dxdxdd

drd ψλ

ψ

λ

ψ

λψ ===

代入 0)(822

2

2

2

2

=Ψ++Ψ

+Ψ

rZeE

hdrd

rdrd µπ

得 0)(8842

2

22

2

22 =+++ ϕµπ

λψλψλ

rZeE

hdd

xdxd

0)(222

22

2

2

2

=Ψ++Ψ

+Ψ

rZeE

hdrd

rdrd

λµπ

0)(118

22 22

2

2

2

2

2

=Ψ++Ψ

+Ψ

⇒r

ZeEEu

hhu

dxd

xdxd

ππ

0)(412 2

2

2

=Ψ+−

+Ψ

+Ψ

⇒r

ZeEEdx

dxdx

d

0)41

4(2

2

2

2

=Ψ−−

+Ψ

+Ψ

⇒E

Zedxd

xdxd

0)412

4(2

2

2

2

=Ψ−−

+Ψ

+Ψ

⇒xE

Zedxd

xdxd λ

0)41

2(2

2

2

2

=Ψ−−+Ψ

+Ψ

⇒Ex

Zedxd

xdxd λ

故E

Zep2

2λ−=

(b)令 )()( 2 xuexx

−=ψ

則 )(21 22 xue

dxdue

dxd xx −−

−=ψ

)(41 22

2

22

2

2

xuedxdue

dxude

dxd xxx −−−

+−=ψ

代入 0)41(22

2

=Ψ−+Ψ

+Ψ

xp

dxd

xdxd

92北科-11

得 )(12)(41 2222

2

22 xue

xdxdue

xxue

dxdue

dxude

xxxxx−−−−−

−++−

0)()41( 2 =−+

−xue

xp x

0)(1)12(22

=−

+−+⇒ xux

pdxdu

xdxud

0)()1()2(22

=−+−+⇒ xupdxdux

dxudx

故 1)( −= pxg

(c) )()()( 2 xexuxyx

ψ==′

故 dxxexyx

x

)()(0

2ψ∫=

Evaluate the inverse of the following Laplace transform

(a) £ })1(s

1{1−

−

s (10%)【92 北科化工】

【範圍】§7-3

【詳解】£ }1

1s

1{1−

⋅−s

=£ ∗− }s

1{1 £ }1

1{−s

tet

*11π

=

Evaluate the inverse of the following Laplace transform

(b) £ }2

tan2

{ 11 s−− −π (10%)【92 北科化工】

【範圍】§7-3

範例 2

範例 2

92北科-12

【詳解】 =)(tf £ =− −− }2

tan2

{ 11 sπ £ }2{tan 11s

−−

由變換後之微分定理

£ ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−= −

sdsdttf 2tan})({ 1 2

2

2

)2(1

1s

s

−

+−= 22 2

2+

=s

=)(ttf £⎭⎬⎫

⎩⎨⎧

+−

221

22

st2sin=

t

ttf 2sin)( =∴

Expand the function in a Fourier integral and determine what the integral representation converge to？

xxf =)( as ππ ≤≤− x and 0)( =xf π>x (15%)【92 北科化工】

【範圍】§12-2 【詳解】 )(xfQ 為奇函數

∴令 ∫∞

=0

wxdwwBxf sin)(2)(π

則 ∫∞

=0

wxdxxfwB sin)()(

ππππ

ww

ww

wxdxx 20 cossin1sin −== ∫

故 ∫∞

−=0 2

wxdwww

ww

xf sin)cossin1(2)( ππππ

1當 ππ

92北科-13

⎪⎩

⎪⎨

⎧

=

−=−=−∫

∞

ππ

ππ

ππππ x

2

x2wxdww

ww

w0 2 收斂至sin)cossin1(2

3當 π>x ：

0sin)cossin1(2收斂至=−∫

∞

0 2wxdww

ww

wπππ

π

Solve the following partial differential equation

0x 2

2

2

2

=∂∂

+∂∂

yUU )0,0( >> yx

0),0( =yU )0( >y )()0,( xFxU = )0( >x where xxF =)( as 20 ≤≤ x

0)( =xF as 2>x (20%)【92 北科化工】

【範圍】§15-2 【詳解】由特徵函數展開法

令 =),( yxU ∫∞

0sin)( wxdwyc

代入得 0)()( 2 =−′′ ycwyc ; )(∞c 有界 wykeyc −=⇒ )(

=∴ ),( yxU ∫∞ −

0sin wxdwke wy

B.C. ∫∞

=0

)(sin)0,( xFwxdwkxU

得 ∫∞

=0

sin)(2 wxdwxFkπ

範例 4

92北科-14

202cos22sin2sin2

wwwwwxdxx −== ∫

∞

ππ

故 wxdwew

w2w2w22yxU0

wy2 sin

cossin),( ∫∞ −−=

π

Evaluate ∫∫ ⋅S

dAnF rr

kyzjxyiFrrrr

++= S: zyx ≤+ 22 , 0≥y , 4≤z (10%)【92 北科化工】

【範圍】§19-3

【答案】 ∫∫ ⋅S

dAnF rr

564

−=

Solve xexyyyy 28126 =−′+′′−′′′ (15%)【92 北科化工】

【範圍】§3-3 【詳解】 齊性解： 08126 23 =−+− mmm 2,2,20)2( 3 =⇒=−⇒ mm

xxxn excxececy32

32

22

1 ++=∴

特解：由逆算子法

xD

eexD

y xxp 322

31

)2(1

=−

=

25

223

22

1541

321 x

Dex

De xx ==

範例 5

範例 6

92北科-15

27

2

1058 xe x=

通解：

xxxxpn exexcxececyyy22

722

32

22

1 1058

+++=+=

92北科-16

92北科光電

Find the general solution of the following equations (1) )1sin(1374 +=+′+′′ − teyyy t . (10%)【92 北科光電】

【範圍】§3-3 【詳解】 齊性解： 0742 =++ mm

im 32 ±−=⇒

)3sin3cos( 212 tctcey tn +=∴−

特解：由待定係數法

令 [ ])1sin()1cos( +++= − tBtAey tp

代入得 3,2 =−= BA

得 [ ])1sin(3)1cos(2 +++−= − ttey tp

通解： ph yyy +=

[ ])sin()cos()sincos( 1t31t2et3ct3ce t21t2 +++−++= −−

【令解】由逆算子法

)1sin(133)2(

1)1sin(1374

122 +++

=+++

= −− teD

teDD

y ttp

[ ] )1sin(32)1(113 2 +++−

= − tD

e t

範例 1

92北科-17

)1sin(42

113 2 +++= − t

DDe t )1sin(

421113 2 +++−

= − tD

e t

)1sin(943D213)1sin(

32113 2 +−

−=+

+= −− t

Det

De tt

)1sin()23()1sin(13

3213 +−=+−−

= −− tDetDe tt

[ ])1sin(3)1cos(2 +++−= − tte t

Find the general solution of the following equations (2) )ln(13413132 )1()2(2)3(3 xyxyyxyx −=−+− (10%)【92 北科光電】

【範圍】§4-1

【詳解】令dtdDxtex t ≡== ,ln,

代入得{ } tyDDDDDD 1341313)1(2)2)(1( −=−+−−−− tyDDD 134)13175( 23 −=−+−⇒ tyDDD 134)134)(1( 2 −=+−−⇒

齊性解： 0)134)(1( 2 =+−− mmm im 32,1 ±=⇒

)3sin3cos( 322

1 tctceecytt

n ++=⇒

[ ])(ln3sin)(ln3cos 3221 xcxcxxc ++=

特解： )134(13175

123 tDDD

yp −−+−=

1)134)(16917

131( +=−+−−= ttD L 1ln += x

通解： [ ] 1ln)(ln3sin)(ln3cos 3221 ++++=+= xxcxcxxcyyy ph

範例 1

92北科-18

The Bernoulli equation is in the form of αyxRyxQyxP ⋅=⋅+′⋅ )()()( , in which α is a constant. Please find the integrating factor ),( yxµ . (Hint: try αµ −⋅= yxfyx )(),( ) (10%)【92 北科光電】

【範圍】§2-6

【詳解】 αyxRyxQdxdyxP )()()( =+

[ ] 0)()()( =−+⇒ dxyxRyxQdyxP α 乘上積分因子 αµ −= yxfyx )(),( 得 [ ] 0)()()()()()( 1 =−+ −− dxxfxRyxfxQdyyxfxP αα 必為正合方程式

故 [ ] [ ])()()()()()( 1 xfxRyxfxQy

yxfxPx

−∂∂

=∂∂ −− αα

αα α −− −=∂

∂⇒ Qfy1

xPfy )()( Qf

dxPfd )1()( α−=⇒

由分離變數法 dxpQ

pfPfd )1()( α−=

積分得 ∫−= dxxpxQf)()()1()Pln( α

故∫−

=dx

xpxQ

exfxP )()()1(

)()(α

∫−

=⇒dx

xPxQ

exP

xf )()()1(

)(1)(

α

故積分因子為∫

=−− dx

xPxQ

exP

yyx )()()1(

)(),(

αα

µ

範例 2

92北科-19

If )()( tftf =+ω , so that )(tf has a period ω . Prove that the Laplace transform of )(tf is

£ ∫ ⋅⋅−=−

−

ω

ω 0)(

11)}({ dttfee

tf sts . (10%)【92 北科光電】

【範圍】§7-4 f(t) 【證明】 0 ω 2ω 3ω 4ω 5ω 6ω

£{ } ∫∞ −=0

)()( dttfetf st ∫ ∫ −− +=ω ω

ω0

2)()( dttfedttfe stst

∫ ++ −ω

ω

3

2)( dttfe st ∫ ∑∫

+ ∞

=

+ −− =++ω

ω

ω

ω

)1(

0

)1()()(

n

nn

n

n

stst dttfedttfe LL

令 ωntu −= 則 dudt = 當 ω)1( += nt → ω=u 當 ωnt = → 0=u

則 £{ } ∫∞ −=0

)()( dttfetf st ∑∫∞

=

+−=0

0

)( )(n

nus duufeω ω

∑ ∫∞

=

−−=0

0)(

n

susn duufeeωω ∫ −−−=

ω

ω 0)(

11 dttfee

sts

Solve the initial value problem )(65 tfyyy =+′+′′ , 0)0()0( =′= yy ,

where ⎩⎨⎧

≥

92北科-20

【範圍】§8-1 【詳解】 )2()0(65 −+−−=+′+′′ tutuyyy 取 Laplace transform

得 )(1)()65( 022 ss ees

sYss −− −=++

)()3)(2(

1)( 02 ss eesss

sY −− −++

=⇒

))(3

131

21

211

61()( 02 ss ee

ssssY −− −

++

+−=⇒

故 )()31

21

61()( 32 tueety tt −− −+−=

)2(31

21

61 )2(3)2(2 −⎥⎦

⎤⎢⎣⎡ +−+ −−−− tuee tt

(1) Define ∫ ⋅⋅>=<π

0)()(, dxxgxfgf , where f, g is in the vector space of

functions continuous on ],0[ π . Prove that )cos(,),2cos(),cos( nxxx K are mutual orthogonal. (7%)

(2) Compute the cosine series of the function xexf −⋅= 2)( , where 20 ≤≤ x . (8%) 【92 北科光電】

【範圍】§12-3

【詳解】(1) ∫=π

0coscoscos,cos nxdxmxnxmx

[ ]∫ =−++=π

00)cos()cos(

21 dxxnmxnm

故 )cos(,),2cos(),cos( nxxx L 互相正交。

(2)令 ∑∞

=

+=1

0 2cos)(

nn x

naaxf π

範例 5

92北科-21

則 [ ]∫∫ −==−− −=−===2

0

220

2

0012

21)(

21 eedxedxxfa xx

xx

xdxnexdxnxfa xn 2cos2

2cos)(

22 2

0

2

0

ππ∫∫ −==

2

022 2

cos

41

2 =

=

−⎥⎦⎤

⎢⎣⎡

+

−=

x

x

x xnen

ππ

[ ]nen

nen

)1(14

8)1cos(4

8 222

222 −−+

=−+

−= −−π

ππ

故 ∑∞

=

−−

+−−

+−=1

22

22

2cos

4)1(181)(

n

n

xnn

eexf ππ

⎥⎦

⎤⎢⎣

⎡=

3241

A

(1) Find the eigenvalues and the corresponding eigenvectors. of A. (4%) (2) Find An for any positive integer n. (6%) 【92 北科光電】

【範圍】§24-3

【詳解】(1) ⎥⎦

⎤⎢⎣

⎡−

−=1

2k1 1:λ ; ⎥

⎦

⎤⎢⎣

⎡=

11

k5 1:λ

(2) 取 ⎥⎦

⎤⎢⎣

⎡−=

5001

D , ⎥⎦

⎤⎢⎣

⎡−

=1112

P , ⎥⎦

⎤⎢⎣

⎡ −=−

2111

31P 1

則可將 A對角化成 DAPP =−1 ，即 1A −= PDP

1n PPD −=nA ⎥⎦

⎤⎢⎣

⎡−

=1112

⎥⎦

⎤⎢⎣

⎡ −n

n

5001)(

⎥⎦

⎤⎢⎣

⎡ −2111

31

⎥⎦

⎤⎢⎣

⎡

⋅+−+−−⋅+−−+−

= nnnnnnnn

521515212512

31

)()()()(

範例 6

92北科-22

)(, CMPA nn×∈ and APP1− is a diagonal matrix D.

Prove that 1−⋅⋅= PePe DA (10%)【92 北科光電】

【範圍】§24-3 【證明】已知 1PDPA −= ，則 ))(())(( 1111n PDPPDPPDPPDPA −−−−= L 1111 PDPPDPPDPPDP −−−−= L 其中 IPP =−1 ⇒ 1nn PPDA −− == 1IDIDPPDIDI LL

1n1nn PDPPPDA −∞

=

−∞

=

∞

=∑∑∑ ===

1n1n1n

A

n1

n1

n1e

!!!

11n PPPDP −−∞

=

⋅⋅=⎟⎠

⎞⎜⎝

⎛= ∑ D

1ne

n1!

範例 7

92北科-23

Derive the following formula by using the residue theorem.

∫∞

∞− − −⋅−⋅

=+ 2222 ])!1[(2

)!22()1( n

nx

dxnn

π (8%)【92 北科光電】

【範圍】陳立工數經典題型 §1-1

【分析】 )1())(()()(

1 +−− +−=+=+

nnn iznizdz

dizdz

d

)2(2)1(2

2

)()1)(()1())(()(

1 +−+− +⋅+−=+−=+

nnn iznnizndz

dizdz

d

)3(33

3

)()2)(1)(()1()(

1 +−+⋅++−=+

nn iznnnizdz

d

)1(11

1

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

1 −+−−−

−

+−+−+++−=+

nnnnn

n

iznnnnnnnizdz

dL

)12(1 ))(22)(32()2)(1)(()1( −−− +−−++−= nn iznnnnn L

121

)(1

)!1()!22()1( −

−

+−−

−= nn

iznn

【詳解】令 nzzf

)1(1)( 2+

= ，則 iz ±= 為『n階極點』

其留數

)(Re is [ ])()()!1(

1lim 11

zfizdzd

nn

n

n

iz−

−= −

−

→

[ ])()()!1(

1lim 11

zfizdzd

nn

n

n

iz−

−= −

−

→

⎥⎦

⎤⎢⎣

⎡+

−−

= −−

→ nn

n

n

iz ziz

dzd

n )1(1)(

)!1(1lim 21

1

範例 8

92北科-24

⎥⎦

⎤⎢⎣

⎡−+

−−

= −−

→ nnn

n

n

iz iziziz

dzd

n )()(1)(

)!1(1lim 1

1

nn

n

iz izdzd

n )(1

)!1(1lim 1

1

+−= −

−

→

121

)(1

)!1()!22()1(

)!1(1lim −

−

→ +−−

−−

= nn

iz iznn

n

12

1

)2()1(

)!1()!22(

)!1(1

−

−−−−

−= n

n

inn

n 121

12 )()1(

)!1()!22(

)!1()2(1

−

−

−

−−−

−= n

n

n inn

n

12

1

12 )()()1(

)!1()!22(

)!1()2(1

−

−

−

−−−

−=

iinn

n nn

n

n

n

n

inn

n )1()1(

)!1()!22(

)!1()2(1 1

12 −−

−−

−=

−

− 1)!1()!22(

)!1()2(1

12 −−−

−= −

inn

nn

212 )!1()!22(

)!1()2(1

ii

nn

nn −−

−= − i1n2

2n221n2 ])![(

)!(−⋅

−= −

由留數定理 ∫∞

∞− − −⋅−⋅

==+ 2222 ])!1[(2

)!22()(Re2)1( n

nisix

dxnn

ππ

Evaluate ∫ ++

C 4dz

izzz

)()]sin()[cos( , where C is any simple closed path about –i.

(7%)【92 北科光電】

【範圍】§30-2

【詳解】令 4izzzzf

)()]sin()[cos()(

++

= ，則 iz −= 為『4階極點』

範例 9

92北科-25

)(Re is − ( ) ( )[ ]zfizdzd

31 4

3

3

iz+=

−→ !lim

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+=−→ 4

43

3

iz izzziz

dzd

31

)()]sin()[cos(

!lim

[ ])sin()cos(!

lim zzdzd

31

3

3

iz+=

−→[ ])cos()sin(

!lim zz

31

iz−=

−→

[ ])cos()sin( ii61

−−−= [ ])cos()sin( ii61

−−=

[ ])cosh()sinh( 11i61

+−=

∫ ++

C 4dz

izzz

)()]sin()[cos( )(Re isi2 −= π [ ])cosh()sinh( 11i

3i

+−=π

[ ])cosh()sinh( 1i13

−=π

92北科-26

92北科自動化

試解下列微分方程式 (其中 dxdyy /=′ 22 / dxydy =′′ ) )3/()5.063( 222 yxyxyyxy +++−=′ (20%)【92 北科自動化】

【範圍】§2-3

【詳解】yx

yxyyxdxdy

+++

−= 222

35.063

0)3()5.063( 222 =++++⇒ dyyxdxyxyyx

令⎩⎨⎧

+=++=

yxyxNyxyyxyxM

2

22

3),(5.063),(

1336)63(

2

22

=++

=−++

=∂∂

−∂∂′

yxyx

Nxyxx

NxN

yu

Q

故積分因子為 xdx

eexI =∫=1

)(

乘回 O.D.E.，得正合方程式 0)3()5.063( 222 =++++ dyeyxdxeyxyyx xx

其解為 c).( =+ x22 ey50yx3

範例 1

92北科-27

試解下列微分方程式 (其中 dxdyy /=′ 22 / dxydy =′′ ) 22 42 xyyxy +=′ 4)2( =y (20%)【92 北科自動化】

【範圍】§2-2. §2-3. §2-6.

【詳解】 ⇒+=′ 22 42xy xyy 4)(2dy 2 +=xy

dxxy

令xyu = ，即 uxy = ，則 xduudxdy +=

代入上式得 42 2 +=+ udx

xduudxu

42 22 +=+⇒ udxduxuu 42 +=⇒ u

dxduxu dxuxudu )4( 2 +=⇒

由分離變數法 x

dxduu

u=

+ 42

xdxdu

uu 2

422 =+

⇒

積分得 ∫∫ =+ xdxdu

uu 2

422

222 lnlnlnlnln2)4ln( cxcxcxu =+=+=+⇒

22 4 cxu =+⇒ 222

4 cxxy

=+⇒

由 B.C. x=2, y=4: 222

2424 c=+ 2=⇒ c

故解答為 222

24 xxy

=+ 422 24 xxy =+⇒

【另解】 ⇒+=′ 22 42xy xyy 0)42( 22 =−+ xydydxxy

令⎩⎨⎧

−=+=xyyxN

xyyxM),(

42),( 22，則

xxyyy

NxN

yM

5)(4 −=

−−−

=∂∂

−∂∂

範例 2

92北科-28

積分因子為 55

)( −−

=∫= xexIdx

x

乘回 O.D.E. 0)42( 22 =−+ xydydxxy

得正合方程式 0)42( 4352

=−+ dyxydx

xxy

其解為 cxx

y=−− 24

2 221

由 B.C. x=2, y=4: c=−− 242

22

24

21 1−=c

故解答為 1221

24

2

−=−−xx

y 422 24 xxy =+⇒

【另解】 ⇒+=′ 22 x4y2yxy 22 42 xydxdyxy =−

令 2yu = ，則dxdyy

dxdu 2= ，代入上式得 242

2xu

dxdux

=−

xuxdx

du 84 =−⇒

1積分因子： 4)4(

)( −−

=∫= xexIdx

x

2通解： cxxdxxuxI +−=⋅= −−∫ 24 48)( 424 cxxu +−=⇒ 422 4 cxxy +−=⇒

由 B.C. x=2, y=4: 422 2244 c+⋅−= 2=⇒ c 故解答為 422 24 xxy +−= 422 24 xxy =+⇒

92北科-29

試解下列微分方程式 (其中 dxdyy /=′ 22 / dxydy =′′ ) xxyyy sin3cos444 +=+′+′′ (20%)【92 北科自動化】

【範圍】§3-3 【詳解】1齊性解： 2,20442 −−=⇒=++ mmm

xxh xececy

22

21

−− +=∴

2特解：由待定係數法，令 xBxAy p sincos +=

代入得 0=A ， 1=B xy p sin=⇒

3通解： xxececyyxy xxph sin)(2

22

1 ++=+=−−

【另解】由逆算子法

)sin3cos4(441

1)sin3cos4(44

122 xxD

xxDD

y p +++−=+

++=

)sin3cos4(916

34)sin3cos4(34

12 xxD

DxxD

+−−

=++

=

)sin3cos4(25

34)sin3cos4(9)1(16

342 xx

DxxD +−−

=+−−

−=

)sin9cos12cos12sin16(251 xxxx −−+−−= xsin=

試解下列微分方程式 (其中 dxdyy /=′ 22 / dxydy =′′ )

xxex3xyy )( +=′

−′′ (20%)【92 北科自動化】

範例 3

範例 4

92北科-30

【範圍】§4-1 【詳解】同乘以 2x

得 xexxyxyx )3( 432 +=′−′′

令 tex = ,dtdDxxt ≡>= ),0(ln

得{ } tet4t3 eee3yD1DD )()( +=−−

{ } tet4t32 eee3y2DDyD2D )()()( +=−=−⇒

(1) 齊性解： ( ) 2m0m02mmmm2 ==⇒=−=− ,2

221t2

21h xcceccy +=+=⇒

(2) 特解：

( )tet4t3

p eee32DD1y )( +−

=

tt et4t3et4t3 eee3D1

21eee3

2D1 )()(

21

+−+−

=

∫∫ +−+= − dteee321dteee3ee

21 tt et4t3et4t3t2t2 )()(

∫∫ +−+= dteee321dteee3e

21 tt et4tet2tt2 )()( 3

∫∫ +−+= tet3t2tett2 deeee321deee3e

21 tt )()(

tt et3ett2 ee21eee

21

−+= )2(tt et3et3t2 ee

21eee

21

−+= )2(

tet2 ee= x2ex=

(3) 通解： 221ph xccyyy +=+=x2ex+

92北科-31

試解下列微分方程式組 (其中 dtdD /= ) 25654)5( 2 ++−=−− ttyxD 0)0( =x

42)2( 2 ++−=−− ttxyD 0)0( =y (20%)【92 北科自動化】

【範圍】ch5 【答案】 5tee4x 2t6 −++= t teey t6 −−= t

範例 5

92北科-32

92北科冷凍空調

Solve the following ordinary differential equations.

(a) yeyxxyy 422

3

8322

++

−=′ (5%)【92 北科冷凍】

【範圍】§2-3

【詳解】 yeyxxy

dxdy

422

3

8322

++

−=

0)83()22( 4223 =+++ dyeyxdxxy y

)83()22( 4223 yeyxy

xyy

+∂∂

=+∂∂

Q

∴此為正合方程式 ),( yxφ∃

⎪⎪⎩

⎪⎪⎨

⎧

+=∂∂

+=∂∂

∋yeyx

y

xyx

422

3

83

22

φ

φ

其解為 cexyx y =++= 432 22φ

範例 1

92北科-33

Solve the following ordinary differential equations. (b) xexyy 224 −+=+′′ (5%)【92 北科冷凍】

【範圍】§3-3 【詳解】 齊性解： imm 2042 ±=⇒=+

xcxcyn 2sin2cos 21 +=∴

特解：由待定係數法

令 xp CeBAxy2−++=

代入得41,0,

41

=== CBA

xp exy2

41

41 −+=∴

通解：

xpn exxcxcyyy2

21 41

412sin2cos −+++=+=

【另解】由逆算子法

xp eD

xD

y 222 241

41 −

++

+=

xexD 222 2

4)2(1)

161

41( −

+−++−= L

xex 241

41 −+=

範例 1

92北科-34

Solve the following ordinary differential equations. (c) 0)642()352( =−+−+− dyyxdxyx (5%)【92 北科冷凍】

【範圍】§2-2

【詳解】聯立⎩⎨⎧

=−−=+−

06420352

yxyx

得交點 )1,1(),( =yx

令座標平移⎩⎨⎧

−=Υ−=Χ

11

yx，即

⎩⎨⎧

+Υ=+Χ=

1y1x

代入 O.D.E.得齊次方程式 0)42()52( =ΥΥ+Χ−ΧΥ−Χ dd

將上式同 Χ÷ 得 ΥΧΥ

+=ΧΧΥ

− dd )42()52(

令ΧΥ

=u ，即 Χ=Υ u ，則 duudd Χ+Χ=Υ

代入上式得 ))(42()52( duududu Χ+Χ+=Χ− 0)24()274( 2 =+Χ+Χ−+⇒ duuduu

0)21()

21

47( 2 =+Χ+Χ−+⇒ duuduu

由分離變數法 0

21

47

21

2=

−+

++

ΧΧ du

uu

ud

0)

41)(2(

21

=−+

++

ΧΧ

⇒ duuu

ud 0)

41

31

232

( =−

++

+ΧΧ

⇒ duuu

d

0)

41

12

2(3 =−

++

+ΧΧ

⇒ duuu

d

範例 1

92北科-35

積分得 cuu ln)41ln(20ln(2ln3 =−+++Χ

cuu ln)41ln()2ln(ln 23 =−+++Χ⇒

cuu ln)41()2(ln 23 =−+Χ⇒ cuu =−+Χ⇒ )

41()2( 23

c=−ΧΥ

+ΧΥ

Χ⇒ )41()2( 23 c=Χ−ΥΧ+Υ⇒ )

41()2( 2

c421 2 =Χ−ΥΧ+Υ⇒ )()(4

k=Χ−ΥΧ+Υ⇒ )4()2( 2

kxyxy =−−−+⇒ )34()32( 2

Solve the following ordinary differential equations. (d) 34222 2)42()42()1( xxyxxyxxxyxx −−=+++′++−′′+

(Hint: xy = is a homogeneous solution) (5%)【92 北科冷凍】

【範圍】§4-2 【分析】 xxhxaa =⇒=+ )(001 為齊性解 【詳解】令 )(xxy φ= ，則 )()( xxxy φφ ′+=′ )()(2 xxxy φφ ′′+′=′′ 代入 O.D.E.得 2)2()1( −−=′+−′′+ xxx φφ

可降為一階 O.D.E. 12

12

++

−=′++

−′

xx

xx

dxd φφ

積分因子：1

)( 12

+=∫=

−++

−

xeexI

xdxxx

通解： ∫ ++

−=′ dxxxxIxI )

12)(()( φ

∫ −−

++

−=′+

⇒ dxexx

xe xx

2)1(2

1φ ∫ −+

++−= dxe

xx x

2)1(1)1(

範例 1

92北科-36

∫∫ −− +−+−= dxexdxexxx

2)1(1

11

∫∫ −−

−

++

++

+−= dxe

xxedxe

xx

xx

11

111

111c

xe

xe xx

++

=′+

⇒−−

φ

)1(1 1 ++=′⇒ xecxφ

21 cxecxx ++=⇒φ

xcexcxxxy x 22

12)( ++==∴ φ

Apply the Laplace transform to solve 0)0()0(;142 =′==−′+′′ yyyyty (15%)【92 北科冷凍】

【範圍】§8-1

【分析】1￡{ } )(ˆ)( sfdsdttf −= 2初值定理： )(ˆlim)0( sysy

s ∞→=

【詳解】 [ ]s

yysdsdys 1ˆ4ˆ2ˆ2 =−−

syy

dsydsys 1ˆ4ˆ2ˆ

2ˆ2 =−−−⇒

sys

dsyds 1ˆ)6(ˆ

2 2 =−+−⇒ 221ˆ)

23(

ˆs

yssds

yd−=−+⇒

1積分因子： 434ln4ln34ln3)

23(

22

3

22

)(ss

ss

sssdss

s eseeeeeesI−−−−−

====∫=

2通解： ∫ +=−=−

ces

sIsysIs2

41

2 )21)(()(ˆ)(

2

41

3311)(ˆ

se

sc

ssy +=⇒

由初值定理： )0()(ˆlim ysyss

=→∞

0)11(lim2

41

22 =+⇒ →∞s

se

sc

s 0=⇒ c

即 31)(ˆs

sy = 221)( tty =⇒

範例 2

92北科-37

Suppose a substance has density ),,( zyxρ , specific heat ),,( zyxc , and thermal conductivity ),,( zyxk . Let ),,,( tzyxT be the temperature of this substance at time t and point ),,( zyx . Consider an imaginary smooth closed surface Σ within the substance, bounding a solid region M. If Ν is the unit outer normal to Σ , from the energy balance we have

∫∫∫∫∫ ∂∂

=Ν•∇Σ M

dVtTcdTk ρσ)( .

(a) Briefly explain the Gauss’s divergence theorem. (5%) (b) Apply the Gauss’s divergence theorem to the above energy balance result to

obtain the heat conduction partial differential equation. (10%) (c) If the thermal conductivity k is constant and the substance is

one-dimensional with length L and has the following conditions, find the solution of ),( txT .

Boundary conditions ⎩⎨⎧

==

BtLTAt0T

),(),(

,

Initial condition )()0,( xxT φ= , Lx0

92北科-38

Tc

ktT 2∇=∂∂

ρ (heat conduction P.D.E.)

(c) 令 ( ) ( ) ( )xstxtx += ,,T ω

代入 P.D.E. 22

xT

ck

tT

∂∂

=∂∂

ρ ( )xs

ck

xck

t 22

′′+∂∂

=∂∂

ρω

ρω

B.C. ( )( )⎩

⎨⎧

==

BtLuAt0u

,,

( ) ( )( ) ( )⎩

⎨⎧

=+=+

BLstLA0st0

,,

ωω

I.C. ( ) )(, x0xu φ= ( ) ( ) )(, xxstL φω =+ 1 steady state： O.D.E. ( )xs0 ′′=

B.C. ( )( )⎩

⎨⎧

==

BLsA0s

( ) AxL

ABxs +−=

2 transient state：

P.D.E. 22

xck

t ∂∂

=∂∂ ω

ρω

B.C. ( )( )⎩

⎨⎧

==

0tL0t0

,,

ωω

I.C. ( ) ( ) ⎟⎠⎞

⎜⎝⎛ +

−−=−= Ax

LABxxsx0x )()(, φφω

( ) xL

nebtx1n

tL

nc

k

n2

22

πωπ

ρ sin, ∑∞

=

−=

由 I.C. ( ) ⎟⎠⎞

⎜⎝⎛ +

−−==∑

∞

=

axL

abxxL

nb0x1n

n )(sin, φπω

得 ( )∫⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ +

−−=

L

0nxdx

Lnax

Labxf

L2b πsin

92北科-39

【答案】 ( ) ( ) ( )xstxtx += ,,T ω xL

neb1n

tL

nc

k

n2

22

ππρ sin∑∞

=

−= ⎟

⎠⎞

⎜⎝⎛ +

−+ Ax

LAB

其中 ( )∫⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ +

−−=

L

0nxdx

Lnax

Labxf

L2b πsin

23)( xxf = for 40

92北科-40

Explain briefly the concepts and procedures to solve a system AXX =′ with A an nn× matrix of real numbers. (10%)【92 北科冷凍】

【範圍】§24-4 【答案】完全抄襲 陳立工數 §24-4 題型1 課文

範例 5

92北科-41

92北科車輛

Solve the differential equation 0t =++dtdytey with the initial value 2)1( =y .

(25%)【92 北科車輛】

【範圍】§2-5

【詳解】 tet

ytdt

dy 11 −=+

積分因子： tetIdt

t =∫=1

)(

通解： ∫ +−=−

= cedtet

tIytI tt1)()( tce

ty t +−=⇒ 1

由 I.C. 22)1( +=⇒=+−= eccey

tee

tty t 21)( ++−=∴

Find the values )1(y and )2(y for the differential equation )(tfydtdy

=+

with the initial value 1)0( =y , where ⎪⎩

⎪⎨

⎧

<≤<

≤=

tt

ttf

2 ,021 ,1

1 ,0)( .

(25%)【92 北科車輛】

範例 1

範例 2

92北科-42

【範圍】§8-1

【詳解】O.D.E. )2()1( −−−=+ tutuydtdy

取 Laplace transform

{ } ss es

es

syss 211)()0()( −− −=Υ+−Υ

ss es

es

ss 2111)()1( −− −+=Υ+⇒

ss ess

esss

s 2)1(

1)1(

11

1)( −−+

−+

++

=Υ⇒

ss ess

esss

2)1

11()1

11(1

1 −−+

−−+

−++

=

[ ] [ ] )2(1)1(1)( )2()1( −−−−−+=⇒ −−−−− tuetueety ttt 1)1( −= ey

12 1)2( −− −+= eey

Let ⎥⎦

⎤⎢⎣

⎡=

2112

A .

(a) (15%) Find the eigenvalues and eigenvectors of the matrix A.

(b) (10%) Find the value of Ae . ( Ae is defined as LL++++!3!2!1

32 AAAI )

【92 北科車輛】

【範圍】§24-3

【詳解】(a) 021

12)det( =⎥

⎦

⎤⎢⎣

⎡−

−=Ι−

λλ

λA 3,1=⇒ λ

範例 3

92北科-43

⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡⇒⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

11

00

1111

:1 12

1

2

11 kx

xxx

λ

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⇒⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

11

00

1111

:3 22

1

2

12 kx

xxx

λ

(b) 取 ⎥⎦

⎤⎢⎣

⎡=

3001

D , ⎥⎦

⎤⎢⎣

⎡−

=1111

P , ⎥⎦

⎤⎢⎣

⎡ −=−

1111

211P

則可將 A對角化成 1−= PDPA

Ae 1−= PPeD ⎥⎦

⎤⎢⎣

⎡−

=1111

⎥⎦

⎤⎢⎣

⎡3

1

00e

e⎥⎦

⎤⎢⎣

⎡ −1111

21

Determine the Fourier series representation of the steady-state output )(ty for

the dynamic equation )(tfydtdy

=+ , where the input ∑∞

= −+=

12 1

)sin(21)(

n ntntf π .

(25%)【92 北科車輛】

【範圍】附錄 §2-9

【詳解】 ⎥⎦

⎤⎢⎣

⎡−

++

=+

= ∑∞

=12 1

)sin(21

11)(

11

np n

tnD

tfD

y π = ∑∞

= −++

12 1

)sin(1

121

n ntn

Dπ

= ∑∞

= −−−

+1

22 1)sin(

11

21

n ntn

DD π = ∑

∞

= −−−−

+1

222 1)sin(

11

21

n ntn

nD ππ

= ∑∞

= −+−

+1

222 )1)(1()cos()sin(

21

n nntnntn

ππππ

範例 4

92北科-44

92北科高分子

0)2( 223 =′−+ yxyxy (10%)【92 北科高分子】

【範圍】§2-4 【詳解】 02 223 =−+ dyxydyxdxy 0)2( 22 =+−⇒ dyxxdyydxy

0232

2 =+⇒ −−

dyxy

dxyy 0225 =+⇒ dyx

yxdy

同 4y÷ ，得 042

2 =+ dyyx

yxyd 0

y42

2 =+⇒dy

yx

yxd

由分離變數法，將上式同 42

yx

÷ ，得 0)(

)(

22

2

=+y

dy

yxyxd

積分得 cy

yx

lnln1 =+− cye xy

lnlnln =+⇒−

cye xy

lnln =⇒−

cye xy

=⇒−

0)sinh()4cosh3( =++ dyyxdxxy (10%)【92 北科高分子】

【範圍】§2-5

範例 1

範例 2

92北科-45

【詳解】 xydxdyyx 4cosh3sinh −=+

令 yu cosh= ，則dxdyyu sinh=′

代入 O.D.E.得 xuux 43 −=+′ 43 −=+′⇒ ux

u

積分因子： 33

)( xexIdx

x =∫=

通解： ∫ −= dxxIuxI )4)(()( cxdxxux +−=−=⇒ ∫ 433 4

3xcxu +−=⇒

3cosh xcxy +−=∴

222 22 +=+′−′′ xyyxyx (10%)【92 北科高分子】

【範圍】§4-1

【詳解】令dtdDxtex t === ,ln,

則{ } 222)1( 2 +=+−− teyDDD

2)1)(2(

11

12

1 2221 −−

+−−

++=⇒DD

eDD

ececy ttt

12

1 2221 +−

++= ttt eD

ecec 12221 +++=ttt teecec

1ln2221 +++= xxxcec

範例 3

92北科-46

342

=+′+′=++′−′

xyxtyxyx

, solve )(),( tytx . (10%)【92 北科高分子】

【範圍】Ch5

【詳解】⎩⎨⎧

=++=−−+

3)4()1()12(

DyxDtyDxD

由 Cramer rule

⎪⎪⎩

⎪⎪⎨

⎧

++

=+

−−+

−−=

+−−+

3412

4)1(12

3)1(

4)1(12

DtD

yDDDD

DDt

xDDDD

⎩⎨⎧

−=−+−=−+

⇒txDD

xDD42)443(

2)443(2

2

⎪⎩

⎪⎨

⎧

−=+−

−=+−⇒

tyDD

xDD

34

32)2)(

32(

32)2)(

32(

⎪⎪⎩

⎪⎪⎨

⎧

+++=

++=⇒

−

−

tececty

ecectx

tt

tt

21)(

21)(

24

32

3

22

32

1

代入 34 =+′+′ xyx

得 12322

32 2

432

32

232

1 +−+−−− tttt ecececec 3244 223

2

1 =+++− tt ecec

⎩⎨⎧

=−=

⇒24

13 7cc

cc

故

⎪⎪⎩

⎪⎪⎨

⎧

+++=

++=

−

−

tececty

ecectx

tt

tt

21)(

21)(

24

32

3

22

32

1

範例 4

92北科-47

Find the eigenvalue and eigenfunction of the of the following problem: 0=+′′ yy λ with 0)4()3( == yy . (10%)【92 北科高分子】

【範圍】§11-1 【詳解】令座標平移 3−= xt

則 O.D.E. 022

=+ ydt

yd λ

B.C. 010==

== ttyy

得⎩⎨⎧

−=∈=

)(sinsin:oneigenfunti)(:eigenvalue

3xntnNnn 22

πππλ

Please find the Inverse Laplace Transform of the following:

(a) 1)3(

53)(Y 2 +++

=s

ss (5%)【92 北科高分子】

【範圍】§7-2

【詳解】1)3(4)3(3

1)3(53)( 22 ++

−+=

+++

=ss

sssY

1)3(14-

1)3(33 22 ++++

+=

sss

故 tettty 3)sin4cos3()( −−=

Please find the Inverse Laplace Transform of the following:

(b) )4(

1)( 2 +=

sssY (5%)【92 北科高分子】

範例 5

範例 6

範例 6

92北科-48

【範圍】§7-1

【詳解】4

41

41

)4(1)( 22 +

−+=

+=

s

s

ssssY

故 tty 2cos41

41)( −=

Using Frobenius theorem (power series) to solve the following equation: 0=+′+′′ yyyx (20%)【92 北科高分子】

【範圍】§9-3 【詳解】 0=xQ 為規則奇點

∴令 ∑∑∞

=

+∞

=

===00 n

rnn

n

nn

r xaxaxy

則 ∑∞

=

−++=′0

1)(n

rnn xarny

∑∞

=

−+−++=′′0

2)1)((n

rnnxarnrny

代入 O.D.E. 0=+′+′′ yyyx

得 ∑∑∑∞

=

+∞

=

−+∞

=

−+ =+++−++00

1

0

1 0)()1)((n

rnn

n

rnn

n

rnn xaxarnxarnrn

∑∑∑∞

=

−+∞

=

−+∞

=

−+ =+++−++⇒1

1

0

1

0

1 0)()1)((n

rnn

n

rnn

n

rnn xaxarnxarnrn

0=n ： 0,00)1( =⇒=+− rrrr 1≥n ： [ ] 0)()1)(( 1 =+++−++ −nn aarnrnrn

12)(1

−+−= nn arn

a (降 1遞迴)

範例 7

92北科-49

令 00 =a

則 2021 )1(1

)1(1

+−=

+−=

ra

ra

22122 )1()1(1

)2(1

++−=

+−=

rra

ra

222123 )1()1()1(1

)3(1

+++−=

+−=

rrra

ra

M

M

{ }LL++++=∴ 332210 xaxaxaaxy r

{ 2222 )2()1(1

)1(11 x

rrx

rxr

+++

+−=

}L++++

− 3222 )3()2()1(1 x

rrr

則⎩⎨⎧

⎥⎦⎤

⎢⎣⎡

+−

+−+⋅=

∂∂ x

rrxyx

ry r

112

)1(1ln 2

222 212

112

)2()1(1 x

rrrr ⎥⎦⎤

⎢⎣⎡

+−

+−

+++

}L+⎥⎦⎤

⎢⎣⎡

+−

+−

+−

+++− 3222 3

122

121

12)3()2()1(

1 xrrrrrr

故 L+−+−=== 321 361

411)0( xxxryy

且 L+−−−−−−++⋅= 3212 )3212(

361)12(

412ln xxxyxy

L++−+⋅= 321 10811

432ln xxxyx

得通解 2211 ycycy +=

92北科-50

Find the Fourier Series of the following function: π+= xxf )( , if ππ

92北科-51

92北科電腦通訊甲、乙、丁組

A system consists of four switches is illustrated in Fig. 1. 1A is the event that switch 1C is connected,

2A is the event that switch 2C is connected,

3A is the event that switch 3C is connected, and

4A is the event that switch 4C is connected, If 1A , 2A , 3A , 4A are independent events, 1.0)( 1 =AP , 2.0)( 2 =AP ,

3.0)( 3 =AP , and 4.0)( 4 =AP , find the probability that electrical current can pass through the system.

(5%)【92 北科電腦】

【範圍】機率 【答案】0.1624

A continuous random variable X has a pdf (probability density function ) of the form 9/2)( xxf x = for 30

92北科-52

【範圍】機率

【答案】(a) 2][ =XE (b) 21XVar =][ (c)

3625511XP =≤− ].[

Let X and Y be continuous random variables with a joint pdf (probability density function) of the form )(),( yxkyxf XY += for 10 ≤≤≤ yx and zero otherwise. (a) Find k such that ),( yxf XY is a joint pdf. (5%)

(b) Find the conditional pdf )( xyf XY . (5%)

(c) Find ][ 2YXE . (5%) 【92 北科電腦】

【範圍】機率 【答案】(a)k=2

(b)⎪⎩

⎪⎨⎧

92北科-53

【範圍】機率 【答案】(a) 0.7696 (b) )2.083()2.083()2.083( Φ−=−Φ−Φ 21 (查表)

Let W be the subspace of 4R consisting of vectors of the form

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

4

3

2

1

xxxx

x . All vectors x in W satisfy 0321 =−+ xxx and 042 =− xx .

(a) Find the dimention of W. (5%) (b) Find a basis for W. (5%) (c) Find an orthonormal basis for W. (5%) 【92 北科電腦】

【範圍】電機線代

【答案】 (b)

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

1110

x

0101

x

xxx

xx

xxxx

x 21

2

21

2

1

4

3

2

1

(a) dim(W)=2

(c) {⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

0101

21 ,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

0121

101 }

範例 5

92北科-54

Let V be 3R and let },,{ 321 XXXS = and },,{ 321 YYYT be bases for 3R ,

where ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

102

1X , ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

021

2X , ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

111

3X and ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

336

1Y , ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=31

4

2Y , ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

255

3Y .

Compute the transition matrix P from the T-basis to the S-basis. (10%)【92 北科電腦】

【範圍】電機線代

【答案】

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

111211122

P

Let ⎥⎦

⎤⎢⎣

⎡−

=4211

A and ⎥⎦

⎤⎢⎣

⎡−

=1201

B .

(a) Which matrix is diagonalizable？ (5%) (b) If A is diagonalizable, find a nonsingular matrix P such that APP 1− is

diagonal; If B is diagonalizable, find a nonsingular matrix Q such that AQQ 1− is

diagonal. (10%) 【92 北科電腦】

【範圍】§24-3 【詳解】(a) A可對角化；而 B為下三角矩陣，特徵值為對角線上元素

1 1,=λ ，即特徵值全為重根，故不可對角化。

範例 6

範例 7

92北科-55

(b) 由 0)det( =− IA λ ⇒ 042

11=

−−−

λλ

⇒ 3,2=λ

1 2=λ ： ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡11

12

1 kxx

；2 3=λ ： ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡21

22

1 kxx

令 ⎥⎦

⎤⎢⎣

⎡=

2111

P ，可使得 ⎥⎦

⎤⎢⎣

⎡==−

30021 DAPP

Find the rank of

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

=

232383191121

A . (10%)【92 北科電腦】

【範圍】§22-2

【詳解】

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

=

232383191121

A ～

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

0700140070121

～

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

000000070121

2)( =Arank

範例 8

92北科-56

92北科電腦通訊丙組

Solve the non-homogeneous equation xexyyy 82 +=+′−′′ (20%)【92 北科電腦】

【範圍】§3-3 【詳解】 齊性解： 1,10122 =⇒=+− mmm

xxh xeCeCy 21 +=∴

特解：由特定係數法，令 xp eCxBAxy2++=

代入得 4,2,1 === CBA

xp exxy242 ++=∴

通解： xxxph exxxeCeCyyy2

21 42 ++++=+=

【另解】由逆算子法 xp eDx

DDy 8

)1(1

121

22 −+

+−=

= 81)21( 2DexD x+++ LL = xexx 242 ++

Find the Laplace transform of the following function with period ωπ2

=T :

tVtv m ωsin)( = when ωπ /0

92北科-57

【範圍】§7-4 【詳解】取 Laplace transform

£{ } ∫ −−−=T st

sT dtetvetv

0)(

11)( ∫ −−

−= ω

π

ωπ

2

0s2)(

1

1 dtetve

st

∫ −− ⋅−

= ωπ

ωπ ω02 sin

1

1 dtetVe

stm

s

s

m

e

V

ωπ2

1−

−=

ωπ

ωωωω

=

=

−

−−+

t

t

st

ttse

022 )cossin(s

)1(1

222

s

s

m es

e

V ωπ

ωπ ω

ω −−

++

−= 22

1ω

ω

ωπ +

−=

− se

Vs

m

【積分公式】 cttsedtetst

st +−−+

=⋅−

−∫ )cossin(ssin 22 ωωωωω

cttsedtetst

st ++−+

=⋅−

−∫ )sincos(scos 22 ωωωωω

Show that if )(tf has the Fourier transform )(ωF and )(tg has the Fourier transform )(ωG , then the convolution of the two functions )()( tgtf ∗ has the Fourier transform )()( ωω GF . (20%)【92 北科電腦】

【範圍】§13-3

【詳解】F { } dtedgtftgtf ti ])()([)()( ωτττ −∞∞−

∞

∞−−=∗ ∫∫

令 τ−= tu ，則 dudt =

上式 duedguf ui )( )()( τωττ +−∞

∞−

∞

∞− ∫∫=

範例 3

92北科-58

])([ dueuf uiω−∞

∞−∫= ])([ ττ τω deg i−

∞

∞−∫⋅

])t([ dtef tiω−∞

∞−∫= ])([ dtetg tiω−

∞

∞−∫⋅

=F }{ f F }{g = )()( ωω GF

Integrate maz )/(1 − (m is a positive integer) in the counterclockwise sense round any simple closed path C enclosing the point az = .

(20%)【92 北科電腦】

【範圍】§28-2

【詳解】1 1−≠m : dzazC

m∫ − )(

0

0

i0

2i0

1m0

rezzrezz

1mzz

θ

πθ

+=+=

+−

=++ )()(

[ ] 01ee1m

r 1n2i1ni1m 0 =−+

= +++

)()( πθ

2 1m −= :

0

0

i

2i

C

1

reazreaz

azdzaz θπθ

+=+=

−=−+

−∫)(

)ln()(

[ ] )ln(ln )( 00 i2i rere θπθ −= + i2ir2ir 00 πθπθ =−−++= ln)(ln

Find the eigenvalues and eigenfunctions of the Sturm-Liouville problem 0=+′′ yy λ , 0)0( =y , 0)( =πy . (20%)【92 北科電腦】

範例 4

範例 5

92北科-59

【範圍】§11-1

【詳解】○1 相異實根：令 ∞

92北科-60

92北科電機丁、戊組

One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black？ (10%)【92 北科電機】

【範圍】機率

【答案】6338

Consider two random variables X and Y. the conditional probability density function of X given Y has a Poisson distribution with rate y:

!

)/(xeyyxf

yx

YX

−

= K,2,1,0=x

The probability density function of variable Y is given by

⎩⎨⎧ ∞

92北科-61

(You may assume !0

nden =∫∞

− µµ µ for a positive integer n) 【92 北科電機】

【範圍】機率

【答案】(a) ⎪⎩

⎪⎨⎧

>==

−

.).(

),,,,,(!),(YX,

wo0

0y3210xxey2

yxfy3x

L

(b) ),,,,(3

)( 1X L3210x2xf x == +

(c) ),,,,,(!

)()(XY 0y3210xxey33xyf

y3x

>==−

L

(d) ][ XYE3

1x +=

Let X be a Gaussian random variable with probability density function

22

2x

2Xe

21xf σπσ

−=)(

Find the probability density function of the random variable 24XZ = . (10%)【92 北科電機】

【範圍】機率

【答案】28

z

2Ze

z221xf σπσ

−=)(

範例 3

92北科-62

Let X, Y be two independent random variables with probability density functions Xf and Yf , respectively. Let the random variable Z = X+Y, show that the probability density function of the random variable Z is YXZ fff ∗= where ∗ denotes the convolution operation. (10%)【92 北科電機】

【範圍】機率 【答案】詳見上課講義

Label the following statements as being true or false:(assume that all matrices in the following items are real, 2 points for each item with correct answer, -2 points for each item with incorrect answer, 0 points for each item with no answer) (a) The rank of a matrix is equal to the number of its nonzero columns. (b) The product of two matrices always has rank equal to the lesser of the ranks

of two matrices. (c) The determinant of a diagonal matrix is the product of its diagonal entries. (d) If E is an elementary matrix, then 1)det( ±=E . (e) Similar matrices always have the same eigenvectors. (10%)【92 北科電機】

【答案】(a) false (b) false (c) true (d) false (e) false

Let D, E be ordered bases for R3 given by ]}6,1,2[],1,5,3[],2,2,1{[=D ]}1,5,0[],8,41,0[],1,4,1{[=E

(a) (10%) Find the transition matrix form D-coordinates to E-coordinates (b) (5%) Let ]1,8,3[ −=v , find Dv][ :the D-coordinates of vector v.

【92北科電機】

範例 4

範例 5

範例 6

92北科-63

【範圍】電機線代

【答案】(a) transition matrix ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−=2202657

2737231

T

(b) Dv][ ],,2[ 11 −=

Consider the following matrix

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=021232120

A

(a) (10%) Find the eigenvalues and the corresponding eigenvectors of A. (b) (5%) Find the orthonormal matrix U such that A = UDUT where D is

diagonal. (c) (5%) Give the definition of the Hermitian matrix. Is A a Hermitian

matrix？ (d) (5%) Give the definition of the positive definite matrix. Is A a positive

definite matrix？ 【92 北科電機】

【範圍】§25-4

【詳解】(a) =− )det( IA λ 0021

232120

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−

λλ

λdet 511 ,,−−=λ

範例 7

92北科-64

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

−−=000

xxx

121242121

11

3

2

1

:,λ ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

01

2

101

xxx

3

2

1

,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−

=000

xxx

521222125

5

3

2

1

:λ ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

121

xxx

3

2

1

(b) 將 eigenvectors 取 Gram – Schmidt正交化，

得

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−=

61

31

21

62

310

61

31

21

U

可將 A正交對角化成 A = UDUT ，其中⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−=

500010001

D

(c) Hermitian matrix : AA =∗

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=021232120

A 為 Hermitian matrix

(d) 511 ,,−−=λ 不全大於 0，故 A非正定

92北科-65

92北科電機甲、乙、丙組

Let xey 31−= be one of the solutions to the differential equation

096 =+′+′′ yyy . (a) Derive the other linearly independent solution 2y from 1y . (10%) (b) Show they are indeed linearly independent. (5%) (c) Find the dimension of the solution space. (5%) 【92 北科電機】

【範圍】§4-2 【詳解】(a)令 )(3 xey xφ−= 則 )()(3 33 xexey xx φφ ′+=′ −− )()(6)(9 333 xexexey xxx φφφ ′′+′−=′′ −−− 代入 O.D.E. 096 =+′+′′ yyy 得 2)(0)( cxx =′⇒=′′ φφ 12)( cxcx +=⇒φ 故 x32

x3x3 xece9xey −−− +== )(φ ，即另一組解 x32 xey−=

(b) 0)31(3

),( 63333

21 ≠=−−= −

−−

−−x

xx

xx

eexe

xeeyyW

故 1y 與 2y 為 L.I. (c) dimension=2

Solve )(22 πδ −=+′+′′ tyyy ; 0)0()0( =′= yy . (15%)【92 北科電機】

【範圍】§8-1

範例 1

範例 2

92北科-66

【詳解】取 Laplace transform sesYss π−=++ )()22( 2

ses

sY π−++

=⇒1)1(

1)( 2

)()sin()( )( ππ π −−=⇒ −− tuetty t

Consider the periodic function ∑∞

−∞=

=n

jnten

tfπ2

2)( , 0≠n .

(a) Find the forced response of )(2502.0 tfyyy =+′+′′ in the form of complex Fourier series. (10%)

(b) Express the derived response in trigonometric form. (5%) 【92 北科電機】

【範圍】附錄 §2-9 【詳解】(a) 特解 (forced response)

jnt2

0nn

2jnt

0nn

22p e25D020D1

n2e

n2

25D020D1y

++=

++= ∑∑

∞

≠−∞=

∞

≠−∞= .. ππ

jnt20n

n2 e25jn020jn

1n

2++

= ∑∞

≠−∞= .)(π

jnt2

0nn

2 e25jn020n1

n2

++−= ∑

∞

≠−∞= .π

(b) )sin(cos.)(25

ntjntnj020n

1n

2y 20n

n2p ++−

= ∑∞

≠−∞= π

範例 3

92北科-67

)sin(cos.)(25.)(25 ntjnt

n00040nnj020n

n2

222

2

0nn

2 ++−−−

= ∑∞

≠−∞= π

222

2

0nn

2 n00040nntn020ntn

n2

.)(25sin.cos)(25

+−+−

= ∑∞

≠−∞= π

222

2

0nn

2 n00040nntn020ntn

n2j

.)(25cos.sin)(25

+−−−

+ ∑∞

≠−∞= π

222

2

0nn

2 n00040nntn020ntn

n2

.)(25sin.cos)(25

+−+−

= ∑∞

≠−∞= π

222

2

0nn

2 n00040nntn020ntn

n2j

.)(25cos.sin)(25

+−−−

+ ∑∞

≠−∞= π

( )奇函數偶函數 j0n

n∑∞

≠−∞=

+= ( 01n

+= ∑∞

=

)2 偶函數

222

2

1n2 n00040n

ntn020ntnn

4.)(25

sin.cos)(25+−+−

=∑∞

= π

Suppose that S consists of all vectors ),,,( xyyx −− in 4R . (a) Show that S is a subspace. (5%) (b) Determine a basis for the subspace S of 4R . (5%) (c) Determine the dimension of the subspace. (5%) 【92 北科電機】

【範圍】電機線代

範例 4

92北科-68

【詳解】(b)

tt

01

10

y

1001

xxyyx

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=−− ),,,(

(a) 由封閉性可證明 S為 4R 之子空間(subspace) (c) dim(S)=2

Let ⎥⎦

⎤⎢⎣

⎡ −=

3120

A and )sin()( xxf = .

(a) Find the eigenvalues of )(Af . (5%) (b) Find )(Af . (10%) 【92 北科電機】

【範圍】§24-3

【詳解】 031

2)det( =⎥

⎦

⎤⎢⎣

⎡−−−

=Ι−λ

λλA 2,1=⇒ λ

⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡⇒⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−−

=1

200

4121

:1 12

1

2

1 kxx

xx

λ

⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡⇒⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−=

11

00

1122

:2 22

1

2

1 kxx

xx

λ

(a) f(A)之特徵值為 )1(s)1( inf = , )2(s)2( inf =

(b)令 ⎥⎦

⎤⎢⎣

⎡=

2001

D , ⎥⎦

⎤⎢⎣

⎡−−

=11

12P , ⎥

⎦

⎤⎢⎣

⎡−−

=21

111P

可將 A對角化成 1−= PDPA

則 11)2(0

0)1()()( −− ⎥

⎦

⎤⎢⎣

⎡== p

ff

pPDPfAf

⎥⎦

⎤⎢⎣

⎡−−

=11

12⎥⎦

⎤⎢⎣

⎡2sin0

0sin1⎥⎦

⎤⎢⎣

⎡−− 2111

⎥⎦

⎤⎢⎣

⎡+−+−−−

=21212212212

sin2sinsinsinsinsinsinsin

範例 5

92北科-69

Consider the quadratic form 3122

21 222 xxxx ++ . Find the matrix Q that

transforms the quadratic form into the standard form (i.e., 233222

211 yyy λλλ ++ )

(20%)【92 北科電機】

【範圍】§25-4

【詳解】 3122

21 222 xxxx ++ [ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

3

2

1

321

xxx

002020201

xxx xAxt=

其中⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=002020201

A

=− )det( IA λ 0002

02020

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

λλ

λ1det 122 −= ,,λ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

000

xxx

202000201

22

3

2

1

:,λ

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

31032

010

xxx

3

2

1

,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=000

xxx

102030202

1

3

2

1

:λ

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

32

03

1

xxx

3

2

1

範例 6

92北科-70

取 transfer matrix

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

=

32

03

1

31032

010

Q

令座標旋轉

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

xxx

Q=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

yyy

可將 quadratic form 3122

21 222 xxxx ++ xAx

t=

轉成 standard form

[ ] 2322213

2

1

321 yy2y2yyy

100020002

yyy −+=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

92北科-71

92北科製造

Solve the following differential equations. In case the power series method is used, at least four (4) non-zero terms must be given. (a) 3168 =+′−′′ yyxy ; 0)0()0( =′= yy . (15%)【92 北科製造】

【範圍】§9-2 【詳解】 3168 +−′=′′ yyxy 33)0(16)0( =+−=′′ yy yxyy ′′+′−=′′′ 88 0)0(8)0( =′−=′′′ yy yxy ′′′= 8)4( 0)0()4( =y

M

M

)4()5( 88 xyyy +′′=

0)0(8)0()5( =′′′= yy

代入 Taylor級數

L+++′′′

+′′

+′

+= 5)5(

4)4(

32

!5)0(

!4)0(

!3)0(

!2)0(

!1)0()0()( xyxyxyxyxyyxy

223)( xxy =⇒

(b) Let 1)( 2 −= xxy be a solution of the ODE 022)1( 2 =+′−′′+ yyxyx . Find its general solution. (15%)【92 北科製造】

【範圍】§4-2

範例 1

範例 1

92北科-72

【詳解】令 )()1( 2 xxy φ−= 則 )()1()(2 2 xxxxy φφ ′−+=′ )()1()(4)(2 2 xxxxxy φφφ ′′−+′+=′′ 代入 022)1( 2 =+′−′′+ yyxyx 得 0)1(2)1(4)1)(1( 2222 =′+−′++′′++ φφφ xxxxxx 0)62()1)(1( 322 =′++′′++⇒ φφ xxxx

0)1)(1)(1(

622

3

=′−++

++′′⇒ φφ

xxxxx

0)1

21

21

2( 2 =′+−

−+

++′

⇒ φφx

xxxdx

d

由分離變數法

0)1

21

21

2( 2 =+−

−+

++

′′

dxx

xxx

dφφ

積分得 12 ln)1ln()1ln(2)1ln(2ln cxxx =+−−+++′φ

1222 ln)1ln()1ln()1ln(ln cxxx =+−−+++′⇒ φ

12

22

ln1

)1()1(ln cx

xx=

+−+′⇒ φ

⎥⎦

⎤⎢⎣

⎡−

++

=−+

+=′⇒ 22

122

2

1 )1(1

)1(1

2)1()1(1

xxc

xxxcφ

21 )

11

11(

2)( c

xxcx +

−+

+−=⇒φ

221 1c

xxc +−

−=

故 )1()()1( 2212 −+−=− xcxcxxy φ

(c) )cos(4)4( xyy −=− . (15%)【92 北科製造】

【範圍】§3-3 【詳解】 齊性解： imm ±±=⇒=− ,1014

範例 1

92北科-73

xcxcececy xxh sincos 4321 +++=⇒−

特解： 由待定係數法

令 xBxxAxyp sincos +=

代入得 1,0 == BA

故 xxyp sin=

通解：

xxxcxcececyyy xxph sinsincos 4321 ++++=+=−

【另解】 )cos4(1

11

1)cos4(1

1224 xDD

xD

yp −−+=−

−=

xxxD

sincos21

12 =+

=

(d) Find )(yx satisfies ODE 0)1( =+′− yy eyxe . (15%)【92 北科製造】

【範圍】§2-3

【詳解】 0)1( =+− yy edxdyxe

0)1( =−+⇒ dyxedxe yy

)1( −∂∂

=∂∂ yy xe

ye

yQ

∴此為正合方程式 ),( yxφ∃

範例 1

92北科-74

⎪⎪⎩

⎪⎪⎨

⎧

−=∂∂

=∂∂

∋1y

y

xey

exφ

φ

)(

)(

2

1

xkyxe

ykxe

y

y

+−=

+=

φ

φ

故 cyxeyx y =−=),(φ 即本題之解為 )()( cyeyx y += −

Prove the Laplace transform of the Dirac delta function at a is )exp( as− , i.e., to prove £ aseat −=− )]([δ Where £[.] denotes the Laplace transform from t to s. (20%)【92 北科製造】

【範圍】§7-1

【分析】ε

εδε

)()(lim)( −−−−=−→

atuatuat0

【詳解】£{ }=− )( atδ £⎭⎬⎫

⎩⎨⎧ −−−−

→ εε

ε

)()(lim0

atuatu

)00(~

11

lim

)(

0 ε

ε

ε

saas es

es

+−−

→

−= as

sa

ese

s −+−

→=

−⋅⋅−=

1

)(10lim

)(

0

ε

ε

【另解】£{ } dtateat st∫∞ − −=−

0)()( δδ

∫ ∫∞ ∞ −−− =−=−=0 0

)()( sasasa edtatedtate δδ

範例 2

92北科-75

Solve the initial value problem ⎪⎩

⎪⎨

⎧

+−=′−+=′+−=′

3213

3212

3211

)()(

3)(

yyytyyyytyyyyty

with the initial

condition ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

151

)0(

3

2

1

yyy

. (20%)【92 北科製造】

【範圍】§24-4

【詳解】

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′

3

2

1

3

2

1

3

2

1

111111

113y

yyy

Ayyy

yy

令 texxx

yy λ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1y

代入原式可得 0)det( =− IA λ ⇒ 0111

11111-3

=−−−−

−

λλ

λ

⇒ 3,2,0=λ

1 0=λ ：⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

011

1

3

2

1

kxxx

，

2 2=λ ：⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

011

2

3

2

1

kxxx

範例 3

92北科-76

2 2=λ ：⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

113

3

3

2

1

kxxx

∴ ttt ekekekyyy

33

22

01

3

2

1

113

011

110

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

由 I.C. =⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧)0(y

3

2

1

y

y

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

151

⎪⎩

⎪⎨

⎧

−===

⇒1

42k

3

2

1

kk

∴ tt eeyyy

32

3

2

1

113

011

4110

2⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

92北科-77

92北科機電整合

Solve the initial value problem xy

yyyxlnln −

=−′ ; 2)2( =y .

(10%)【92 北科機電整合】

【範圍】§2-1

【詳解】令 xyu lnln −= ，則xy

yu 1−′

=′ ，即x

uyy 1

+′=′

代入 O.D.E. xy

yyyxlnln −

=−′ xyy

yxlnln

11-−

=′

⇒

可得u

ux 111 =−+′ udx

duxu

ux 11 =⇒=′⇒

由分離變數法得 x

dxudu =

再積分得 ∫∫ = xdxudu cxu +=⇒ ln

21 2

[ ] cxxy +=−⇒ lnlnln21 2

B.C. 令 2,2x == y 代入，得 [ ] c+=− 2ln2ln2ln21 2

2ln−=⇒ c

故解答為 [ ] 2lnlnlnln21 2 −=− xxy

範例 1

92北科-78

Find the general solution of differential equation 116 2)4( +=+ tyy . (10%)【92 北科機電整合】

【範圍】§3-3 【詳解】 齊性解： )2(444 216016 ππ kiemm +=−=⇒=+

)

21

4(

2ππ ki

em+

=⇒ )3,2,1,0( =k (複數方根,詳見 ch18.)

ππππ47

45

43

4 2,2,2,2iiii

eeeem =⇒

)(),(),(),( i12i12i12i12m −−−+−+=⇒

)(),( i12i12m ±−±=⇒

)sincos()sincos( t2ct2cet2ct2cey 43t2

21t2

n +++=∴−

特解：由待定係數法，令 CBtAtyp ++=2

代入得161,0,

161

=== CBA ，故161

161 2 += typ

通解：

)sincos( t2ct2ceyyy 21t2

pn +=+=

)sincos( t2ct2ce 43t2 ++ −

161

161 2 ++ t

【另解】由逆算子法

161

161)1(

161 22

4 +=++= tt

Dyp

範例 2

92北科-79

Solve the initial value problem )(4 tfyy =+′′ , 1)0( =y , 0)0( =′y with

⎩⎨⎧

≥

92北科-80

得 L+++′′′

+′′

+′

+= 5)5(

4)4(

32

!5)0(

!4)0(

!3)0(

!2)0(

!1)0()0()( xyxyxyxyxyyxy

LL+−+−−= 543!5