100 Math 1

目錄

100 99–11. 台灣大學機械工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–12. 台灣大學工程科學研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–113. 台灣大學應用力學研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–164. 台灣大學生物機電工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–235. 台灣大學化學工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–316. 台灣大學環境工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–427. 台灣大學電機工程研究所(C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–498. 台灣大學土木工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–599. 台灣大學物理研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–6510. 台灣科技大學機械工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–7011. 台灣科技大學化學工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–7712. 清華大學動力機械工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–8313. 清華大學工程科學研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–9014. 中山大學材料工程研究所 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–10015. 台聯系統 A 卷 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99–108

I

100試詳解 99–1

1. 100 台灣大學機械工程研究所

1. (1) Solve x3y′′′ + xy′ − y = 0

(2) Solve (y − x√x2 + y2) dx+ (x− y

√x2 + y2)dy = 0. (15%)

《喻超凡 100台大機械》

《解》 �(1) 令 y = xm 代回 ODE 中可得

m(m− 1)(m− 2) +m− 1 = 0 ⇒ m = 1 , 1 , 1

故

y(x) = c1x+ c2x ln x+ c3x(ln x)2

(2) ODE 可改寫成

d{xy − 1

3(x2 + y2)

32} = 0

故 ODE 的通解為

xy − 1

3(x2 + y2)

32 = c

2. (15%) Knowing L {J0(t)} =1√s2 + 1

and y(0) = 1, y′(0) = 0; use Laplace

transform to solve ty′′ + y′ + ty = 0. 《喻超凡 100台大機械》

《解》 � 對 ODE 兩端取 Laplace 轉換可得

L {ty(t) + y′(t) + ty(t)} = 0

即

− d

ds{s2Y (s) − sy(0) − y′(0)} + sY (s) − y(0) − d

dsY (s) = 0

其中 L {y(t)} = Y (s) , 故

−2sY (s) − s2dY (s)

ds+ y(0) + sY (s) − y(0) − d

dsY (s) = 0

99–2 喻超凡叢書 http://www.superyu.idv.tw 喻超凡數學工作室叢書

整理可得

(s2 + 1)dY (s)

ds+ sY (s) = 0

即dY (s)

ds+

s

s2 + 1Y (s) = 0

故dY

Y= − s

s2 + 1ds

對上式兩端積分可得

ln |Y (s)| = −1

2ln |s2 + 1| + c∗

兩端取指數可得

Y (s) =c√

s2 + 1

再由初值定理可知

lims→∞

sY (s) = lims→∞

cs√s2 + 1

= c = y(0) = 1

故 c = 1 , 因此 Y (s) =1√s2 + 1

, 則 ODE 的解為

y(t) = L−1{Y (s)} = L

−1{ 1√s2 + 1

} = J0(t)

3. (10%) 《喻超凡 100台大機械》

(1) Write the 3×3 matrix of the geometric transformation representing the z-axis

counterclockwise rotation (i.e., the axis of rotation perpendicular to the x-y

plane).

(2) Find the eigenvalues and eigenvectors of this 3 × 3 transformation matrix.

《解》 �(1) 因 ⎡

⎢⎣x′

y′

z′

⎤⎥⎦ =

⎡⎢⎣

cos θ sin θ 0

− sin θ cos θ 0

0 0 1

⎤⎥⎦⎡⎢⎣x

y

z

⎤⎥⎦

1. 台灣大學機械工程研究所 99–3

故旋轉換矩陣為

A =

⎡⎢⎣

cos θ sin θ 0

− sin θ cos θ 0

0 0 1

⎤⎥⎦

(2) 由

det(A− λI) = (1 − λ)(λ2 − 2 cos θλ+ 1) = 0

可得 A 的特徵值為 λ = 1 、 cos θ ± i sin θ 。將 λ = 1 代回 (A− λI)X = 0 中可得⎡⎢⎣

cos θ − 1 sin θ 0

− sin θ cos θ − 1 0

0 0 0

⎤⎥⎦⎡⎢⎣x1

x2

x3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦

可得對應的特徵向量為

X = c1

⎡⎢⎣

0

0

1

⎤⎥⎦ (c1 �= 0)

將 λ = cos θ + i sin θ 代回 (A− λI)X = 0 中可得⎡⎢⎣−i sin θ sin θ 0

− sin θ −i sin θ 0

0 0 0

⎤⎥⎦⎡⎢⎣x1

x2

x3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦


X = c2

⎡⎢⎣

1

i

0

⎤⎥⎦ (c2 �= 0)

且 λ = cos θ − i sin θ 對應的特徵向量為

X = c3

⎡⎢⎣

1

−i0

⎤⎥⎦ (c3 �= 0)


4. Consider an elastic string of length L, fixed at its ends on the x axis at x = 0 and

x = L. Its displacement function satisfies :

∂2y

∂t2= c2

∂2y

∂x2for 0 < x < L , t > 0

y(x , 0) = f(x) for 0 ≤ x ≤ L,∂y

∂t(x , 0) = g(x) for 0 ≤ x ≤ L.

(1) (8%) For zero initial velocity, which of the functions listed below gives the

correct displacement (Justification of your answer is required to ge credit.)

(a) y(x , t) =2

L

∞∑n=1

{∫ L

0

f(ξ) cosnπξ

Ldξ } sin

nπx

Lcos

nπct

L

(b) y(x , t) =2

L

∞∑n=1

{∫ L

0

f(ξ) sinnπξ

Ldξ } sin

nπx

Lcos

nπct

L

(c) y(x , t) =2

L

∞∑n=1

{∫ L

0

f(ξ) cosnπξ

Ldξ } sin

nπx

Lsin

nπct

L

(d) y(x , t) =2

L

∞∑n=1

{∫ L

0

f(ξ) sinnπξ

Ldξ } sin

nπx

Lsin

nπct

L

(2) (6%) For zero initial velocity and f(x) = 95 sin6πx

L, determine the displace-

ment function.

(3) (6%) For zero initial displacement and g(x) = 95 sin6πx

L, the displacement is

y(x , t) =95

3πcsin

3πx

Lsin

3πct

L. What is the displacement if

f(x) = g(x) = 95 sin6πx

L?

(4) (5%) For zero initial velocity and f(x) = H(x − 1

3L) − H(x − 2

3L), sketch

y(x ,L

6c).

(5) (5%) Following (4), sketch y(x ,L

2c). 《喻超凡 100台大機械》

《解》 �


(1) 因 y(0 , t) = y(L , t) = 0, 故特徵函數為{

sinnπx

L

}∞

n=1, 因此令

y(x , t) =∞∑

n=1

an(t) sinnπx

L

代回原 PDE 中可得

∞∑n=1

a′′n(t) sinnπx

L= c2

∞∑n=1

an(t)(−n2π2

L2) sin

nπx

L

即 ∞∑n=1

{a′′n(t) + (nπc

L)2an(t)} sin

nπx

L= 0

故可得

a′′n(t) + (nπc

L)2an(t) = 0

其解為

an(t) = An cosnπct

L+Bn sin

nπct

L

故 PDE 的解為

y(x , t) =∞∑

n=1

{An cosnπct

L+Bn sin

nπct

L} sin

nπx

L(1)

由 I.C.

y(x , 0) = f(x) =∞∑

n=1

An sinnπx

L

可得

An =2

L

∫ L

0

f(ξ) sinnπξ

Ldξ

再由∂y

∂t(x , 0) = 0 =

∞∑n=1

Bnnπc

Lsin

nπx

L

故 Bn = 0 , 則

y(x , t) =2

L

∞∑n=1

{∫ L

0

f(ξ) sinnπξ

Ldξ} cos

nπct

Lsin

nπx

L

選 (b) 。

(2) 因 g(x) = 0, 故 Bn = 0, 且

y(x , 0) = 95 sin6πx

L=

∞∑n=1

An sinnπx

L

故 A6 = 95 , 其他 An = 0, 因此

y(x , t) = 95 cos6πct

Lsin

6πx

L


(3) 題目給的結果有誤 , 不過不影響求解 , 因

y(x , 0) = f(x) = 95 sin6πx

L=

∞∑n=1

An sinnπx

L

故 A6 = 95 , 其他 An = 0, 且

∂y

∂t(x , 0) = 95 sin

6πx

L=

∞∑n=1

Bnnπc

Lsin

nπx

L

故 B66πc

L= 95, 則 B6 =

95L

6πc, 且其他 Bn = 0 。則

y(x , t) = {95 cos6πct

L+

95L

6πcsin

6πct

L} sin

6πx

L

(4) 令 y(x, , t) = φ(x+mt) 代入 PDE 中 , 可得

c2 −m2 = 0 ⇒ m = ±c

故 PDE 的通解為

y(x , t) = φ(x+ ct) + ψ(x− ct) (1)

代入起始條件 , 可得

y(x , 0) = f(x) ⇒ f(x) = φ(x) + ψ(x) (2)

∂y

∂t(x , 0) = g(x) ⇒ g(x) = cφ′(x) − cψ′(x) (3)

做1

c

∫ x

a

{(3) 式} dξ 可得

1

c

∫ x

a

g(ξ) dξ = φ(x) − φ(a) − ψ(x) + ψ(a) (4)

(2) 式 + (4) 式可得

f(x) +1

c

∫ x

a

g(ξ) dξ = 2φ(x) − φ(a) + ψ(a)

故可得

φ(x) =1

2{f(x) +

1

c

∫ x

a

g(ξ)dξ + φ(a) − ψ(a)} (5)

(2) 式 − (4) 式可得

f(x) − 1

c

∫ x

a

g(ξ) dξ = 2ψ(x) + φ(a) − ψ(a)


故可得

ψ(x) =1

2{f(x) − 1

c

∫ x

a

g(ξ)dξ − φ(a) + ψ(a)} (6)

將 (5) 、 (6) 式代回 (1) 式 , 即可得到一維波 D’Alembert 之解為

y(x , t) = φ(x+ ct) + ψ(x− ct)

=1

2{f(x+ ct) +

1

c

∫ x+ct

a

g(ξ)dξ + φ(a) − ψ(a)}

+1

2{f(x− ct) − 1

c

∫ x−ct

a

g(ξ)dξ − φ(a) + ψ(a)}

=1

2{f(x+ ct) + f(x− ct)} +

1

2c

∫ x+ct

a

g(ξ)dξ − 1

2c

∫ x−ct

a

g(ξ)dξ

=1

2{f(x+ ct) + f(x− ct)} +

1

2c

∫ x+ct

x−ct

g(ξ)dξ

現 f(x) = H(x− 1

3L) −H(x− 2

3L) 、 g(x) = 0, 故

y(x , t) =1

2{H(x+ ct− L

3) −H(x+ ct− 2L

3) +H(x− ct− L

3) −H(x− ct− 2L

3)}

因此

y(x ,L

6c) =

1

2{H(x+

L

6− L

3) −H(x+

L

6− 2L

3) +H(x− L

6− L

3)

−H(x− L

6− 2L

3)}

=1

2{H(x− L

6) −H(x− L

2) +H(x− L

2) −H(x− 5L

6)}

=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

0 ; 0 ≤ x < L/6

1

2; L/6 ≤ x < 5L/6

0 ; 5L/6 ≤ x ≤ L

(5)

y(x ,L

2c) =

1

2{H(x+

L

2− L

3) −H(x+

L

2− 2L

3) +H(x− L

2− L

3)

−H(x− L

2− 2L

3)}

=1

2{H(x+

L

6) −H(x− L

6) +H(x− 5L

6) −H(x− 7L

6)}

=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

1

2; 0 < x < L/6

0 ; L/6 ≤ x < 5L/6

1

2; 5L/6 ≤ x < L


5. (15%) Write down the answers to the following questions : ( Derivations are nor

required. )

(1) Let ψ(r , θ) =cos θ

r2. Evaluate the line integral

∮C

ψ−→n d over a circle C of

radius 2 centered at the origin. ( −→n denotes unit vector normal to the contour

of the circle C. )

(2) Let φ(r , θ) satisfy the 2 - D Laplace equation (i.e., ∇2φ = 0 ); and

φ = cos2 θ − sin2 θ,∂φ

∂r= 2 cos 2θ along the contour of a unit circular disk S.

Evaluate the surface integral

∫∫S

∇φ · ∇φ dA over the circular disk S.

(3) Let−→F =

x

x2 + y2

−→i +

y

x2 + y2

−→j be a 2 - D vector field. Evaluate

∮C

−→F · d−→r

over a closed contour C given by :

《喻超凡 100台大機械》

《解》 �(1) 因 C : x2 + y2 = 4, 故 −→n = −→e r = cos θ

−→i + sin θ

−→j , 且 d = 2 dθ, 則∮

C

ψ−→n d =

∫C

(cos θ

r2)(cos θ

−→i + sin θ

−→j )2 dθ

=

∫ 2π

0

1

22(cos2 θ

−→i + cos θ sin θ

−→j )2 dθ

=π

2

(2) ∫∫S

∇ · (φ∇φ) dA =

∫∫S

(∇φ · ∇φ+ φ∇2φ) dA ( 因在 S 中 ∇2φ = 0 )

=

∫∫S

∇φ · ∇φ dA


=

∮C

φ∇φ · −→n ds

=

∮C

φ(∂φ

∂r−→e r +

1

r

∂φ

∂θ−→e θ) · −→e r dθ

(令 S 的邊界為 C, C 的單位法向量 −→n = −→e r , ds = dθ)

=

∫C

φ∂φ

∂rdθ

=

∫ 2π

0

(cos2 θ − sin2 θ)(2 cos 2θ) dθ

= 2

∫ 2π

0

cos 2θ cos 2θ dθ = 2π

(3) 設 (x0 , y0) 為曲線 C 上的任一點, 因曲線 C 沒有通過原點, 故 (x0 , y0) �= (0 , 0), 則∫C

−→F · d−→r =

∫C

(x dx

x2 + y2+

y dy

x2 + y2)

=

∫C

1

x2 + y2

1

2d(x2 + y2)

=1

2

∫C

d(ln |x2 + y2|)

=1

2ln |x2 + y2|

∣∣∣(x0 , y0)

(x0 , y0)

= 0


6. (15%) Let z = x+iy denote the complex variable, z = x−iy the complex conjugate

of z, and f(z) is complex function. Answer the following questions. ( Derivations

are not required )

(1) Evaluate

∮C

z

z − (i/2)d z over C : |z| = 1.

(2) Let f(z) be analytic on the upper-half of z-plane and |f(z)| → 0 as |z| → ∞.

If on the real axis, f(z) takes the formx− i

x2 + 1, find the function f(z).

hint : Apply Cauchy integral formula.

(3) If the Laurent series expansion of f(z) =z

(z − i)(z + 1)2about z = −1 is

denoted by∞∑

n=−∞an(z + 1)n , find

∞∑n=−∞

an =? 《喻超凡 100台大機械》

《解》 �

(1) C : |z| = 1, 故 z = eiθ, θ = 0 ∼ 2π , 則 z = e−iθ =1

z, d z = d(

1

z) = − 1

z2dz , 因此

∮C

z

z − (i/2)d z =

∮C

1

z1

z− i

2

(− 1

z2)dz =

∮C

2

z2(iz − 2)dz

令 f(z) =2

z2(iz − 2), 在 C 內具有 z = 0 的 2 階 pole, 且

Resf(0) = limz→0

d

dz(

2

iz − 2) = − i

2

因此 ∮C

z

z − (i/2)d z =

∮C

2

z2(iz − 2)dz = 2π iResf(0) = 2π i(− i

2) = π

(2) 令

f(z) =

⎧⎨⎩

z − i

z2 + 1; z �= i

α ; z = i

因 f(z) 在上半面為可解析, 令 C : |z − i| = ρ (ρ→ 0+) , 故

α = f(i) =1

2πi

∮C

f(z)

z − idz =

1

2πi

∮C

1

z2 + 1dz = Resf(i) =

1

2i

(3)∞∑

n=−∞an(z + 1)n

∣∣∣z=0

=∞∑

n=−∞an = f(0) = 0

2. 台灣大學工程科學研究所 99–11

2. 100 台灣大學工程科學研究所

1. (20%) Given the Bessel’s equation x2y′′ + xy′ + (x2 − μ2)y = 0 and the solution

can be expressed as y(x) = AJμ(x) + BYμ(x). Find the general solution ( in

terms of the Bessel function) of the equation my′′ + ke−αty = 0 with the indicated

substitution x =2

α

√k

me−αt/2. 《喻超凡 100台大工科》

《解》 � 因

dy

dt=dy

dx

dx

dt=dy

dx(−√k

me−αt/2)

d2y

dt2=

d

dt{dydx

(−√k

me−αt/2)}

=d2y

dx2

dx

dt(−√k

me−αt/2) +

dy

dx

d

dt{−√k

me−αt/2}

=d2y

dx2(

√k

me−αt/2)2 +

dy

dx{α

2

√k

me−αt/2}

=α2

4x2 d

2y

dx2+α2

4xdy

dx

代回 ODE 中可得

m{α2

4x2 d

2y

dx2+α2

4xdy

dx} + ke−αty = 0

即

{α2

4x2 d

2y

dx2+α2

4xdy

dx} +

k

me−αty = 0

或α2

4x2 d

2y

dx2+α2

4xdy

dx} +

α2

4x2y = 0

整理可得

x2 d2y

dx2+ x

dy

dx+ x2y = 0

故

y(x) = c1J0(x) + c2Y0(x) = c1J0(2

α

√k

me−αt/2) + c2Y0(

2

α

√k

me−αt/2)


2. (20%) If f(x) is a periodic function with period of T , prove that

1

T

∫ T/2

−T/2

f 2(x) dx ≥ a20 +

1

2

k∑n=1

(a2n + b2n)

where w0 =2π

T, a0 =

1

T

∫ T/2

−T/2

f(x) dx ,

an =2

T

∫ T/2

−T/2

f(x) cosnw0x dx ; bn =2

T

∫ T/2

−T/2

f(x) sinnw0x dx

《喻超凡 100台大工科》

《解》 � 因 f(x) 為週期 T 的函數 , 則

f(x) = a0 +∞∑

n=1

(an cos2nπx

T+ bn sin

2nπx

T)

又

< f(x) , f(x) > =

∫ T/2

−T/2

f 2(x) dx

=⟨a0 +

∞∑n=1

(an cos2nπx

T+ bn sin

2nπx

T) ,

a0 +∞∑

m=1

(am cos2mπx

T+ bm sin

2mπx

T)⟩

= a20 < 1 , 1 > +

∞∑n=1

[a2

n < cos2nπx

T, cos

2nπx

T>

+b2n < sin2nπx

T, sin

2nπx

T>]

= a20 T +

∞∑n=1

(a2n

T

2+ b2n

T

2)

故1

T

∫ T/2

−T/2

f 2(x) dx = a20 +

1

2

∞∑n=1

(a2n + b2n)

即1

T

∫ T/2

−T/2

f 2(x) dx ≥ a20 +

1

2

k∑n=1

(a2n + b2n)


3. (20%) Consider the 1D wave equation∂2u

∂t2=∂2u

∂x2and the boundary conditions

are specified as∂u

∂x

∣∣∣x=0

= 0 and∂u

∂x

∣∣∣x=L

= 0. If the initial conditions are given by

u(x , 0) = 1 + 2 cos3πx

L+ 5 cos

7πx

Land

∂u

∂t

∣∣∣t=0

= 0. Find the solution of u(x , t).


《解》 � 因∂u

∂x(0 , t) = 0 、

∂u

∂x(L , t) = 0 , 故特徵函數為

{1 , cos

nπx

L

}∞

n=1, 則令

u(x , t) = a0(t) +∞∑

n=1

an(t) cosnπx

L

代回 PDE 中可得

a′′0(t) +∞∑

n=1

a′n](t) cosnπx

L=

∞∑n=1

an(t)(−n2π2

L) cos

nπx

L

即

a′′0(t) +∞∑

n=1

{a′′n(t) + (nπ

L)2an(t)} cos

nπx

L= 0

即 ⎧⎨⎩a′′0(t) = 0

a′′n(t) + (nπ

L)2an(t) = 0

⇒⎧⎨⎩a0(t) = A0 +B0t

an(t) = An cosnπt

L+Bn sin

nπt

L

故

u(x , t) = A0 +B0t+∞∑

n=1

{An cosnπt

L+Bn sin

nπt

L} cos

nπx

L

且

u(x , 0) = 1 + 2 cos3πx

L+ 5 cos

7πx

L= A0 +

∞∑n=1

An cosnπx

L

故 A0 = 1 、 A3 = 2 、 A7 = 5, 其他 An = 0, 且

∂u

∂t(x , 0) = 0 = B0 +

∞∑n=1

Bnnπ

Lcos

nπx

L

故 B0 = 0 、 Bn = 0, 因此

u(x , t) = 1 + 2 cos3πt

Lcos

3πx

L+ 5 cos

7πt

Lcos

7πx

L


4. (20%) For an Eigenvalues problem Ax = λx. prove that the eigenvalues are real

if A is a Hermitian matrix. 《喻超凡 100台大工科》

《証》 � 設 λ 為 Hermitian 矩陣 A 的特徵值且其對應的特徵向量為 x , 故滿足

Ax = λx (1)

將 (1) 式兩端取共軛轉置 , 可得

(Ax)∗ = (λx)∗

即

x∗A∗ = λ x∗ (2)

因 A 為 Hermitian 矩陣 , 故滿足 A∗ = A , 將其代入 (2) 式中 , 可得

x∗A = λ x∗

兩端後乘 x 可得

x∗Ax = λ x∗x ⇒ x∗λx = λx∗x

故整理可得

(λ− λ)x∗x = 0

因 x∗x �= 0 , 則 λ = λ 即 λ 為實數。

5. (20%) Consider the following two initial values problems :

(1) my + cy + ky = δ(t) I.C. y(0) = y(0) = 0, δ(t) : Dirac Delta function.

(2) my + cy + ky = r(t) I.C. y(0) = y(0) = 0 .

IF h(t) is the solution of problem (1), show that the solution of problem (2)

can be expressed by :

y(t) =

∫ t

0

h(t− τ)r(τ) dτ



《解》 �(1) 對

my + cy + ky = δ(t)

兩端取 L–T 可得

m{s2Y (s) − sy(0) − y(0)} + c{sY (s) − y(0)} + kY (s) = 1

其中 L {y(t)} = Y (s), 故可得

Y (s) =1

ms2 + cs + k

則

h(t) = L−1{Y (s)} = L

−1{ 1

ms2 + cs+ k}

(2) 對

my + cy + ky = r(t)

兩端取 L–T 可得

m{s2Y (s) − sy(0) − y(0)} + c{sY (s) − y(0)} + kY (s) = R(s)

其中 L {r(t)} = R(s), 故可得

Y (s) =1

ms2 + cs+ kR(s)

則

y(t) = L−1{Y (s)} = L

−1{ 1

ms2 + cs+ kR(s)}

= h(t) ∗ r(t) =

∫ t

0

h(t− τ)r(τ)d τ


3. 100 台灣大學應用力學研究所

1. (30%) Suppose A is a real symmetric square matrix such that A2 = A.

(a) (5%) Find the determinant of A (i.e.; detA ).

(b) (5%) Find the eigenvalues of A.

(c) (5%) Let x be a column vector and the symbol |x| denote it vector length.

Find (Ax)T (x − Ax) (i.e.; the inner product of two column vectors Ax and

(x−Ax).

(d) (5%) Show that |Ax| ≤ 1 for any column vector x with |x| = 1.

(e) (10%) Suppose A here denote a real symmetric 3× 3 matrix. Let t, m, and n

be its orthonormal eigenvectors such that

Am = m , An = n , At = 0

|m| = 1 , |n| = 1 , |t| = 1 , tTm = tTn = mTn = 0

Let P = [m |n | t ] and Λ =

⎡⎢⎣

1 0 0

0 1 0

0 0 0

⎤⎥⎦ be the 3 × 3 matrices containing the

eigenvectors and eigenvalues of A. Suppose t = (t1 , t2 , t3)T . Find PΛP−1 in

terms of t1, t2, t3. 《喻超凡 100台大應研》

《解》 �(a) 因 A2 = A , 故 det(A2) = {det(A)}2 = det(A), 因此 det(A) = 0 或 det(A) = 1 。

(b) 設 λ 為 A 的特徵值, 且對應的特徵向量為 v, 則

Av = λv

前乘 A 可得

A2v = λAv ⇒ Av = λAv ⇒ (λ− 1)Av = 0

故可得 Av = 0 或 λ− 1 = 0, 則 A 的特徵值為 λ = 0 或 λ = 1 。

(c) 因 A 為實對稱矩陣, 故 AT = A, 又 A2 = A, 則 ATA = A2 = A, 因此

(Ax)T (x−Ax) = xTAT (x− Ax) = xTATx− xTATAx = xTAx− xTAx = 0

3. 台灣大學應用力學研究所 99–17

(d) 因 |x| = 1, 故 xTx = 1, 又

|Ax|2 =< Ax , Ax >= (Ax)T (Ax) = xTATAx = xTA2x = xTAx (1)

由 Rayleigh’s 商數定理可知

λmin ≤ xTAx

xTx= xTAx < λmax (2)

因 A 最大的特徵值 λ = 1, 代回 (1) 、 (2) 兩式可得

|Ax|2 = xTAx ≤ 1 ⇒ |Ax| ≤ 1

(e) 因 P = [m | n | t ], 故

PΛP−1 = [m | n | t ]

⎡⎢⎣

1 0 0

0 1 0

0 0 0

⎤⎥⎦⎡⎢⎣mT

nT

tT

⎤⎥⎦ = [m | n | 0 ]

⎡⎢⎣mT

nT

tT

⎤⎥⎦ = mmT + nnT

與 t 無關。

2. (a) (8%) Given a 1st order system ODE :

y′(x) = Ay(x) , x > 0

y(0) =

[−1

1

], where A =

[2 −1

0 2

]. Solve y(x).

(b) (6%) Given a 1st order ODE :

y′(x) =y ln y

x, y(0) = −1

Solve y(x).

(c) (8%) Find the general solution of the following 2nd order ODE :

x2y′′ − xy′ + y = 0

(d) (10%) Given a 2nd order ODE :

y′′(x) + k2y(x) = f(x) , x > 0

where k is a real number, and f(x) is an arbitrary piecewisely continuous real

function. Solve y(x). 《喻超凡 100台大應研》


《解》 �(a) 由 det(A− λI) = 0, 可得 λ = 2 、 2 , 將 λ = 2 代回 (A− λI)V = 0 中可得[

0 −1

0 0

] [v1

v2

]=

[0

0

]

故可得

V = c1

[1

0

](c1 �= 0) 取 V1 =

[1

0

]

令 (A− λI)V2 = V2, 可得 [0 −1

0 0

] [v1

v2

]=

[1

0

]

故 v1 = 0 、 v2 = −1, 因此

V2 =

[0

−1

]

則

y(x) = c1

[1

0

]e2x + c2(

[0

−1

]+ x

[1

0

]) e2x

再由

y(0) =

[−1

1

]= c1

[1

0

]+ c2

[0

−1

]

可得 c1 = −1 、 c2 = −1, 因此

y(x) = −[

1

0

]e2x − (

[0

−1

]+ x

[1

0

]) e2x =

[−e2x − xe2x

e2x

]

(b) 分離變數可得dy

y ln y=

1

xdx

兩端積分可得

ln | ln y| = ln |x| + c∗1

兩端取指數可得 ln y = c1x , 代入 y(0) = −1 , 無解。

(c) 令 y = xm 代回 ODE 中可得,

m(m− 1) −m+ 1 = 0 ⇒ m2 − 2m+ 1 = 0

故 m = 1 、 1 , 則

y(x) = c1x+ c2x ln x

(d) 對 ODE 兩端取 L–T , 可得

s2Y (s) − sy(0) − y′(0) + k2Y (s) = F (s)


其中 L {y(x)} = Y (s) 、 L {f(x)} = F (s), 整理可得

Y (s) =s y(0)

s2 + k2+

y′(0)

s2 + k2+

F (s)

s2 + k2

故

y(x) = L−1{Y (s)}

= y(0) cos kx+y′(0)

ksin kx+

1

kf(x) ∗ sin kx

= y(0) cos kx+y′(0)

ksin kx+

1

k

∫ x

0

f(τ) ∗ sin k(x− τ) dτ

3. The height of a hill(in meter) is given by

h(x , y) = 10(2xy − 3x2 − 4y2 − 18x+ 28y + 12)

where y is the distance(in km) north, x the distance east of Heaven city.

(a) (3%) Where is the top of the hill located ?

(b) (2%) How high is the hill ?

(c) (3%) How steep is the slope at a point 1 km north and 1 km east of Heaven

city, and in what direction ?

(d) (2%) Calculate ∇ · ∇h and ∇×∇h. 《喻超凡 100台大應研》

《解》 �(a) 由 ⎧⎪⎪⎨

⎪⎪⎩∂h

∂x= 10(−18 − 6x+ 2y) = 0

∂h

∂y= 10(28 + 2x− 8y) = 0

可求得臨界點為 x = −2 、 y = 3 , 故 hill 最高點為在 x = −2 km 、 y = 3 km 處。

(b) hill 最高為 h(−2 , 3) = 720 m

(c) 因

∇h =∂h

∂x

−→i +

∂h

∂y

−→j = {10(−18 − 6x+ 2y)}−→i + {10(28 + 2x− 8y)}−→j


因 ∇h(1 , 1) = −220−→i + 220

−→j , 故在 (1 , 1) 處的最大 slope 為

|∇h(1 , 1)| = 220√

2

方向為−−→i +

−→j√

2

(d) ∇ · ∇h =∂2h

∂x2+∂2h

∂y2= −60 − 80 = −140

∇×∇h = 0

4. (10%) Evaluate the integral

∮C

{2xy dx + (ex + x2)dy} along the curve C, where

C is the boundary of the triangle with vertices (0 , 0), (1 , 0) and (1 , 1), along

clockwise direction. 《喻超凡 100台大應研》

《解》 � 設 C 所圍的區域為 D , 由 Green’s 定理可知∮C

{2xy dx+ (ex + x2)dy} = −∫∫

D

{ ∂∂x

(ex + x2) − ∂

∂y(2xy)} dxdy

= −∫ x=1

x=0

∫ y=x

y=0

ex dy dx

= −∫ x=1

x=0

xex dx = −(xex − ex)∣∣∣x=1

x=0= 1


5. The one dimensional wave equation

∂2u

∂t2= c2

∂2u

∂x2

describes the transverse displacement, u(= u(x , t)), of an elastic stretched string,

where c =√T/ρ is the wave speed, with T the tension in the string, and ρ the

density of the string.

(a) (15%) Solve the equation subject to boundary conditions u(0 , t) = u( , t) = 0

and initial conditions u(x , 0) = f(x),∂u

∂t(x , 0) = g(x). Here is the length

of the string.

(b) (5%) Describe the physics of the problem, including the equation, the bound-

ary conditions, the initial conditions and the solution. State also the

assumptions for deriving the wave equation. 《喻超凡 100台大應研》

《解》 �(a) 因 u(0 , t) = u( , t) = 0, 故令

u(x , t) =∞∑

n=1

an(t) sinnπx


∞∑n=1

a′′n(t) sinnπx

= c2

∞∑n=1

an(t)(−n2π2

2) sin

nπx

整理可得 ∞∑n=1

{a′′n(t) + (nπc

)2an(t)} sin

nπx

= 0

即

a′′n(t) + (nπc

)2an(t) = 0

故

an(t) = An cosnπc t

+Bn sin

nπc t

因此

u(x , t) =∞∑

n=1

{An cosnπc t

+Bn sin

nπc t

} sin

nπx

(1)

再由

u(x , 0) = f(x) =∞∑

n=1

An sinnπx


可得

An =2

∫ �

0

f(x) sinnπx

dx

又

ut(x , 0) = g(x) =∞∑

n=1

Bn(nπc

) sin

nπx

故

Bn(nπc

) =

2

∫ �

0

g(x) sinnπx

dx

將 An 、 Bn 代回 (1) 式中即為所求。

(b) PDE 為波動方程式或雙曲線型的 PDE, 兩端固定, 且初始變位為 f(x), 質點初始速度

為 g(x), 方程式在推導時的相關假設 :

(1) 弦的單位長質量為 ρ , 且變形為 u(x , t) 。

(2) 弦為完全彈性體 , 不承受 bending 。

(3) 重力影響不計。

(4) 只考慮橫向運動 (u 方向) 。

4. 台灣大學生物機電工程研究所 99–23

4. 100 台灣大學生物機電工程研究所

1. 解下列微分方程式 2(3x+ 1)2y′′ + 21(3x+ 1)y′ + 18y = 0. (10%)

《喻超凡 100台大生機》

《解》 � 令 y = (3x+ 1)m, 代回 ODE 中可得

2 · 32m(m− 1) + 21 · 3m+ 18 = 0

即

6m2 + 15m+ 6 = 0 ⇒ m = −2 , −1

2

則

y(x) = c1(3x+ 1)−2 + c2(3x+ 1)−12

2. 以 Laplace Transformation 解下列微分方程式 (15%)

y′′ + 6y′ + 8y = e−3t − e−5t , y(0) = 0 , y′(0) = 0


《解》 � 對 ODE 兩端取 L–T 可得

s2Y (s) − sy(0) − y′(0) + 6{sY (s) − y(0)} + 8Y (s) =1

s+ 3− 1

s+ 5

其中 L {y(t)} = Y (s) , 整理可得

(s2 + 6s+ 8)Y (s) =1

s+ 3− 1

s+ 5=

2

(s+ 3)(s+ 5)

即

Y (s) =2

(s+ 2)(s+ 4)(s+ 3)(s+ 5)

=1

3(s+ 2)− 1

s+ 3+

1

s+ 4− 1

3(s+ 5)


故

y(t) = L−1{Y (s)} =

1

3e−2t − e−3t + e−4t − 1

3e−5t

3. 解下列微分方程式 yex dx+ (2y + ex)dy = 0. (10%) 《喻超凡 100台大生機》

《解》 � 原式可改寫成 d(yex + y2) = 0 , 故 ODE 的通解為

yex + y2 = c

4. 求下列微分方程式之通解 y′′′ − 6y′′ + 12y′ − 8y =√

2x e2x. (15%)


《解》 �(1) 齊次解 : 令 y = emx 代回 ODE 中可得

m3 − 6m2 + 12m− 8 = 0 ⇒ m = 2 , 2 , 2

故

yh(x) = c1e2x + c2xe

2x + c3x2e2x

(2) 特解 :

yp(x) =1

(D − 2)3

√2x e2x = e2x 1

D3

√2x

= e2x

∫∫∫ √2x dxdxdx =

8√

2

105e2xx7/2

(3) 通解 : y(x) = yh(x) + yp(x)


5. 參考圖一所示的矩形金屬板,金屬板上 (x , y)點處的溫度為 T (x , y) = 5+2x2 +y2。

(1) 此金上屬板上溫度最低點位於何處 ? (1%)

(2) 金屬板上 (4 , 2)點處的昆蟲為了能儘速逃離此高溫點,試問昆蟲應朝那個方向跑

? 假設昆蟲不會飛。 (3%)

(3) 試求此隻昆蟲由 (4 , 2) 點跑至溫度最低點的最佳路徑向量方程式。假設此最佳

路徑的位置向量方程式為 −→r (t) = x(t)−→i + y(t)

−→j 。 (6%)


《解》 �(1) (0 , 0) 處溫度最低。

(2) 因

∇T (x , y) = 4x−→i + 2y

−→j

應沿著溫度最低方向逃跑, 故應朝

− ∇T (4 , 2)

|∇T (4 , 2)| = −16−→i + 4

−→j

4√

17= − 1√

17(4−→i +

−→j )

(3) 因d−→rdt

=dx

dt

−→i +

dy

dt

−→j ‖ 4x

−→i + 2y

−→j

故dx

dt= 4mx ,

dy

dt= 2my

即

x(t) = c1e4mt , y(t) = c2e

2mt


再由 t = 0 時 x(0) = 4 = c1 、 y(0) = 2 = c2 , 故

−→r (t) = x(t)−→i + y(t)

−→j = 4e2t −→i + 2et −→j

6. 參考圖二所示的金屬板。

(1) 試寫出可描述此金屬板穩態溫度分佈函數 u(r , θ) 的偏微分方程式及其邊界條

件。 (3%)

(2) 試求出此金屬板的穩態溫度分佈函數 u(r , θ) 。 (12%) 《喻超凡 100台大生機》

《解》 �(1) 方程式為

∇2u(r , θ) =∂2u

∂r2+

1

r

∂u

∂r+

1

r2

∂2u

∂θ2= 0 (1)

且

u(1

2, θ) = u0 , ur(0 , θ) = 0 , u(r , 0) = 0 , u(r ,

π

4) = 0

(2) 因 u(r , 0) = 0 , u(r ,π

4) = 0 , 故令

u(r , θ) =∞∑

n=1

an(r) sin(4nθ)

代回 (1) 式可得

∞∑n=1

a′′n(r) sin(4nθ) +1

r

∞∑n=1

a′n(r) sin(4nθ) +1

r2

∞∑n=1

an(r)(−16n2) sin(4nθ) = 0

即 ∞∑n=1

{r2a′′n(r) + r a′n(r) − 16n2an(r)} sin(4nθ) = 0


故

r2a′′n(r) + r a′n(r) − 16n2an(r) = 0

上式為 Euler-Cauchy 方程式, 可解得

an(r) = Anr4n +Bnr

−4n

因此

u(r , θ) =∞∑

n=1

{Anr4n +Bnr

−4n} sin(4nθ) (2)

由

u(1

2, θ) = u0 =

∞∑n=1

{An(1

2)4n +Bn(

1

2)−4n} sin(4nθ)

可得

An(1

2)4n +Bn(

1

2)−4n =

8

π

∫ π4

0

u0 sin(4nθ) dθ = −2 u0(−1 + cosnπ)

nπ(3)

再由

ur(1 , θ) = 0 =∞∑

n=1

{4nAn − 4nBn} sin(4nθ)

故

4nAn − 4nBn = 0 (4)

聯立 (3) 式、 (4) 式可解得

An = Bn = −21+4nu0(−1 + cosnπ)

(1 + 28n)nπ

將 An 、 Bn 代回 (2) 式即為所求。

7. 參考圖三所示的電路。

(1) 設 I = [i1(t) i2(t)]T 。試寫出可描述此電路之電流 i1(t) 與 i2(t) 的微分方程式

系統 , 即dI

dt= AI +BE 。 (3%)

(2) 若 R1 = 8Ω , R2 = 3Ω , L1 = 1H , L2 = 1H , E(t) = 100 sin t V, i1(0) = 0

與 i2(0) = 0 。試以矩陣理論的參數變異 (variation of parameters) 法求解此系

統方程式。 (12%) 《喻超凡 100台大生機》

《解》 �


(1) 由圖三可知 ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

i1 = i2 + i3

i1R1 + L1di2dt

+ L2di1dt

= E

i3R2 − L1di2dt

= 0

整理可得 ⎧⎪⎪⎨⎪⎪⎩L2di1dt

= E − i1 R1 − i3 R2 = E − i1(R1 +R2) + i2R2

L1di2dt

= i3R2 = (i1 − i2)R2

故寫成dI

dt= AI +BE (1)

的型式時

A =

[−(R1 +R2)/L2 R2/L2

R2/L1 −R2/L1

], B =

[1/L

0

]

(2) 現 L1 = L2 = 1, R1 = 8 、 R2 = 3 、 E = 100 sin t 代回 (1) 式中可得

A =

[−11 3

3 −3

], B =

[1

0

]

由 det(A− λI) = 0, 可得 λ = −2 、 −12, 將 λ = −2 代回 (A− λI)X = 0 中可得[−9 3

3 −1

] [x1

x2

]=

[0

0

]

可解得對應的特徵向量為

X1 =

[1

3

]

將 λ = −12 代回 (A− λI)X = 0 中可得[1 3

3 9

] [x1

x2

]=

[0

0

]



X2 =

[3

−1

]

故 ODE 的故聯立 ODE 的基本矩陣 (fundamental matrix) 為

Q = [X1e−2t X2e

−12t] =

[e−2t 3e−12t

3e−2t −e−12t

]

則 Ih = QC, 其中 C 為 2 × 1 任意常數矩陣, 即 Q′ = AQ 。

(b) 求特解 Ip : 令 Ip = QY , 其中

Y =

[y1(t)

y2(t)

]

故 I ′p = Q′Y +QY ′, 代回 (1) 式 (原 ODE) 中可得

Q′Y +QY ′ = AQY +BE ⇒ QY ′ = (Q′ − AQ)Y +BE ⇒ QY ′ = BE

故

Y =

∫Q−1BE dt =

∫1

10

[e2t 3e2t

3e12t −e12t

] [100 sin t

0

]dt

=

∫ [10e2t sin t

30e12t sin t

]dt =

⎡⎣ 4e2t sin t− 2e2t cos t

72

29e12t sin t− 6

29e12t cos t

⎤⎦

因此

Ip = QY =

[e−2t 3e−12t

3e−2t −e−12t

]⎡⎣ 4e2t sin t− 2e2t cos t72

29e12t sin t− 6

29e12t cos t

⎤⎦

=

⎡⎣

332

29sin t− 76

29cos t

276

29sin t− 168

29cos t

⎤⎦

故

I = Ih + Ip = c1

[e−2t

3e−2t

]+ c2

[3e−12t

−e−12t

]+

⎡⎣

332

29sin t− 76

29cos t

276

29sin t− 168

29cos t

⎤⎦

再由

I(0) =

[i1(0)

i2(0)

]= c1

[1

3

]+ c2

[3

−1

]+

⎡⎣ −76

29

−168

29

⎤⎦ =

[0

0

]


故 c1 = 2 、 c2 =6

29, 即故

I = 2

[e−2t

3e−2t

]+

6

29

[3e−12t

−e−12t

]+

⎡⎣

332

29sin t− 76

29cos t

276

29sin t− 168

29cos t

⎤⎦

8. 試求由力−→F = (−16y + sin x2)

−→i + (4ey + 3x2)

−→j 沿著圖四所示之簡單封閉曲線 C

所做的功 , 其中 C 由 C1 、 C2 及 C3 所組成。 (10%) 《喻超凡 100台大生機》

《解》 � ∮C

−→F · d−→r =

∮C

(−16y + sin x2)dx+ (4ey + 3x2)dy

=

∫∫R

{ ∂∂x

(4ey + 3x2) − ∂

∂y(−16y + sin x2)}dxdy

=

∫∫R

(6x+ 16)dxdy = (6x+ 16)A ( A 為 R 所圍的面積 )

= 16 · π4

= 4π

5. 台灣大學化學工程研究所 99–31

5. 100 台灣大學化學工程研究所

1. Please find the general solution of the following ordinary differential equation

(O.D.E.s):

(a) (6%) (x sin θ)dθ + (x3 − 2x2 cos θ + cos θ)dx = 0

(b) (6%) (xD4 +D3)y = 150x4 . 《喻超凡 100台大化工》

《解》 �(a) 令 {

M = x3 − 2x2 cos θ + cos θ

N = x sin θ

故 ⎧⎪⎪⎨⎪⎪⎩∂M

∂θ= 2x2 sin θ − sin θ

∂N

∂x= sin θ

因∂M

∂θ− ∂N

∂xN

=2x2 sin θ − sin θ − sin θ

x sin θ= 2x− 2

x

則積分因子為

I = exp{∫

(2x− 2

x) dx} = ex2−2 ln |x| =

ex2

x2

乘回 ODE 中可得

ex2

xsin θ dθ + xex2 − 2ex2

cos θ +ex2

x2cos theta)dx = 0

即

d(−ex2

xcos θ +

1

2ex2

) = 0


−ex2

xcos θ +

1

2ex2

= c

(b) 原式可改寫成

x4y(4) + x3y′′′ = 150x7

令 x = et, 故 t = ln x, 故

x3y′′′ = Dt(Dt − 1)(Dt − 2)y , x3y(4) = Dt(Dt − 1)(Dt − 2)(Dt − 3)y


其中 Dt =d

dt, 代回 ODE 中可得

Dt(Dt − 1)(Dt − 2)(Dt − 3)y +Dt(Dt − 1)(Dt − 2)y = 150e7t

即

Dt(Dt − 1)(Dt − 2)2y = 150e7t (1)

(1) 齊次解 : 令 y = emt 代回 (1) 式中可得

m(m− 1)(m− 2)2 = 0 ⇒ m = 0 , 1 , 2 , 2

故

yh = c1 + c2et + c3e

2t + c4t e2t = c1 + c2x+ c3x

2 + c4x2 ln x

(2) 特解 :

yp =1

Dt(Dt − 1)(Dt − 2)2150e7t =

150

7 × 6 × 52e7t =

1

7x7

(3) 通解 : y(x) = yh + yp = c1 + c2x+ c3x2 + c4x

2 ln x+1

7x7

2. (10%) Solving the following ODE by Laplace transform.

y′′ + 3y′ + 2y = r(t) , r(t) =

{1 if 1 < t < 2

0 otherwise

y(0) = y′(0) = 0. 《喻超凡 100台大化工》

《解》 � 因

r(t) =

{1 if 1 < t < 2

0 otherwise= H(t− 1) −H(t− 2)

故

L {r(t)} =1

se−s − 1

se−2s

對 ODE 兩端取 L–T

L {y′′ + 3y′ + 2y} = L {r(t)}可得

s2Y (s) − sy(0) − y′(0) + 3{sY (s) − y(0)} + 2Y (s) =1

se−s − 1

se−2s

其中 L {y(t)} = Y (s), 整理可得

(s2 + 3s+ 2)Y (s) =1

se−s − 1

se−2s


即

Y (s) =1

s(s+ 1)(s+ 2)e−s − 1

s(s+ 1)(s+ 2)e−2s

= { 1

2s− 1

s + 1+

1

2(s+ 2)} − { 1

2s− 1

s+ 1+

1

2(s+ 2)}e−2s

因此

y(t) = L−1{Y (s)}

= {1

2− e−(t−1) +

1

2e−2(t−1)}H(t− 1) − {1

2− e−(t−2) +

1

2e−2(t−2)}H(t− 2)

3. (a) (6%) Find the inverse Laplace transform of F (s).

F (s) =s2 + 2s+ 3

(s2 + 2s+ 2)(s2 + 2s+ 5)

(b) (6%) Find the Laplace transform of f(t)

f(t) = eat sinh t

t( a is a constant)

《喻超凡 100台大化工》

《解》 �(a)

L−1{F (s)} = L

−1{ s2 + 2s+ 3

(s2 + 2s+ 2)(s2 + 2s+ 5)}

= L−1{ (s+ 1)2 + 2

[(s+ 1)2 + 1] [(s+ 1)2 + 4]}

= e−tL

−1{ s2 + 2

(s2 + 1)(s2 + 4)}

= e−tL

−1{ 1

3(s2 + 1)+

2

3(s2 + 4)}

= e−t(1

3sin t+

1

3sin 2t)

(b)

L {f(t)} = L {eat sinh t

t} =

∫ ∞

s

1

(s− a)2 − 1ds


=

∫ ∞

s

1

(s− a− 1)(s− a + 1)ds

=

∫ ∞

s

{ 1

2(s− a− 1)− 1

2(s− a+ 1)} ds

=1

2{ln |s− a− 1| − ln |s− a+ 1|}

∣∣∣∞s

=1

2ln |s− a+ 1

s− a− 1|

4. For the following ODE

x(x− 1)y′′ + (3x− 1)y′ + y = 0

(a) (5%) Is x = 0 an ordinary point, regular singular point, or irregular singular

point ?

(b) (10%) Find power series solutions near x = 0. 《喻超凡 100台大化工》

《解》 �(a) ODE 可改寫成

y′′ +3x− 1

x(x− 1)y′ +

1

x(x− 1)y = 0

故

P (x) =3x− 1

x(x− 1), Q(x) =

1

x(x− 1)

因 x = 0 為 P (x) 及 Q(x) 的奇點, 但 x = 0 為 xP (x) =3x− 1

x− 1及

x2Q(x) =x

x− 1的常點, 故 x = 0 為 ODE 的規則奇點。

(b) 因 x = 0 為 ODE 的規則奇點 , 令

y =∞∑

n=0

anxn+r (a0 �= 0 , 0 < |x| < 1)

代回 ODE 中 , 可得

x2∞∑

n=0

an(n+ r)(n+ r − 1)xn+r−2 − x∞∑

n=0

an(n+ r)(n+ r − 1)xn+r−2

+3x∞∑

n=0

an(n + r)xn+r−1 −∞∑

n=0

an(n+ r)xn+r−1 +∞∑

n=0

anxn+r = 0


整理可得

−r2a0xr−1 +

∞∑n=0

[(n+ r + 1)2an − (n + r + 1)2an+1]xn+r = 0

故可得 {r2a0 = 0

(n+ r + 1)2an − (n+ r + 1)2an+1 = 0 n = 0 , 1 , 2 · · ·因 a0 �= 0 , 故可得 r2 = 0 , 則 r = 0 , 0 , 且

(n+ r + 1)2an − (n + r + 1)2an+1 = 0

即 an+1 = an , 故

a1 = a0 , a2 = a1 = a0 , · · · an = a0

則

y1 =∞∑

n=0

anxn+r∣∣∣r=0

= a0

∞∑n=0

xn = a01

1 − x

y2 =∂y

∂r

∣∣∣r=0

= a0

∞∑n=0

xn+r ln x∣∣∣r=0

= a0

∞∑n=0

xn ln x = a0ln x

1 − x

故

y(x) = c1y1 + c2y2 = c11

1 − x+ c2

ln x

1 − x

5. (10%) Prove that let A be a real, symmetric matrix, then there is a real, orthogonal

matrix that diagonalize A. 《喻超凡 100台大化工》

《解》 � 由 Schur’s 定理可知, 存在一個正交矩陣 U ( U 滿足 U−1 = UT ) , 使得 U−1AU 為

上三角矩陣, 即

U−1AU = UTAU =

⎡⎢⎢⎢⎣d11 d12 · · · d1n

0 d22 · · · d2n

· · · · · · · · · · · ·0 · · · · · · dnn

⎤⎥⎥⎥⎦

因 A 為實對稱矩陣, 故 AT = A, 對上式兩端取轉置, 可得

(U−1AU)T = UTAU =

⎡⎢⎢⎢⎣d11 0 · · · 0

d12 d22 · · · 0

· · · · · · · · · · · ·d1n · · · · · · dnn

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣d11 d12 · · · d1n

0 d22 · · · d2n

· · · · · · · · · · · ·0 · · · · · · dnn

⎤⎥⎥⎥⎦


比較可得 dij = 0 ( i �= j) , 故存在一個正交矩陣, 使得

U−1AU =

⎡⎢⎢⎢⎣d11 0 · · · 0

0 d22 · · · 0

· · · · · · · · · · · ·0 · · · · · · dnn

⎤⎥⎥⎥⎦

得証。

6. If a vector field−→F = 3x2−→i +2yz

−→j +y2−→k , C : the path connecting (0 , 1 , 2) and

(1 , −1 , 7).

(a) (6%) Please show that−→F is a conservative vector field. Since

−→F is a conser-

vative vector field, there exists a potential function φ. Please also show the

relationship between−→F and φ.

(b) (8%) Please calculate φ and

∫C

−→F · d−→r . 《喻超凡 100台大化工》

《解》 �(a) 因

∇×−→F =

∣∣∣∣∣∣∣∣

−→i

−→j

−→k

∂

∂x

∂

∂y

∂

∂z

3x2 2yz y2

∣∣∣∣∣∣∣∣= 0

故−→F 為保守場, 且 φ 為位能函數, 則 ∇φ =

−→F 。

(b) 由 ∇φ =−→F , 則可得 ⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

∂φ

∂x= 3x2

∂φ

∂y= 2yz

∂φ

∂z= y2

⇒

⎧⎪⎨⎪⎩φ = x3 + f1(y , z)

φ = y2z + f2(x , z)

φ = y2z + f3(x , y)

比較可得 φ(x , y , z) = x3 + y2z + c , 且

∫C

−→F · d−→r = φ(x , y , z)

∣∣∣(1 ,−1 , 3)

(0 , 1 , 2)= (1 + 3) − 2 = 2


7. (15%) Solving the following partial differential equation (PDE) by separation of

variables.∂T

∂t=∂2T

∂x2

I.C. T (x, 0) = f(x).

B.C. (1)∂T (0, t)

∂x= 0 (2)

∂T ( , t)

∂x= −T ( , t). 《喻超凡 100台大化工》

《解》 �(1) 令 T (x , t) = X(x)Y (t) 故

∂2T

∂x2= X ′′(x)Y (t) ,

∂T

∂t= X(x)Y (t)

代回 PDE 中

X(x)Y (t) = X ′′(x)Y (t)

整理可得X ′′

X=Y

Y= −λ (λ 為常數 )

故 {X ′′(x) + λX(x) = 0

Y (t) + λ Y (t) = 0(1)

(2) 由 B.C.∂T (0 , t)

∂x= X ′(0)Y (t) = 0 ⇒ X ′(0) = 0

再由∂T ( , t)

∂x= −T ( , t) ⇒ X ′()Y (t) = −X()Y (t)

可得 X ′() = −X() , 現解

X ′′(x) + λX(x) = 0 , X ′(0) = 0 , X ′() = −X() (2)

Sturm–Liouville 邊界值問題

(i) λ < 0 , 令 λ = −p2 (0 < p <∞) 代入 (2) 式 , 可得

X ′′(x) − p2X(x) = 0


其通解為

X(x) = c1 sinh px+ c2 cosh px

故

X ′(x) = c1p cosh px+ c2p sinh px

代入邊界條件 X ′(0) = 0, 可得 c1 = 0, 再由 X ′() = −X(), 可得

c2p sinh p+ c2 cosh p = 0

因 p sinh p+ cosh p �= 0, 故 c2 = 0 , 因此 X(x) = 0 為 trivial 解。

(ii) λ = 0 , 代入 (2) 式 , 可得 X ′′(x) = 0 , 其通解為

X(x) = c1 + c2x 且 X ′(x) = c2

代入邊界條件 X ′(0) = 0, 可得 c2 = 0, 再由 X ′() = −X(), 可得 c1 = 0 , 因此

X(x) = 0 為 trivial 解。

(iii) λ > 0 , 令 λ = p2 (0 < p <∞) 代入 (2) 式 , 可得

X ′′(x) + p2X(x) = 0

其通解為

X(x) = c1 sin px+ c2 cos px

故

X ′(x) = c1p cos px− c2p sin px

代入邊界條件 X ′(0) = 0, 可得 c1 = 0 , 再由 X ′() = −X(), 可得

−c2p sin p = −c2 cos p

即

c2(p sin p− cos p) = 0

令 c2 �= 0, 可得

p sin p− cos p = 0

即

tan p =1

p(3)

設 (3) 式的解為 pn (n = 1 , 2 , 3 · · ·) , 則

X(x) = c2 cos pnx = Xn (n = 1 , 2 , 3 · · ·)


再將 λ = p2 = p2n 代回 (1) 式中可得

Y (t) + p2nY (t) = 0

上式為一階線性 ODE, 其解為

Y (t) = de−p2nt = Yn

因此

Tn(x , t) = Yn(t)Xn(x) = de−p2ntc2 cos pnx = Ane

−p2nt cos pnx

(3) 由重疊原理知

T (x , t) =∞∑

n=1

Tn(x , t) =∞∑

n=1

Ane−p2

nt cos pnx

代入初始條件

T (x , 0) = f(x) =∞∑

n=1

An cos pnx

故

An =< f(x) , cos pnx >

< cos pnx , cos pnx >(4)

由 (4) 式可求出 An , 再將 An 代入 T (x , t) 中即為所求。

8. You are given the following matrix of 《喻超凡 100台大化工》

A =

[0 4

−4 0

]

(a) (4%) Find the eigenvalues and eigenvectors of A.

(b) (4%) If A is similar to D (a diagonal matrix with eigenvalues as the diagonal

components), what are the transition matrix P and it inverse P−1 ?

(c) (4%) Use the results from (a) and (b) to solve the following system of ODEs :{y′1 = 4y2 + 9t

y′2 = −4y1 + 5

《解》 �


(a) 由 det(A − λI) = λ2 + 16 = 0, 可得 A 的特徵值為 λ = ±4i , 將 λ = 4i 代回

(A− λI)X = 0 中可得 [−4i 4

−4 −4i

] [x1

x2

]=

[0

0

]

可得對應的特徵向量

X = c1

[1

i

](c1 �= 0)

將 λ = −4i 代回 (A− λI)X = 0 中可得[4i 4

−4 4i

] [x1

x2

]=

[0

0

]

可得對應的特徵向量

X = c2

[1

−i]

(c2 �= 0)

(b) 令

P =

[1 1

i −i]

⇒ P−1 = − 1

2i

[−i −1

−i 1

]

則

P−1AP = D =

[4i 0

0 −4i

]

(c) 原式可改寫成 [y1

y2

]′=

[0 4

−4 0

] [y1

y2

]+

[9t

5

]

即

Y ′ = AY +B (1)

其中

Y =

[y1

y2

], A =

[0 4

−4 0

], B =

[9t

5

]

再令 Y = PZ, 其中 Z =

[z1

z2

], 代回 (1) 式可得

PZ ′ = APZ +B ⇒ Z ′ = P−1APZ + P−1B = DZ + P−1B

即 [z1

z2

]′=

[4i 0

0 −4i

] [z1

z2

]+

1

2

[9t− 5i

9t+ 5i

]

故 ⎧⎪⎨⎪⎩z′1 = 4iz1 +

1

2(9t− 5i)

z′2 = −4iz2 +1

2(9t+ 5i)


可解得 ⎧⎪⎨⎪⎩z1 =

29

32+

9t

8i+ c1e

4ti

z2 =29

32− 9t

8i+ c2e

−4ti

因此 ODE 的通解為

Y =

[y1

y2

]= PZ =

[1 1

i −i

]⎡⎢⎣29

32+

9t

8i+ c1e

4ti

29

32− 9t

8i+ c2e

−4ti

⎤⎥⎦

=

⎡⎢⎣

29

16+ c1e

4ti + c2e−4ti

−9t

4+ ic1e

4ti − ic2e−4ti

⎤⎥⎦

=

⎡⎢⎣

29

16+ α cos 4t+ β sin 4t

−9t

4− α sin 4t+ β cos 4t

⎤⎥⎦ ( 其中 α = c1 + c2 、 β = (c1 − c2)i )


6. 100 台灣大學環境工程研究所

1. Solve the following ordinary differential equation (20%)

(1) y′′ − y′ − 2y = 5 cosx

(2) A damped oscillation equation : my′′ + cy′ + ky = 0 ; discuss the solution, if

(a) m = 1, c = 2, k = 2. (b) if m = 1, c = 4, k = 2. 《喻超凡 100台大環工》

《解》 �(1)

(a) 齊次解 : 令 y = emx 代回 ODE 中可得

m2 −m− 2 = 0

可解得 m = 2 、 −1 , 故

yh(x) = c1e2x + c2e

−x

(b) 特解 :

yp =1

D2 −D − 25 cosx = 5

1

−1 −D − 2cosx

= −5D − 3

(D + 3)(D − 3)cosx = −5

D − 3

D2 − 9cosx

= −51

−1 − 9(− sin x− 3 cosx) = −1

2(sin x+ 3 cosx)

(c) 通解 : y(x) = yh(x) + yp(x)

(2)

(a) 因 m = 1 、 c = 2 、 k = 2 , 故 ODE 為

y′′ + 2y′ + 2y = 0

令 y(x) = emx 代回上式, 可得

m2 + 2m+ 2 = 0 ⇒ m = −1 ± i

故

y(x) = e−x(c1 cosx+ c2 sin x)

6. 台灣大學環境工程研究所 99–43

(b) m = 1 、 c = 4 、 k = 2 故 ODE 為

y′′ + 4y′ + 2y = 0

令 y(x) = emx 代回上式, 可得

m2 + 4m+ 2 = 0 ⇒ m = −2 ±√

2

故

y(x) = c1e(−2+

√2)x + c2e

(−2−√2)x

2. The following is the Lucas series (15%)

1 , 3 , 4 , 7 , 11 , 18 , 29 , 47 , 76 , · · ·

a0 = 1 , a1 = 3 , a2 = 4 , a3 = 7 , a4 = 11 , a5 = 18 , · · ·(a) Please find all mathematical rules.

(b) Please find limn→∞

an+1

an

= ?

(c) If F (x) =1

1 − x= 1 + x+ x2 + x3 + · · ·+ xn + · · ·

Please find 1 − 1 + 1 − 1 + 1 − 1 + · · · = ? 《喻超凡 100台大環工》

《解》 �(a) 由數列可知

an + an+1 = an+2 (1)

且 a0 = 1 、 a1 = 3, 令 an = Mn, 代回 (1) 式可得

Mn +Mn+1 = Mn+2

令 Mn �= 0, 故可得

M2 −M − 1 = 0

可解得 M =1

2(1 ±

√5), 故

an = c1(1 +

√5

2)n + c2(

1 −√5

2)n


由 ⎧⎨⎩a0 = 1 = c1 + c2

a1 = 3 = c1(1 +

√5

2) + c2(

1 −√5

2)

故 c1 =1

2(1 +

√5) 、 c2 =

1

2(1 −

√5) , 則

an = (1 +

√5

2)n+1 + (

1 −√5

2)n+1 (n = 0 , 1 , 2 , · · ·)

(b)

limn→∞

an+1

an= lim

n→∞

(1 +

√5

2)n+2 + (

1 −√5

2)n+2

(1 +

√5

2)n+1 + (

1 −√5

2)n+1

= limn→∞

(1 +

√5

2) + (

1 −√5

2)(

1 −√5

1 +√

5)n+1

1 + (1 −√

5

1 +√

5)n+1

=1 +

√5

2

(c) 因

1 − 1 + 1 − 1 + · · · =∞∑

n=0

(−1)n · 1 =∞∑

n=0

(−1)nan

又

limn→∞

an = limn→∞

1 = 1 �= 0

故

1 − 1 + 1 − 1 + · · · =∞∑

n=0

(−1)n · 1

發散。

3. Given A =

⎡⎢⎢⎢⎣

2 1 0 0

0 2 1 0

0 0 2 1

0 0 0 2

⎤⎥⎥⎥⎦

(a) find eigenvalues and eigenvectors of A. (10%)

(b) find all solution of (A− 2I)X = (1 0 0 0)T . (10%)

(c) find eigenvalues and eigenvectors of (A− 2I)100. (10%)

《喻超凡 100台大環工》


《解》 �(a) 由 det(A− λI) = 0, 可得特徵值為 λ = 2 、 2 、 2 、 2 , 將 λ = 2 代回 (A− λI)X = 0,

可得 ⎡⎢⎢⎢⎣

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣x1

x2

x3

x4

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

0

0

0

0

⎤⎥⎥⎥⎦

對應的特徵向量為

X = c1

⎡⎢⎢⎢⎣

1

0

0

0

⎤⎥⎥⎥⎦ (c1 �= 0)

(b) 由 (A− 2I)X = (1 0 0 0)T , 可得

⎡⎢⎢⎢⎣

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣x1

x2

x3

x4

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

1

0

0

0

⎤⎥⎥⎥⎦

可解得 x2 = 1 、 x3 = 0 、 x4 = 0 、 x1 = c, 故

X =

⎡⎢⎢⎢⎣x1

x2

x3

x4

⎤⎥⎥⎥⎦ = c

⎡⎢⎢⎢⎣

1

0

0

0

⎤⎥⎥⎥⎦+

⎡⎢⎢⎢⎣

0

1

0

0

⎤⎥⎥⎥⎦

(c) 因

B = (A− 2I) =

⎡⎢⎢⎢⎣

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

⎤⎥⎥⎥⎦

故由 det(B − λI) = 0, 可得 B 的特徵向量為 λ = 0 、 0 、 0 、 0 , 且對應的特徵向量為

X = c1

⎡⎢⎢⎢⎣

1

0

0

0

⎤⎥⎥⎥⎦ (c1 �= 0)

故 B100 = (A− 2I)100 的特徵值為 0 、 0 、 0 、 0 , 對應的特徵向量仍為 X 不變。


4. Consider the boundary-value problem y′′ + λy = 0, y(0) = y(2π), y′(0) = y′(2π).

Show that except for the case λ = 0, there are two independent eigenfunctions

corresponding to each eigenvalue. (15%) 《喻超凡 100台大環工》

《解》 �

(a) λ < 0 ; 令 λ = −p2 (0 < p <∞) , 故 ODE 可改寫成

y′′(x) − p2y(x) = 0

ODE 的通解為

y(x) = c1epx + c2e

−px (1)

且

y′(x) = p c1epx − p c2e

−px (2)

將邊界條件 y(0) = y(2π) 代回 (1) 式, 可得

c1 + c2 = c1e2pπ + c2e

−2pπ

即

c1(1 − e2pπ) + c2(1 − e−2pπ) = 0 (3)

將邊界條件 y′(0) = y′(2π) 代回 (2) 式, 可得

p c1 − p c2 = p c1e2pπ − p c2e

−2pπ

即

c1(1 − e2pπ) − c2(1 − e−2pπ) = 0 (4)

聯立 (3) 式及 (4) 式, 即可求出 c1 、 c2 , 因∣∣∣∣ 1 − e2pπ 1 − e−2pπ

1 − e2pπ −1 + e−2pπ

∣∣∣∣ �= 0

故由 Cramer’s 法則可得 c1 = c2 = 0 , 將 c1 = c2 = 0 代回 (1) 式 , 可得 y(x) = 0 ,

故 ODE 無非零解 (為 trivial solution) 。

(b) λ = 0 , 故 y′′(x) = 0 , 則 ODE 的通解為

y(x) = c1x+ c2 且 y′(x) = c1 (5)



c2 = c1 2π + c2

故 c1 = 0 , 再將邊界條件 y′(0) = y′(2π) 代回 (5) 式可得, c1 = c1 , 自動滿足, 因

此 c2 為任何數, 故令 c2 �= 0 , 則 λ = 0 為 ODE 的特徵值 , 且對應的特徵函數為

y(x) = c2 (c2 �= 0) 。

(c) λ > 0 ; 令 λ = p2 (0 < p <∞) , 故 ODE 可改寫成

y′′(x) + p2y(x) = 0

ODE 的通解為

y(x) = c1 cos px+ c2 sin px (6)

且

y′(x) = −p c1 sin px+ p c2 cos px (7)


c1 = c1 cos 2pπ + c2 sin 2pπ

即

c1(1 − cos 2pπ) − c2 sin 2pπ = 0 (8)

將邊界條件 y′(0) = y′(2π) 代回 (7) 式, 可得

p c2 = −p c1 sin 2pπ + p c2 cos 2pπ

即

c1 sin 2pπ + c2(1 − cos 2pπ) = 0 (9)

由 Cramer’s 法則可知 , 若 (8) 式、 (9) 式中的 c1 、 c2 要具有非全為 0 的解時∣∣∣∣ 1 − cos 2pπ − sin 2pπ

sin 2pπ 1 − cos 2pπ

∣∣∣∣ = 0

故可得

(1 − cos 2pπ)2 + sin2 2pπ = 0

或

1 − 2 cos 2pπ + cos2 2pπ + sin2 2pπ = 0 則 2 − 2 cos 2pπ = 0

即 cos 2pπ = 1 , 故 2pπ = 2nπ , 即

p = n (n = 1 , 2 , 3 · · ·)


故 ODE 的特徵值為

λ = p2 = n2 = n2 (n = 1 , 2 , 3 · · ·)

對應的特徵函數為

y(x) = c1 cosnx+ c2 sinnx (c1 、 c2 不全為0)

即每一個特徵值均對應到兩個線性獨立的特徵函數。

5. Find the Taylor series of the following functions about the point z = a and deter-

mine the radius of convergence. (20%)

(1) ln z, a = 1.

(2) cos z2 , a = 0. 《喻超凡 100台大環工》

《解》 �(1) 令 t = z − 1, 故對 z = 1 展開, 即對 t = 0 展開, 則

ln z = ln(1 + t) =∞∑

n=0

(−1)n

n+ 1tn+1 =

∞∑n=0

(−1)n

n+ 1(z − 1)n+1

且 |z − 1| < 1, 收斂半徑為 1 。

(2) 因 cos z =∞∑

n=0

(−1)n

(2n)!z2n, 故 cos z2 =

∞∑n=0

(−1)n

(2n)!z4n , 且 |z| <∞, 故收斂半徑為 ∞ 。

7. 台灣大學電機工程研究所 (C) 99–49

7. 100 台灣大學電機工程研究所 (C)第 1 ∼ 13 複選題 (答案可能有一個或多個, 完全答對才給分, 答錯不倒扣) , 考生應作答於

「答案卡」, 第 14 ∼ 15 計算題 , 考生需出計算過程及答案。

1. (10%) Here is an augmented matrix in which ∗ denotes an arbitrary number and

# denotes a nonzero number.⎡⎢⎢⎢⎣

# ∗ ∗ ∗ ∗ ∗0 # ∗ ∗ 0 ∗0 0 # ∗ ∗ #

0 0 0 0 # ∗

⎤⎥⎥⎥⎦

Which of the following statements is (are) true ?

(A) The given augmented matrix is consistent and the solution is unique.

(B) The given augmented matrix is inconsistent and the solution is unique.

(C) The given augmented matrix is consistent and the solution is not unique.

(D) The given augmented matrix is inconsistent and the solution is not unique.

(E) None of the above. 《喻超凡 100台大電機 C》

《解》 � 令 A 為 4 × 5 矩陣且 B 為 4 × 1 的矩陣, 已知

[A|B] =

⎡⎢⎢⎢⎣

# ∗ ∗ ∗ ∗ ∗0 # ∗ ∗ 0 ∗0 0 # ∗ ∗ #

0 0 0 0 # ∗

⎤⎥⎥⎥⎦

故 rank (A) = rank [A|B] = 4 , 則方程式具有無窮多解。故選 (C)

2. (5%) Assume A is an m× n matrix . If rankA = n, then

(A) The columns of A are linearly independent.

(B) Ax = b has at least one solution for every b in Rm.

(C) Every column of its reduced row echelon form contains a pivot position.

(D) Ax = b has at most one solution for every b in Rm.

(E) None of the above statements is true. 《喻超凡 100台大電機 C》


《解》 �(A) True ; rank (Am×n) = n , 則 A 之行向量為線性獨立。

(B) False ; 若 rank (A) �= rank [A|b] 則 Ax = b 無解。

(C) True ; 性質。

(D) True ; 若 rank (A) = rank [A|b] = n , 則 Ax = b 具有唯一解 ,

若 rank (A) = n �= rank [A|b] , 則 Ax = b 無解 , 故 Ax = b 最多只有一解。

故選 (A)(C)(D)

3. (5%) For the four vectors below ,which of the following statements is (are) true ?⎡⎢⎣

3

2

0

⎤⎥⎦

⎡⎢⎣

8

0

6

⎤⎥⎦

⎡⎢⎣

1

2

3

⎤⎥⎦

⎡⎢⎣

2

1

6

⎤⎥⎦

(A) They are linearly independent.

(B) They span R3

(C) They are linearly dependent.

(D) They do not span R3


《解》 � 因 rank

⎡⎢⎣

3 8 1 2

2 0 2 1

0 6 3 6

⎤⎥⎦ = 3 , 故選 (B)(C)

4. (5%) Let V be a subspace of Rn with dimension k. Which of the following state-

ments is (are) true ?

(A) Every linearly independent subset of V contains at least k vectors.

(B) Any finite subset of V containing more than k vectors is linearly dependent.

(C) n ≥ k .

(D) Any finite subset of V containing less thank k vectors is linearly independent.



《解》 �(A) False ; 應為 at most k vectors

(B) True ; 性質。

(C) True ; dim (Rn) ≥ dim (V )

(D) False ; 不一定。

故選 (B)(C)

5. (5%) Suppose that s, t, and u are vectors in Rn such that s is orthogonal to u and

u is orthogonal to t. Then

(A) s is orthogonal to t

(B) For any orthogonal n× n matrix P , we have that Pu is orthogonal to both s

and t

(C) For any orthogonal n× n matrix P , we have that Ps is orthogonal to u

(D) s+ t is orthogonal to u

(E) None of the preceding statements are true. 《喻超凡 100台大電機 C》

《解》 �

(A) False ; 不一定 , 如 s =

[1

0

], u =

[0

1

], t =

[2

0

]

(B) False ; 沒有此性質。

(C) False ; 沒有此性質。

(D) True ; 因 < s , u >= 0 、 < t , u >= 0, 故 < s+ t , u >=< s , u > + < t , u >= 0

選 (D)


6. (5%) Let M be an inner product space

(A) Since the inner product is an integral of a product of functions, M is a function

space

(B) c < u , v >=< cu , cv > for all vectors u and v in M and for every scalar c

(C) < T (u) , v >=< u , T (v) > for all vectors u and v in M and for every linear

operator T on M

(D) If M is finite-dimensional, then M contains an orthonormal basis

(E) None of the preceding statements are true. 《喻超凡 100台大電機 C》

《解》 �(A) True ; 定義。

(B) False ; < cu , cv >= cc < u , v >

(C) False ; < T (u) , v >=< u , T ∗(v) >

(D) True ; 性質。

選 (A)(D)

7. (5%) An m× n matrix P is invertible if

(A) The reduced row echelon form of P is In

(B) The column of P are linearly independent

(C) The column of P span Rm

(D) The rows of P are linearly independent

(E) None of the preceding statements is true. 《喻超凡 100台大電機 C》

《解》 � P 應為 n× n, 即當 m = n 之下 , 選 (A)(B)(C)(D)


8. (5%) Suppose that S is an arbitrary n× n matrix. Then

(A) det(S) is the sum of its diagonal entries

(B) detS2 = 2 detS

(C) detS is a vector in Rn

(D) detS = detST


《解》 �(A) False ;

(B) False ; det(S2) = (det(S))2

(C) False ; det(S) ∈ R


故選 (D)

9. (5%) Let A be a subset of R3 containing two or more vectors. Then :

(A) The span of any two vectors in A is a plane in R3.

(B) If A contains more than three vectors, then A is linearly independent

(C) The span of any two nonzero vectors in A is a plane in R3.

(D) Every vector in A is in the span of A.


《解》 �(A) False ; 不一定成立, 必需 2 向量為線性獨立。

(B) False ; A 中超過 3 個向量, 則必為線性相依。

(C) False ; 理由同 (A) 。



故選 (D)

10. (8%) The solution for f(t) of the following equation

f(t) = 3t2 − e−t −∫ t

0

f(ρ)et−ρdρ

Has the following form : f(t) = g(t) − k1ek2t, where k1, k2 are constants and g(t)

is a function containing only polynomial terms like kntn. Which of the following

item(s) is (are) true :

(A) k1 + k2 = 3 (B) k2 + g(0) = 1 (C) g(1)+ g(0) = 4 (D) g(1)+ g(−1) = 8

(E) none of above. 《喻超凡 100台大電機 C》

《解》 � 原式為

f(t) = 3t2 − e−t − f(t) ∗ et (1)

對 (1) 式兩端取 L–T , 可得

F (s) =6

s3− 1

s+ 1− 1

s− 1F (s)

其中 L {f(t)} = F (s) , 整理可得

F (s) =(s− 1)(−s3 + 6s+ 6)

s4(s+ 1)=

6

s3− 6

s4+

1

s− 2

s+ 1

故

f(t) = L {F (s)} = 3t2 − t3 + 1 − 2e−t

因此 g(t) = 3t2 − t3 + 1 、 k1 = 2 、 k2 = −1 , 故選 (C)(D)

11. (7%) The function f(x) = |x| can be expressed as a Fourier series on the interval

−π ≤ x ≤ π, where f(x) = |x| = a0 +∞∑

n=1

[an cos(nx) + bn sin(nx)]

Which of the following item(s) is (are) true :

(A) a0 = −π (B) a2 + b3 = 0 (C) b2 + b4 = 0 (D) a1 + a3 = −−4π

9(E) none of above . 《喻超凡 100台大電機 C》


《解》 � 因 f(x) = |x| 為偶函數, 故

a0 =1

π

∫ π

0

x dx =π

2

an =2

π

∫ π

0

x cos(nx) dx =2[(−1)n − 1]

πn2

且 bn = 0 , 故選 (B)(C)

12 、13 We are going to solve of the following differential equation system :⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

dx1

dt= 3x1 + x2 − x3

dx2

dt= x1 + 3x2 − x3

dx3

dt= 3x1 + 3x2 − x3

with initial values as x1(0) = 4, x2(0) = 7 and x3(0) = 7. The solution has the

form as ⎧⎪⎨⎪⎩x1 = A11e

λ1t + A12eλ2t + A13e

λ3t

x2 = A21eλ1t + A22e

λ2t + A23eλ3t

x3 = A31eλ1t + A32e

λ2t + A33eλ3t

where |λ1| ≤ |λ2| ≤ |λ3| and Aij are constants. Which of the following item(s) is

(are) true :

(12.) (5%) (A) λ1 +λ2 = 3 (B) 2λ1 +λ3 = 3 (C) λ2 +λ3 = 3 (D) λ1 +λ3 = 3

(E) none of above.

(13.) (5%) (A) x1(1) + 2x2(1) = 6e (B) x1(1) + x3(1) = 2 cos(1)

(C) x2(1) + x3(1) = 8e (D) x1(1) − x2(1) = 2 cos(1) (E) none of above.

《喻超凡 100台大電機 C》

《解》 � 令

A =

⎡⎢⎣

3 1 −1

1 3 −1

3 3 −1

⎤⎥⎦ , X =

⎡⎢⎣x1

x2

x3

⎤⎥⎦


由 det(A− λI) = 0 , 可得 λ = 1 、 2 、 2 , 將 λ = 1 代回 (A− λI)V = 0 中可得⎡⎢⎣

2 1 −1

1 2 −1

3 3 −2

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦


V = c1

⎡⎢⎣

1

1

3

⎤⎥⎦ (c1 �= 0) 取 V1 =

⎡⎢⎣

1

1

3

⎤⎥⎦

將 λ = 2 代回 (A− λI)V = 0 中可得⎡⎢⎣

1 1 −1

1 1 −1

3 3 −3

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦


V = c2

⎡⎢⎣

1

−1

0

⎤⎥⎦+ c3

⎡⎢⎣

1

0

1

⎤⎥⎦ ( c2 、 c3 不全為 0)

取

V2 =

⎡⎢⎣

1

−1

0

⎤⎥⎦ , V3 =

⎡⎢⎣

1

0

1

⎤⎥⎦

則

X = α1 V1 et + α2 V2 e

2t + α3 V3 e2t

= α1

⎡⎢⎣

1

1

3

⎤⎥⎦ et + α2

⎡⎢⎣

1

−1

0

⎤⎥⎦ e2t + α3

⎡⎢⎣

1

0

1

⎤⎥⎦ e2t

由

X(0) =

⎡⎢⎣x1(0)

x2(0)

x3(0)

⎤⎥⎦ =

⎡⎢⎣

4

7

7

⎤⎥⎦ = α1

⎡⎢⎣

1

1

3

⎤⎥⎦+ α2

⎡⎢⎣

1

−1

0

⎤⎥⎦+ α3

⎡⎢⎣

1

0

1

⎤⎥⎦

可解得 α1 = −4 、 α2 = −11 、 α3 = 19 , 故⎡⎢⎣x1(t)

x2(t)

x3(t)

⎤⎥⎦ = −4

⎡⎢⎣

1

1

3

⎤⎥⎦ et − 11

⎡⎢⎣

1

−1

0

⎤⎥⎦ e2t + 19

⎡⎢⎣

1

0

1

⎤⎥⎦ e2t =

⎡⎢⎣−4et − 11e2t + 19e2t

−4et + 11e2t + 0e2t

−12et + 0e2t + 19e2t

⎤⎥⎦


故第 12 題選 (A)(D) , 第 13 題選 (E)

14. (19%) The initial value problem

1

5

d2x

dt2+

6

5

dx

dt+ 2x = 5 cos(4t) , x(0) =

1

2, x′(0) = 0

has the solution with the form :

x(t) =1

102{f(t)[k1 cos(t) + k2 sin(t)] + k3 cos(k4t) + k5 sin(k4t)}

k1, k2, k3, k4, k5 are constants. f(t) is a function of t . Find k1, k2, k3, k4, k5 and

f(t). 《喻超凡 100台大電機 C》

《解》 � 由原式改寫成d2x

dt2+ 6

dx

dt+ 10x = 25 cos(4t) (1)

(1) 齊性解 : 令 x = emt 代回 (1) 式可得

m2 + 6m+ 10 = 0 ⇒ m = −3 ± i

故

xh = e−3t(c1 cos t+ c2 sin t)

(2) 特解 :

xp =1

D2 + 6D + 1025 cos(4t) =

1

−16 + 6D + 1025 cos(4t)

=25

6

(D + 1)

(D − 1)(D + 1)cos(4t) =

25

6

D + 1

D2 − 1cos(4t)

=25

6

1

−16 − 1(−4 sin 4t+ cos 4t) =

1

102(100 sin 4t− 25 cos 4t)

(3) 通解 :

x(t) = xh(t) + xp(t) = e−3t(c1 cos t+ c2 sin t) +1

102(100 sin 4t− 25 cos 4t)

由 ⎧⎪⎨⎪⎩x(0) =

1

2= c1 − 25

102

x′(0) = 0 = c2 − 3c1 +400

102


可得 c1 =38

51、 c2 = −86

51, 故

x(t) = e−3t(38

51cos t− 86

51sin t) +

1

102(100 sin 4t− 25 cos 4t)

15. (6%) The differential equation

x2 d2y

dx2− x

dy

dx+ y = ln(x)

has the solution with the form :

y(x) = c1x+ c2g1(x) + p1 + g2(x)

c1 and c2 are arbitrary constants P1 is a constant. g1(x) and g2(x) are function of

x. Find p1 and g1(x) 《喻超凡 100台大電機 C》

《解》 � 令 x = et, 故 t = ln x, 故

xy′ = Dy , x2y′′ = D(D − 1)y , x3y′′′ = D(D − 1)(D − 2)y

其中 D =d

dt, 代回 ODE 中可得 [ D

D(D − 1)y −Dy + y = t

整理可得

(D − 1)2y = t (1)

(1) 齊性解 : 令 y = emt 代回 (1) 式中可

(m− 1)2 = 0 ⇒ m = 1 , 1

故

yh = c1et + c2te

t = c1x+ c2x ln x

(2) 特解 :

yp =1

(D − 1)2t =

1

1 − 2D +D2t = (1 + 2D + · · ·)t = t+ 2 = 2 + lnx

(3) 通解 : y(x) = yh + yp = c1x+ c2 x ln x+ 2 + ln x , 故 p1 = 2, g1(x) = x ln x 。

8. 台灣大學土木工程研究所 99–59

8. 100 台灣大學土木工程研究所

1. (20%) Give the following matrix A =

[2 4

3 1

](a) (5%) Find the eigenvalues of A.

(b) (5%) Find the eigenvectors of A.

(c) (5%) Diagonalize A, i.e., find matrices P and D such that P−1AP = D, where

D is a diagonal matrix.

(d) (5%) Find A6 ( By using the results from (c)). 《喻超凡 100台大土木》

《解》 �(a) 由

det(A− λI) = (−1)2(λ2 − 3λ− 10) = 0

可得特徵值 λ = 5 、 −2 。

(b) 將 λ = 5 代回 (A− λI)X = 0 中可得[−3 4

3 −4

] [x1

x2

]=

[0

0

]


X = c1

[4

3

](c1 �= 0)

將 λ = −2 代回 (A− λI)X = 0 中可得[4 4

3 3

] [x1

x2

]=

[0

0

]


X = c2

[1

−1

](c2 �= 0)

(c) 令

P =

[4 1

3 −1

]

則

P−1AP = D =

[5 0

0 −3

]


(d) A6 = PD6P−1 = P

[56 0

0 (−3)6

]P

2. (15%) Let L {f(t)} =

∫ ∞

0

f(t)e−st dt be the Laplace transformation of f(t).

(a) (5%) Show that L {t} =1

s2.

(b) (5%) Show that L {f ′′(t)} = s2L {f(t)} − sf(0) − f ′(0).

(c) (5%) Find L {cos(wt+ θ)}, where w and θ are constants.

《喻超凡 100台大土木》

《解》 �(a)

L {t} =

∫ ∞

0

t e−st dt = (−1

st e−st − 1

s2e−st)

∣∣∣∞0

=1

s2

(b) 設 f(t) 、 f ′(t) 、 f ′′(t) 在 t ≥ 0 為連續, 則

L {f ′′(t)} =

∫ ∞

0

f ′′(t)e−st dt

= {e−stf ′(t) + se−stf(t)}∣∣∣∞0

+ s2

∫ ∞

0

f(t)e−st dt

= −f ′(0) − sf(0) + s2L {f(t)}

(c)

L {cos(wt+ θ)} = L {coswt cos θ − sinwt sin θ}=

s cos θ

s2 + w2− w sin θ

s2 + w2

3. (15%) Given a vector−→F = y2−→i + x2−→j + z2−→k , and a cylindrical surface S of

x2 + y2 ≤ 4, 0 ≤ z ≤ 5. Using the Gauss divergence theorem, evaluate the surface

integral. 《喻超凡 100台大土木》

《解》 � 令 S 為 x2 + y2 ≤ 4 , 0 ≤ z ≤ 5 , 所圍的區域為 D , 則∫∫S

(−→F · −→n ) dA =

∫∫∫D

(∇ · −→F ) dxdydz


=

∫∫∫D

2 z dxdydz = 2 z · V = 2 · 5

2· π · 22 · 5

= 100π

4. (a) (10%) Find the general solution of 《喻超凡 100台大土木》

dy(x)

dx= x3(y − x)2 + x−1y , x > 0

(b) (5%) If we further require y(1) = −2, what is the solution of y(x).

《解》 �

(a) 令 u = (y − x) , 故du

dx=dy

dx− 1, 代回 ODE 中可得

du

dx+ 1 = x3u2 +

1

x(u+ x)

整理可得du

dx− 1

xu = x3u2

兩端同除 u2, 可得

u−2du

dx− 1

xu−1 = x3

令 v =1

u代回上式可得

dv

dx+

1

xv = −x3

積分因子為 I = e∫

1x

dx = eln |x| = x, 故

Iv =

∫x(−x3) dx = −1

5x5 + c

即

v =1

u= −1

5x4 +

c

x

ODE 的通解為1

y − x= −1

5x4 +

c

x(1)

(b) 將 x = 1 、 y = −2 代回 (1) 式中可得1

−3= −1

5+ c, 故 c = − 2

15, 因此可得

1

y − x= −1

5x4 − 2

15x


5. (a) (5%) Find the Fourier series expansion for the following function

f(x) = x+ x2 , − π < x < π

(b) (3%) Express the Parseval’s relation in terms of the Fourier coefficients a0, ak,

bk, k = 1 , 2 , · · ·.(c) (7%) Can you prove that the Fourier coefficients in (a) satisfy the Parseval’s

relation. (Hint :∞∑

k=1

1

k4=π4

90and

∞∑k=1

1

k2=π2

6) 《喻超凡 100台大土木》

《解》 �(a) 令

f(x) = a0 +∞∑

k=1

(ak cos kx+ bk sin kx)

其中

a0 =1

2π

∫ π

−π

(x+ x2) dx =π2

3

ak =1

π

∫ π

−π

(x+ x2) cos kx dx =4 cos kπ

k2

bk =1

π

∫ π

−π

(x+ x2) sin kx dx = −2 cos kπ

k

(b)1

2π

∫ π

−π

f 2(x) dx = a20 +

∞∑k=1

(a2

k

2+b2k2

)

(c) 因1

2π

∫ π

−π

f 2(x) dx =1

2π

∫ π

−π

(x2 + x)2 dx =π2

3+π4

5(1)

a20 = (

π2

3)2 =

π4

9(2)

∞∑k=1

(a2

k

2+b2k2

) =∞∑

k=1

1

2(4 cos kπ

k2)2 +

∞∑k=1

1

2(−2 cos kπ

k)2

= 8∞∑

k=1

1

k4+ 2

∞∑k=1

1

k2

= 8 · π4

90+ 2 · π

2

6=

4π4

45+π2

3(3)


由 (1) 、 (2) 、 (3) 式可知,

a20 +

∞∑k=1

(a2

k

2+b2k2

) =π4

9+

4π4

45+π2

3=π4

5+π2

3=

1

2π

∫ π

−π

f 2(x) dx

6. (20%) For the one-dimensional heat conduction equation :

ut = uxx + f(u) , x ∈ R , t > 0 (1)

the solution with u(x , t) = q(x− ct) is called a wave solution, where c ∈ R is the

wave propagation speed. Let ξ = x− ct, and q′ =dq

dξand q′′ =

d2q

dξ2.

(a) (5%) Derive an ordinary differential equation for q(ξ), that Eq.(1) has a wave

solution u = q(x− ct).

(b) (15%) Let

f(u) = u(1 − u)(u− 0.3)

Derive the wave solution u(x − ct) of Eq.(1) with the above f(u). ( Hint

q′ = kq(q − 1), and c and k are to be determined ). Write k = ? and c = ?

《喻超凡 100台大土木》

《解》 �(a) 因 ξ = x− ct, 故

∂u

∂t=du

dξ

∂ξ

∂t= −cdu

dξ

∂u

∂x=du

dξ

∂ξ

∂x=du

dξ,∂2u

∂x2=

∂

∂x(∂u

∂x) =

d2u

dξ2


−cdudξ

=d2u

dξ2+ f(u)

整理可得d2u

dξ2+ c

du

dξ+ f(u) = 0 (1)

(b) 由 hint 可知du

dξ= ku (u− 1)

即du

u(u− 1)= k dξ ⇒ (

1

u− 1− 1

u)du = k dξ


兩端積分可得

ln |u− 1| − ln |u| = k ξ + α∗

兩端取指數可得u− 1

u= α ek ξ ( α 為任意常數 )

即

u =1

1 − α ek ξ(2)

將 (2) 式代回d2u

dξ2+ c

du

dξ+ u(1 − u)(u− 0.3) = 0 (3)

整理可得α ekξ(0.7 − c k − k2 + α ekξ(0.3 + c k − k2))

(−1 + α ekξ)3= 0

即

(0.7 − c k − k2) + α ekξ(0.3 + c k − k2) = 0

比較係數可得 {0.7 − c k − k2 = 0

0.3 + c k − k2 = 0

可解得 {c = −0.282843

k = −0.707107或

{c = 0.282843

k = 0.707107

故 (3) 式的解為

u =1

1 − α ek ξ=

1

1 − α ek (x−ct)

9. 台灣大學物理研究所 99–65

9. 100 台灣大學物理研究所

1. (15%) Which of the following is (are) correct ? (複選題)

(a) Tr(AB) = Tr(BA) holds for arbitrary finite or infinite dimensional matrices

A, B.

(b) For all 3×3 matrices A, it is possible to express det(A) as a function of Tr(A),

Tr(A2), Tr(A3).

(c) The 3 functions 1, x, x2 are linear independent.

(d) The matrix U = exp(A) is unitary whenever A is anti-Hermitian.

(e) Arbitrary vectors A,B, C in 3 dimensional Euclidean space satisfy the relation

A × (B × C) = B(A · C) + C(B · A), where ”×” denotes the cross product,

and ”·” denotes the inner product. 《喻超凡 100台大物理》

《解》 �(a) False ; 要有限維的矩陣。

(b) False ; 設 A 的特徵值為 λ1 、 λ2 、 λ3

λ1λ2λ3 = α(λ31 + λ3

2 + λ33) + β(λ2

1 + λ22 + λ2

3) + γ(λ1 + λ2 + λ3)

不一定成立。故

det(A) = αTr(A3) + β Tr(A2) + γ Tr(A)

不一定成立。

(c) True ; 因 W (1 , x , x2) =

∣∣∣∣∣∣∣1 x x2

0 1 2x

0 0 2

∣∣∣∣∣∣∣ = 2 �= 0 , 故 {1 , x , x2} 為線性獨立。

(d) True ; 因

UU∗ = (exp(A))(exp(A))∗ = (exp(A))(exp(A∗)) = exp(A+ A∗) = I

故 A+ A∗ = 0 , 即 A = −A∗ , 故 A 為反厄米特矩陣。

(e) False ; 應該為 A× (B × C) = B(A · C) − C(B · A)


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

M =

[3/5 4/5

4/5 −3/5

]

(b) (10%) Find the value of Tr(f(M)) for f(x) = x2 + sin x.

《喻超凡 100台大物理》

《解》 �(a) 由

det(A− λI) = (−1)2(λ2 − 1) = 0

可得特徵值為 λ = ±1, 將 λ = 1 代回 (A− λI)X = 0 中可得[−2/5 4/5

4/5 −8/5

] [x1

x2

]=

[0

0

]


X = c1

[2

1

](c1 �= 0)

將 λ = −1 代回 (A− λI)X = 0 中可得[8/5 4/5

4/5 2/5

] [x1

x2

]=

[0

0

]


X = c2

[1

−2

](c2 �= 0)

(b) f(M) = M2 + sinM 的特徵值為 1 + sin 1 、 (−1)2 + sin(−1), 故

Tr(f(M)) = (1 + sin 1) + [(−1)2 + sin(−1)] = 2

3. (15%) Find the general solution for y(x) to the differential equation

d2y

dx2− 2

dy

dx+ y = exp(x)



《解》 �(1) 齊次解 : 令 y(x) = emx 代回 ODE 中可得

m2 − 2m+ 1 = 0 ⇒ m = 1 , 1

故

yh(x) = c1ex + c2x e

x

(2) 特解 :

yp(x) =1

D2 − 2D + 1ex =

x2ex

2

(3) 通解 : y(x) = yh(x) + yp(x)

4. (15%) Solve the differential equation

dy

dx=

y2

1 + x2

for the solution with y(0) = 1. 《喻超凡 100台大物理》

《解》 � 分離變數可得dy

y2=

1

1 + x2dx

兩端積分可得

−1

y= tan−1 x+ c ⇒ y(x) = − 1

tan−1 x+ c

再由 y(0) = 1 = −1

c, 可得 c = −1, 故

y(x) = − 1

tan−1 x− 1

5. (15%) For all non-negative integers n, evaluate the integral∫ ∞

−∞xn exp(−x2/2) dx



《解》 �(1) n = 2m+ 1 (m = 1 , 2 , 3 · · ·) 時∫ ∞

−∞xn exp(−x2/2) dx =

∫ ∞

−∞x2m+1 exp(−x2/2) dx = 0

(2) n = 2m 時∫ ∞

−∞xn exp(−x2/2) dx =

∫ ∞

−∞x2m exp(−x2/2) dx

u = x2m−1 dv = xe−x2/2dx

dv = (2m− 1)x2m−2dx v = −e−x2/2

= −x2m−1e−x2/2∣∣∣∞−∞

+ (2m− 1)

∫ ∞

−∞x2m−2e−x2/2 dx

· · ·= (2m− 1)(2m− 3) · · ·3 · 1

∫ ∞

−∞e−x2/2 dx

= (2m− 1)(2m− 3) · · ·3 · 1 ·√

2π

=(2n)!

2nn!

√2π

6. (15%) Define F (k) by the equality∫ ∞

−∞F (k) exp(ikx) dk =

1

1 + x2

Find F (k). 《喻超凡 100台大物理》

《解》 � 令 f(x) =1

1 + x2, 故

F (k) =1

2π

∫ ∞

−∞

1

1 + x2e−ikx dx (1)

令 g(z) =e−ikz

1 + z2, 則 g(z) 具有 z = ±i 的一階 pole, 且

Resf(i) =ek

2i, Resf(−i) =

e−k

−2i

(1) k > 0 時

F (k) =1

2π

∫ ∞

−∞

1

1 + x2e−ikx dx =

1

2π(−2π i) Resf(−i) = −i · e

−k

−2i=e−k

2


(2) k = 0 時

F (k) =1

2π

∫ ∞

−∞

1

1 + x2dx =

1

2πtan−1 x

∣∣∣∞−∞

=1

2

(3) k < 0 時

F (k) =1

2π

∫ ∞

−∞

1

1 + x2e−ikx dx =

1

2π2π iResf(i) = i · e

k

2i=ek

2

(4) 結論 : F (k) =e−|k|

2(−∞ < k <∞)


10. 100台灣科技大學機械工程研究所

1. Find the general solution of the differential equation (20%)

y′ = (−2x+ y)2 − 7

《喻超凡 100台科大機械》

《解》 � 令 u = (−2x+ y), 故du

dx= −2 +

dy

dx, 代回原 ODE 中可得

du

dx+ 2 = u2 − 7 ⇒ du

dx= u2 − 9

分離變數可得du

u2 − 9= dx

兩端積分 ∫{ 1

6(u− 3)− 1

6(u+ 3)} du =

∫dx

可得1

6ln |u− 3

u+ 3| = x+ c1

兩端取指數, 可得 ODE 的通解為

−2x+ y − 3

−2x+ y + 3= c e6x

2. Solve the initial value problem (20%)

y′′ − 4y′ + 4y = f(t) ; y(0) = y′(0) = 0 with f(t) =

{t for 0 ≤ t < 3

t+ 2 for t ≥ 3


《解》 � 因

f(t) = t{H(t) −H(t− 3)} + (t+ 2)H(t− 3)

10. 台灣科技大學機械工程研究所 99–71

故

L {f(t)} = L {t+ 2H(t− 3)} =1

s2+

2e−3s

s

對 ODE 兩端取 L–T 可得

s2Y (s) − sy(0) − y′(0) − 4{sY (s) − y(0)} + 4Y (s) = L {f(t)} =1

s2+

2e−3s

s


(s2 − 4s+ 4)Y (s) =1

s2+

2e−3s

s

即

Y (s) =1

s2(s− 2)2+

2e−3s

s(s− 2)2

=1

4(s− 2)2− 1

4(s− 2)+

1

4s2+

1

4s+ { 1

(s− 2)2− 1

2(s− 2)+

1

2s}e−3s

故

y(t) = L−1{Y (s)}

=1

4te2t − 1

4e2t +

t

4+

1

4+ {(t− 3)e2(t−3) − 1

2e2(t−3) +

1

2}H(t− 3)

3. Use an orthogonal transformation to transform the real quadratic form

Q = 4x1x2 + 4x2x3 + 4x3x1

into the canonical form. Write down the canonical form and the transformation

matrix. (20%) 《喻超凡 100台科大機械》

《解》 � 令 Q = XTAX, 其中

A =

⎡⎢⎣

0 2 2

2 0 2

2 2 0

⎤⎥⎦ , X =

⎡⎢⎣x1

x2

x3

⎤⎥⎦

由

det(A− λI) = (−1)3(λ3 − 12λ− 16) = 0


可得 A 的特徵值為 λ = 4 、 −2 、 −2 , 將 λ = 4 代入 (A− λI)V = 0 中可得⎡⎢⎣−4 2 2

2 −4 2

2 2 −4

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦

故對應的特徵向量為

V = c1

⎡⎢⎣

1

1

1

⎤⎥⎦ (c1 �= 0) 取 V1 =

⎡⎢⎣

1

1

1

⎤⎥⎦

λ = −2 代入 (A− λI)V = 0 中可得⎡⎢⎣

2 2 2

2 2 2

2 2 2

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦

故對應的特徵向量為

V = c2

⎡⎢⎣

1

−1

0

⎤⎥⎦+ c3

⎡⎢⎣

1

1

−2

⎤⎥⎦ ( c2 、 c3 不全為 0 )

故取正交的特徵向量為

V2 =

⎡⎢⎣

1

−1

0

⎤⎥⎦ , V3 =

⎡⎢⎣

1

1

−2

⎤⎥⎦

令

S =[ V1

||V1||V2

||V2||V3

||V3||]

=

⎡⎢⎢⎢⎢⎢⎣

1√3

1√2

1√6

1√3

− 1√2

1√6

1√3

0 − 2√6

⎤⎥⎥⎥⎥⎥⎦

則 S 為正交矩陣 , 滿足 ST = S−1, 再令 X = SY , 其中 Y = [y1 y2 y3]T , 故

Q = XTAX = (SY )TASY = Y TSTASY = Y TS−1ASY

= Y TDY = [y1 y2 y3]

⎡⎢⎣

4 0 0

0 −2 0

0 0 −2

⎤⎥⎦⎡⎢⎣y1

y2

y3

⎤⎥⎦

= 4y21 − 2y2

2 − 2y23


4. Verify Green’s theorem for the case where the vector field

−→F (x , y) = (xy + y2)

−→i + x2−→j

and the region is bounded by the curves y = x and y = x2.

(Note : Calculate each side of Green’s theorem, respectively, and show that both

results are identical.) (20%) 《喻超凡 100台科大機械》

《解》 �(1) 線積分 : 令 C = C1 +C2 (如圖) , 其中 C1 : y = x2, 且 x = 0 → 1 、 C2 : y = x , 且

x = 1 → 0 , 故∮C

−→F · dr =

∮C

(xy + y2) dx+ x2 dy

=

∫C1

(xy + y2) dx+ x2 dy +

∫C2

(xy + y2) dx+ x2 dy

=

∫ x=1

x=0

{[x(x2) + (x2)2] dx+ x2 d(x2)} +

∫ x=0

x=1

{(x · x+ x2)dx+ x2 dx}

=

∫ x=1

x=0

(x3 + x4 + 2x3)dx+

∫ x=0

x=1

3x3 dx

=19

20− 1 = − 1

20

(2) 二重積分 : 令 C 所圍的區域為 D , 故

∫∫D

{∂x2

∂x− ∂

∂y(xy + y2)} dxdy =

∫ x=1

x=0

∫ y=x

y=x2

(2x− x− 2y) dydx

= − 1

20


(3) 由 (1) 、 (2) 可知∮C

−→F · dr =

∮C

(xy + y2) dx+ x2 dy =

∫∫D

{∂x2

∂x− ∂

∂y(xy + y2)} dxdy

故 Green’s 定理得証。

5. Starting from the separation of variables, solve the boundary-value problem

∂2w

∂r2+

1

r

∂w

∂r+

1

r2

∂2w

∂θ2= 0 ; 1 ≤ r ≤ 2 , 0 ≤ θ ≤ π

with w(r , 0) = w(r , π) = w(1 , θ) = 0, and w(2 , θ) = sin3 θ. (20%)


《解》 �(1) 令 w(r , θ) = R(r)Θ(θ) 則

∂w

∂r= R′(r)Θ(θ) ,

∂2w

∂r2= R′′(r)Θ(θ) ,

∂2w

∂θ2= R(r)Θ′′(θ)


−r2R′′(r) + rR′(r)

R(r)=

Θ′′(θ)Θ(θ)

= −λ (λ 為常數 )

{Θ′′(θ) + λΘ(θ) = 0

r2R′′(r) + rR′(r) − λR(r) = 0(1)

(2) 因 w(r , θ) = R(r)Θ(θ) , 故由 B.C.

w(r , 0) = R(r)Θ(0) = 0 ⇒ Θ(0) = 0

w(r , π) = R(r)Θ(π) = 0 ⇒ Θ(π) = 0

現解

Θ′′(θ) + λΘ(θ) = 0 , Θ(0) = Θ(π) = 0 (2)

Sturm–Liouville 邊界值問題

(i) λ < 0 令 λ = −p2 (0 < p <∞) , 代入 (2) 式可得

Θ′′(θ) − p2Θ(θ) = 0


其通解為

Θ(θ) = c1 sinh pθ + c2 cosh pθ

代入邊界條件 Θ(0) = 0 , 可得 c2 = 0 , 再由 Θ(π) = 0 = c1 sinh pπ , 可得 c1 = 0 ,

故 Θ(θ) = 0 , 則 w(r , θ) = R(r)Θ(θ) = 0 無非零解。

(ii) λ = 0 代入 (2) 式 , 可得 Θ′′(θ) = 0 , 其通解為

Θ(θ) = c1 + c2θ

代入邊界條件 Θ(0) = 0 , 可得 c1 = 0 , 再由 Θ(π) = c2π , 可得 c2 = 0, 故 Θ(θ) = 0,

則 w(r , θ) = R(r)Θ(θ) = 0 無非零解。

(iii) λ > 0 令 λ = p2 (0 < p <∞) , 代入 (2) 式可得

Θ′′(θ) + p2Θ(θ) = 0

其通解為

Θ(θ) = c1 sin pθ + c2 cos pθ

代入邊界條件 Θ(0) = 0 , 可得 c2 = 0 , 再由 Θ(π) = 0 = c1 sin pπ , 令 c1 �= 0 , 可

得 sin pπ = 0 , 故

p = n (n = 1 , 2 , 3 , · · ·)則

Θ(θ) = c1 sinnθ = Θn (n = 1 , 2 , 3 , · · ·)將 λ = p2 = n2 代回 (1) 式中可得

r2R′′(r) + rR′(r) − n2R(r) = 0

上式為 Euler - Cauchy ODE , 其通解為

R(r) = d1rn + d2r

−n = Rn

因此

wn(r , θ) = Rn Θn = (d1rn + d2r

−n) · c1 sin nθ = (Anrn +Bnr

−n) sinnθ

(3) 由重疊原理知

w(r , θ) =∞∑

n=1

wn(r , θ) =∞∑

n=1

(Anrn +Bnr

−n) sinnθ

由邊界條件可得

w(1 , θ) = 0 =∞∑

n=1

(An +Bn) sinnθ


故

An +Bn = 0 (1)

再由

w(2 , θ) = sin3 θ =3

4sin θ − 1

4sin 3θ =

∞∑n=1

(An2n +Bn2−n) sinnθ

故 ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

A1 · 2 +B1 · 2−1 =3

4

A3 · 23 +B3 · 2−3 = −1

4

An · 2n +Bn · 2−n = 0 (n �= 1 , 3)

(2)

解 (1) 、 (2) 兩式可得

A1 =1

2, B1 = −1

2, A3 = − 2

63, B3 =

2

63

其他 An = Bn = 0, 故

w(r , θ) = (r

2− 1

2r) sin θ + (−2r3

63+

2

r363) sin 3θ

11. 台灣科技大學化學工程研究所 99–77

11. 100台灣科技大學化學工程研究所

1. 解釋名詞 : 何謂 partial differential equation ? Linear differential equation ?

Solution of differential equation ? (12%) 《喻超凡 100台科大化工》

《解》 �(1) partial differential equation : 描述多變數函數及其偏導數與變數之間的關係式 , 稱為

偏微分方程式。

(2) Linear differential equation : 在微分方程式中未知函數及其導數均滿足

(a) 次數為一次。

(b) 無互乘之項。

(c) 無非線性函數。

則稱此微分方程式為線性微分方程式。

(3) Solution of differential equation : 凡是能夠滿足 ODE 的函數 , 即稱為 ODE 的解。

2. Find the solution of DE : y′′ − 3y′ + 2y = cos(e−x). (20 %)

《喻超凡 100台科大化工》

《解》 �(1) 齊次解 : 令 y = emx 化回 ODE 中可得

m2 − 3m+ 2 = 0 ⇒ m = 1 , 2

故

yh(x) = c1ex + c2e

2x

(2) 特解 :

yp(x) =1

(D − 1)(D − 2)cos(e−x)


= e2x

∫e(1−2)x

∫e−x cos(−x) dx

= e2x

∫e−x{− sin(e−x)} dx

= e2x{− cos(e−x}

(3) 通解 : y(x) = yh(x) + yp(x) = c1ex + c2e

2x − e2x cos(e−x)

3. Solve the initial value problem : y′′ − 4y′ + 13y = 4δ(t − 2), y(0) = 0, y′(0) = 1.

(16%) 《喻超凡 100台科大化工》

《解》 � 對 ODE 兩端取 L–T 可得

s2Y (s) − sy(0) − y′(0) − 4{sY (s) − y(0)} + 13Y (s) = 4e−2s


(s2 − 4s+ 13)Y (s) = 1 + 4e−2s

即

Y (s) =1 + 4e−2s

s2 − 4s+ 13=

1 + 4e−2s

(s− 2)2 + 32

故

y(t) = L−1{Y (s)} =

1

3e2t sin 3t+

4

3e2(t−2) sin 3(t− 2) ·H(t− 2)

4. Evaluate the line integral with respect to arc length :

∫yz dz, with C the

parabola: z = y2, x = 1 for 0 ≤ y ≤ 2. (12%) 《喻超凡 100台科大化工》

《解》 �

∫C

yz dz =

∫ y=2

y=0

y(y2)d(y2) =

∫ y=2

y=0

2y4 dy =2y5

5

∣∣∣20

=64

5


5. For systems of linear differential equations, X ′ = AX, A =

⎡⎢⎣

3 0 −2

0 2 0

−2 0 0

⎤⎥⎦.

Please find out if A is diagonalizable(show clearly the processes), also find the

fundamental matrix, and the general solution. (16%) 《喻超凡 100台科大化工》

《解》 � 由

det(A− λI) = −(λ− 2)(λ− 3λ− 4) = 0

可得特徵值為 λ = 2 、 −1 、 4 , 將 λ = 2 代回 (A− λI)V = 0 中可得⎡⎢⎣

−1 0 −2

0 0 0

−2 0 −2

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦


V = c1

⎡⎢⎣

0

1

0

⎤⎥⎦ (c1 �= 0) 取 V1 =

⎡⎢⎣

0

1

0

⎤⎥⎦

將 λ = −1 代回 (A− λI)V = 0 中可得⎡⎢⎣

4 0 −2

0 3 0

−2 0 1

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦


V = c2

⎡⎢⎣

1

0

2

⎤⎥⎦ (c2 �= 0) 取 V2 =

⎡⎢⎣

1

0

2

⎤⎥⎦

將 λ = 4 代回 (A− λI)V = 0 中可得⎡⎢⎣

−1 0 −2

0 −2 0

−2 0 −4

⎤⎥⎦⎡⎢⎣v1

v2

v3

⎤⎥⎦ =

⎡⎢⎣

0

0

0

⎤⎥⎦



V = c3

⎡⎢⎣

2

0

−1

⎤⎥⎦ (c3 �= 0) 取 V3 =

⎡⎢⎣

2

0

−1

⎤⎥⎦

令

S = [V1 V2 V3] =

⎡⎢⎣

0 1 2

1 0 0

0 2 −1

⎤⎥⎦

故

S−1AS = D =

⎡⎢⎣

2 0 0

0 −1 0

0 0 4

⎤⎥⎦

再令 X = SY , 其中 Y = [y1 y2 y3]T , 代回 X ′ = AX 中可得

SY ′ = ASY ⇒ Y ′ = S−1ASY = DY

即 ⎧⎪⎨⎪⎩y′1 = 2y1

y′2 = −y2

y′3 = 4y3

⇒

⎧⎪⎨⎪⎩y1 = α e2t

y2 = β e−t

y3 = γ e4t

則 ODE 的解為

X = SY = [V1 V2 V3]

⎡⎢⎣α e2t

β e−t

γ e4t

⎤⎥⎦

= αV1e2t + β V2e

−t + γ V3e4t

= α

⎡⎢⎣

0

e2t

0

⎤⎥⎦+ β

⎡⎢⎣

e−t

0

2 e−t

⎤⎥⎦+ γ

⎡⎢⎣

2e4t

0

−e4t

⎤⎥⎦

且 fundamental matrix 為 ⎡⎢⎣

0 e−t 2e4t

e2t 0 0

0 2 e−t −e4t

⎤⎥⎦


6. For a thin, homogeneous bar of length L, given the initial temperature throughout

the bar is f(x), the temperature at both ends at all time are zero. i.e.

∂θ

∂t= a2 ∂

2θ

∂x2; 0 < x < L , t > 0

θ(0 , t) = θ(L , t) = 0, and θ(x , 0) = f(x). Please determine the temperature

distribution θ(x , t) on the bar. (16%) 《喻超凡 100台科大化工》

《解》 � 因 θ(0 , t) = θ(L , t) = 0, 故特徵函數為{

sinnπx

L

}∞

n=1, 因此令

θ(x , t) =∞∑

n=1

an(t) sinnπx

L


∞∑n=1

a′n(t) sinnπx

L= a2

∞∑n=1

an(t)(−n2π2

L2) sin

nπx

L

即 ∞∑n=1

[a′n(t) + (

nπa

L)2an(t)

]sin

nπx

L= 0

故可得

a′n(t) + (nπa

L)2an(t) = 0

其解為 an(t) = Ane−(nπa

L)2t , 故 PDE 的解為

θ(x , t) =∞∑

n=1

Ane−(nπa

L)2t sin

nπx

L(1)

由 I.C.

θ(x , 0) = f(x) =∞∑

n=1

An sinnπx

L

可得

An =2

L

∫ L

0

f(x) sinnπx

Ldx

將 An 代回 (1) 式, 即為所求。

7. Please find all z such that ez = 2 + i. (8%) 《喻超凡 100台科大化工》


《解》 � 因 (2 + i) =√

5 e(2kπ+θ)i, 其中 θ = tan−1 1

2, 故

z = ln(2 + i) = ln(√

5 e(2kπ+θ)i) = ln√

5 + (2kπ + θ)i (k = 0 , ±1 , ±2 · · ·)

12. 清華大學動力機械工程研究所 99–83

12. 100清華大學動力機械工程研究所

1. Find the inverse Laplace transform of the following function (10%)

F (s) =e−2s

s2 + 2s+ 3+ 2e−3s

《喻超凡 100清大動機》

《解》 �

L {F (s)} = L−1{ e−2s

(s+ 1)2 + 2+ 2e−3s}

=1√2e−(t−2) sin[

√2 (t− 2)]H(t− 2) + 2 δ(t− 3)

2. Find the Laplace transform of the following function (10%)

g(t) =

⎧⎨⎩

0 0 ≤ t < 1

t− 1 1 ≤ t < 2

1 t ≥ 2

《喻超凡 100清大動機》

《解》 � 因

g(t) = (t− 1){H(t− 1) −H(t− 2)} +H(t− 2) = (t− 1)H(t− 1) − (t− 2)H(t− 2)

故

L {g(t)} = L {(t− 1)H(t− 1) − (t− 2)H(t− 2)} =e−s

s2− e−2s

s2


3. Find the most general solution to the ODE (10%)

3dy

dx− y = xe

x3

What is the unique solution satisfying y(0) = 1 ? 《喻超凡 100清大動機》

《解》 � 原式改寫成dy

dx− 1

3y =

x

3e

x3

積分因子 I = e−x3 , 故

Iy =

∫e−

x3x

3e

x3 dx =

x2

6+ c

則 ODE 的通解為

y(x) =x2

6e

x3 + c e

x3

4. Find the least-square solution of Ax = b, with A =

⎡⎢⎣

1 1

2 −1

−1 2

⎤⎥⎦, b =

⎡⎢⎣

1

1

1

⎤⎥⎦.

(10%) 《喻超凡 100清大動機》

《解》 � 最小二乘方的解為

x = (ATA)−1AT b =1

3

[2

2

]

5. Find the expression for the plane tangential to xy3z2 = 4 at the point (1 , 1 , 2).

(10%) 《喻超凡 100清大動機》

《解》 � 令 φ = xy3z2 = 4, 故

∇φ = y3z2−→i + 3xy2z2−→j + 2xy3z−→k


則 (1 , 1 , 2) 處的法向量為

−→n = ∇φ = 4−→i + 12

−→j + 4

−→k

故 (1 , 1 , 2) 處的切平面為

4(x− 1) + 12(y − 1) + 4(z − 2) = 0

即 4x+ 12y + 4z = 24

6. Solve the following problem of partial differential equation (10%)

∂T

∂t=∂2T

∂x2+ f(x , t)

T (0 , t) = g(t) , T (L , t) = h(t) , T (x , 0) = p(x). 《喻超凡 100清大動機》

《解》 � 令 T (x , t) = φ(x , t) +h(t) − g(t)

Lx+ g(t) 代回 PDE 中可得

∂φ

∂t+h′(t) − g′(t)

Lx+ g′(t) =

∂2φ

∂x2+ f(x , t)

即∂φ

∂t=∂2φ

∂x2+ f(x , t) − h′(t) − g′(t)

Lx− g′(t) (1)

再由

T (0 , t) = g(t) = φ(0 , t) + g(t) ⇒ φ(0 , t) = 0

T (L , t) = h(t) = φ(L , t) + h(t) − g(t) + g(t) ⇒ φ(L , t) = 0

T (x , 0) = p(x) = φ(x , 0) +h(0) − g(0)

Lx+ g(0)

即

φ(x , 0) = p(x) − h(0) − g(0)

Lx− g(0)

因 φ(0 , t) = φ(L , t) = 0, 故特徵函數為 {sin nπxL

}, 令

φ(x , t) =∞∑

n=1

an(t) sinnπx

L

且

f(x , t) − h′(t) − g′(t)L

x− g′(t) =∞∑

n=1

qn(t) sinnπx

L


代回 (1) 式中可得

∞∑n=1

a′n(t) sinnπx

L=

∞∑n=1

an(t)(−n2π2

L2) sin

nπx

L+

∞∑n=1

qn(t) sinnπx

L

整理可得 ∞∑n=1

{a′n(t) + (nπ

L)2an(t) − qn(t)} sin

nπx

L= 0

故

a′n(t) + (nπ

L)2an(t) = qn(t) (2)

(2) 式為一階線性 ODE , 積分因子為 I = e(nπL

)2t , 故

Ian(t) =

∫e(

nπL

)2tqn(t) dt+ An

即

an(t) = e−(nπL

)2t

∫ t

0

e(nπL

)2τqn(τ) dτ + Ane−(nπ

L)2t

故

φ(x , t) =∞∑

n=1

{e−(nπL

)2t

∫ t

0

e(nπL

)2τqn(τ) dτ + Ane−(nπ

L)2t} sin

nπx

L

再由

φ(x , 0) = p(x) − h(0) − g(0)

Lx− g(0) =

∞∑n=1

An sinnπx

L

故

An =2

L

∫ L

0

{p(x) − h(0) − g(0)

Lx− g(0)} sin

nπx

Ldx

因此 T (x , t) = φ(x , t) +h(t) − g(t)

Lx+ g(t)

7. Expand f(x) = x2 for 0 < x < L. (15%)

(a) in a sine series.

(b) in a cosine series.

(c) in a Fourier series. 《喻超凡 100清大動機》

《解》 �(a) Fourier sine series : 將 f(x) 視為 −L < x < L 中且週期為 2L 的奇函數 , 故令

f(x) =∞∑

n=1

bn sinnπx

L


其中

bn =2

L

∫ L

0

x2 sinnπx

Ldx =

2L2[−2 + (2 − n2π2) cosnπ]

n3π3

故

f(x) =∞∑

n=1

bn sinnπx

L=

∞∑n=1

2L2[−2 + (2 − n2π2) cosnπ]

(nπ)3sin

nπx

L

(b) Fourier cosine series : 將 f(x) 視為 −L < x < L 中且週期為 2L 的偶函數 , 故令

f(x) = a0 +∞∑

n=1

an cosnπx

L

其中

a0 =1

L

∫ L

0

x2 dx =L2

3

an =2

L

∫ L

0

x2 cosnπx

Ldx =

4L2 cos nπ

n2π2=

(−1)n4L2

(nπ)2

故

f(x) = a0 +∞∑

n=1

an cosnπx

L=L2

3+

∞∑n=1

(−1)n4L2

(nπ)2cos

nπx

L

(c) Fourier series : 將 f(x) 視為 0 < x < L 中且週期為 L 的函數 , 故令

f(x) = a0 +∞∑

n=1

(an cos2nπx

L+ bn sin

2nπx

L)

其中

a0 =1

L

∫ L

0

x2 dx =L2

3

an =2

L

∫ L

0

x2 cos2nπx

Ldx =

L2

n2π2

bn =2

L

∫ L

0

x2 sin2nπx

Ldx = −L2

nπ

故

f(x) = a0 +∞∑

n=1

(an cos2nπx

L+ bn sin

2nπx

L)

=L2

3+

∞∑n=1

{ L2

n2π2cos

2nπx

L− L2

nπsin

2nπx

L}


8. Evaluate the following integrals, where C is the circle x2 + y2 = 4 (counterclock-

wise)

(a)

∮C

z dz (8%)

(b)

∮C

z2(e−z + e1z ) dz (8%) 《喻超凡 100清大動機》

《解》 �(a) 由 Green’s 定理可知∮

C

z dz = 2i

∫∫D

∂z

∂zdxdy = 2i

∫∫D

dxdy = 2i π 22 = 8π i

(b) 令 f(z) = z2e1z , 則 f(z)具有 z = 0的本性奇點,同時 f(z)對 z = 0展開的 Laurent’s

級數為

f(z) = z2(1 +1

z+

1

2!

1

z2+

1

3!

1

z3+ · · ·)

故 Resf(0) =1

3!=

1

6, 因此

∮C

z2(e−z + e1z ) dz =

∮C

z2 e1z dz = 2π iResf(0) =

2πi

6=πi

3

9. Show that coshx sin y is a harmonic function and find its conjugate harmonic

function. (9%) 《喻超凡 100清大動機》

《解》 �(a) 因

∇2(coshx sin y) =∂2

∂x2(coshx sin y) +

∂2

∂y2(coshx sin y)

= cosh x sin y − cosh x sin y = 0

故 cosh x sin y 為 harmonic 函數。


(b) 設 v(x , y) 為 cosh x sin y 的 conjugate harmonic 函數, 故⎧⎪⎪⎨⎪⎪⎩∂v

∂x= − ∂

∂y(coshx sin y) = − cosh x cos y

∂v

∂y=

∂

∂x(coshx sin y) = sinh x sin y

故 {v = − sinh x cos y + f(y)

v = − sinh x cos y + g(x)

比較可得

v(x , y) = − sinh x cos y + c


13. 100清華大學工程科學研究所

1. Find the transient current if R = 6Ω, L = 1H , C = 0.04F ,

E = 600(cos t + 4 sin t)V ; (L, R, C, E, are measured in henrys, ohms, farads,

volts, respectively.) Initial current and charge are assumed to be zero. (10%)

《喻超凡 100清大工科》

《解》 � 設電荷為 q(t) , 電流為 i(t), 故

Ld2q

dt2+R

dq

dt+

1

Cq = E(t)

即d2q

dt2+ 6

dq

dt+ 25q = 600(cos t+ 4 sin t) (1)

且 q(0) = 0,dq

dt(0) = 0

(1) 齊次解 : 令 q(t) = emt 代回 (1) 式可得

m2 + 6m+ 25 = 0 ⇒ m = −3 ± 4i

故

qh(t) = e−3t(c1 cos 4t+ c2 sin 4t)

(2) 特解 :

qp(t) =1

D2 + 6D + 25600(cos t+ 4 sin t)

= 6001

−12 + 6D + 25(cos t+ 4 sin t)

= 600D − 4

6(D + 4)(D − 4)(cos t+ 4 sin t)

13. 清華大學工程科學研究所 99–91

= 100D − 4

D2 − 16(cos t+ 4 sin t)

=100

−17(− sin t+ 4 cos t− 4 cos t− 16 sin t)

= 100 sin t

(3) 通解 :

q(t) = qh(t) + qp(t) = e−3t(c1 cos 4t+ c2 sin 4t) + 100 sin t

再由 {q(0) = 0 = c1

q′(0) = 0 = −3c1 + 4c2 + 100⇒ c1 = 0 , c2 = −25

故

q(t) = −25e−3t sin 4t+ 100 sin t

且

i(t) =dq

dt= 75e−3t sin 4t− 100e−3t cos 4t+ 100 cos t

2. (10%) Consider a general nonhomogeneous linear ODE

y(n) + pn−1(x)y(n−1) + pn−2(x)y

(n−2) + · · · + p1(x)y′ + p0(x)y = r(x)

The particular solution yp(x) can be solved by the method of variation of param-

eters. That is

yp(x) =n∑

k=1

yk(x)

∫Wk(x)

W (x)r(x) dx

(Where the yk’s are n linearly independent homogeneous solutions. W is the

Wronskian of y1, · · ·, yn, and Wk is identical to Wi but with the kth column

replaced by a column of zeros-except for the bottom element, which is 1.) Try to

solve the following equation using the method of variation of parameters.

y′′ +3

4x−2y′ − 3

4x−3y = 9x5/2


《解》 �(1) 齊次解 : 令 y = emx 代回 ODE 中可得

m(m− 1)(m− 2) +3

4m− 3

4= 0


故 m =1

2、 1 、

3

2, 則

yh(x) = c1x1/2 + c2x+ c3x

3/2

(2) 令 y1 = x1/2 、 y2 = x 、 y3 = x3/2 , 故令

yp(x) = φ1(x)y1(x) + φ2(x)y2(x) + φ3(x)y3(x)

且

W (y1 , y2 , y3) =

∣∣∣∣∣∣∣∣∣∣

x1/2 x x3/2

1

2x−1/2 1

3

2x1/2

−1

4x−3/2 0

3

4x−1/2

∣∣∣∣∣∣∣∣∣∣=

1

4

W (y2 , y3) =

∣∣∣∣∣∣x x3/2

13

2x1/2

∣∣∣∣∣∣ =1

2x3/2 , W (y1 , y3) =

∣∣∣∣∣∣x1/2 x3/2

1

2x−1/2 3

2x1/2

∣∣∣∣∣∣ = x

W (y1 , y2) =

∣∣∣∣∣∣x1/2 x

1

2x−1/2 1

∣∣∣∣∣∣ =1

2x1/2

故

φ′1(x) =

9x5/2 W (y2 , y3)

W (y1 , y2 , y3)= 18x4 ⇒ φ1(x) =

18

5x5

φ′2(x) = −9x5/2 W (y1 , y3)

W (y1 , y2 , y3)= −36x7/2 ⇒ φ1(x) = −8x9/2

φ′3(x) =

9x5/2 W (y1 , y2)

W (y1 , y2 , y3)= 18x3 ⇒ φ1(x) =

9

2x4

因此

yp(x) = x1/2(18

5x5) + x(−8x9/2) + x3/2(

9

2x4) =

1

10x11/2

(3) 通解 : y(x) = yh(x) + yp(x)

3. Solve the following equation by Laplace Transform method.

y′ + y = f(t) , y(0) = 3

where f(t) =

{0 0 ≤ t < π

2 cos t t ≥ π.(10%) 《喻超凡 100清大工科》


《解》 � 因 f(t) = 2 cos tH(t− π) , 故

L {f(t)} = L {2 cos tH(t− π)} = 2e−πsL {cos(t+ π)}

= 2e−πsL {− cos t} = 2e−πs(− s

s2 + 1)

對 ODE 兩端取 L–T 可得

sY (s) − y(0) + Y (s) = L {f(t)}

整理可得

(s+ 1)Y (s) = 3 − 2e−πs(s

s2 + 1)

即

Y (s) =3

s+ 1− 2s

(s+ 1)(s2 + 1)e−πs =

3

s+ 1+ (

1

s+ 1− s+ 1

s2 + 1)e−πs

故

y(t) = L−1{Y (s)}

= 3e−t + {e−(t−π) − cos(t− π) − sin(t− π)}H(t− π)

4. (10%) Find the basis of solution y(x) of the following differential equation. Show

the details of your work

xy′′ + (2x+ 1)y′ + (x+ 1)y = 0


《解》 � 原式可改寫成

{xD2 + (2x+ 1)D + (x+ 1)}y = 0 其中 D =d

dx

故可因式分解成

(xD + x+ 1)(D + 1)y = 0

令 z = (D + 1)y, 代回上式可得

(xD + x+ 1)z = 0 ⇒ xz′ + (x+ 1)z = 0


分離變數可得dz

z= −x+ 1

xdx

兩端積分可得

ln |z| = −x− ln |x| + c∗1

兩端取指數可得 z = c1e−x

x, 代回 z = (D + 1)y 中, 可得

y′ + y = c1e−x

x

上式積分因子為 I = ex, 故

Iy =

∫ex c1

e−x

xdx = c1 ln |x| + c2

因此 ODE 的通解為

y(x) = c1e−x ln |x| + c2e

−x

5. Find a unit vector normal to surface S given bye cos(xy) = ez − 1 at the point

(1 , π , 0). (10%) 《喻超凡 100清大工科》

《解》 � 令 φ = cos(xy) − ez + 1 = 0 , 則

∇φ = −y sin(xy)−→i − x sin(xy)

−→k − ez−→k

且

∇φ(1 , π , 0) = −−→k

故單位法向量 −→n = ± ∇φ(1 , π , 0)

|∇φ(1 , π , 0)| = ±−→k

6. Let−→F = (x − y)

−→i + (y − z)

−→j + (z − x)

−→k . Evaluate the surface integral of

−→F

over the unit sphere defined by x2 + y2 + z2 = 1. (10%) 《喻超凡 100清大工科》

《解》 � 設曲面 S : x2 + y2 + z2 = 1 , 所圍的區域為 D, 則∫∫ −→F · −→n dA =

∫∫∫D

(∇ · −→F ) dxdydz =

∫∫∫D

3 dxdydz = 3 · 4π

3= 4π


7. Define the Fourier transform of f(x) to be f(w) =1√2π

∫ ∞

−∞f(x)e−iwx dx.

(a) (5%) Calculate the Fourier transform of f(x) =

{a− |x| |x| < a

0 otherwise.

(b) (5%) Consider the one-dimensional diffusion equation :

∂

∂tu(x , t) = D

∂2

∂x2u(x , t) for −∞ < x <∞

with the initial condition u(x , 0) = f(x). Use the Fourier transform to show

that the solution of the diffusion equation takes the form

u(x , t) =

∫ ∞

−∞K(x− ξ , t)f(ξ) dξ

Find K(x− ξ), which is called the kernel.

( Hint : Caussian integral

∫ ∞

−∞exp(

−x2

2σ2) dx =

√2πσ2 )


《解》 �(a) 因 f(x) 為偶函數, 故

F{f(x)} = f(w) =1√2π

∫ ∞

−∞f(x)e−iwx dx

=1√2π

∫ ∞

−∞f(x)(coswx− i sinwx) dx

=2√2π

∫ ∞

0

f(x) coswxdx

=

√2

π

∫ a

0

(a− x) coswxdx =

√2

π

1 − cos aw

w2

(b) 對 PDE 兩端取 F–T 可得

dU(w , t)

dt= D(−w2)U(w , t)

其中 F{u(x , t)} = U(w , t), 整理可得

dU

dt+ w2DU = 0


故

U(w , t) = Ae−w2Dt

再由 U(w , 0) = F{f(x)} = F (w) = A, 因此

U(w , t) = F (w)e−w2Dt

因

F−1{e−w2Dt} =1√2π

∫ ∞

−∞e−w2Dteiwx dw

=1√2π

∫ ∞

−∞e−Dt(w2−iwx+( ix

2√

Dt)2e−

x2

4Dt dw

=1√2πe−

x2

4Dt

∫ ∞

−∞e−Dt(w− ix

2√

Dt)2dw

=1√2πe−

x2

4Dt

√π

Dt

則

u(x , t) = F−1{F (w)e−w2Dt} =1√2π

F−1{F (w)} ∗ F−1{e−w2Dt}

=1√2π

f(x) ∗ 1√2πe−

x2

4Dt

√π

Dt

=1

2π

∫ ∞

−∞e−

(x−ξ)2

4Dt

√π

Dtf(ξ) dξ

故 K(x− ξ) =1

2πe−

(x−ξ)2

4Dt

√π

Dt

8. (a) (5%) Find and classify all local maxima, local minima and saddles for

f(x , y , z) = exp(2x2 + xz − 5z2)

(b) (5%) Consider a forced vibration system which is described by the equations⎧⎪⎪⎨⎪⎪⎩d2x1

dt2+ 2x1 − x2 = A sin(wt)

d2x2

dt2− x1 + 2x2 = B sin(wt)

where A, B and w are constant. To seek a particular solution, we assume

x1(t) = q1 sinwt and x2(t) = q2 sin(wt). Find q1 and q2.



《解》 �(a) 因 f 中沒 y 的變數, 故 f(x , y , z) = f(x , z) , 由⎧⎪⎪⎨

⎪⎪⎩∂f

∂x= exp(2x2 + xz − 5z2) (4x+ z) = 0

∂f

∂z= exp(2x2 + xz − 5z2) (x− 10z) = 0

可得 x = z = 0, 故臨界點為 (0 , 0) , 又

∂2f

∂x2(0 , 0) = 4 ,

∂2f

∂z2= −10 ,

∂2f

∂x∂z(0 , 0) = 1

因

D =

∣∣∣∣ fxx(0 , 0) fxz(0 , 0)

fxz(0 , 0) fzz(0 , 0)

∣∣∣∣ =∣∣∣∣ 4 1

1 −10

∣∣∣∣ = −41 < 0

故 (x , z) = (0 , 0) 為 saddles point 。

(b) 將 x1(t) = q1 sinwt 、 x2(t) = q2 sinwt 代回 ODE 中可得{−w2q1 + 2q1 − q2 = A

−w2q2 − q1 + 2q2 = B

故可解得

q1 = −−2A− B + Aw2

3 − 4w2 + w4, q2 = −−A− 2B +Bw2

3 − 4w2 + w4

9. (12%) Along the circumference of the circle r = b a solution T (r , θ) of Laplace’s

equation is required to take on the value T0 when 0 < θ < π and the value −T0

when π < θ < 2π. Determine an expression for T valid when r > b. (Show the

details of your work)

∇2T =1

r

∂

∂r(r∂T

∂r) +

1

r2

∂2T

∂θ2


《解》 � 因區域為圓外的區域 , 故 T (r , θ) = T (r , θ + 2π) , 則特徵函數為{1 , cos nθ , sin nθ

}∞

n=1

因此由特徵函數展開法知 , 令

T (r , θ) = A0(r) +∞∑

n=1

{An(r) cosnθ +Bn(r) sinnθ}



A′′0(r) +

∞∑n=1

{A′′n(r) cosnθ +B′′

n(r) sinnθ}

+1

r

[A′

0(r) +∞∑

n=1

{A′n(r) cosnθ +B′

n(r) sinnθ}]

+1

r2

∞∑n=1

{(−n2)An(r) cosnθ + (−n2)Bn(r) sinnθ} = 0

{A′′0(r) +

1

rA′

0(r)} +∞∑

n=1

{[A′′

n(r) +1

rA′

n(r) − n2

r2An(r)] cosnθ

+[B′′n(r) +

1

rB′

n(r) − n2

r2Bn(r)] sinnθ

}= 0

故可得 ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

A′′0(r) +

1

rA′

0(r) = 0

A′′n(r) +

1

rA′

n(r) − n2

r2An(r) = 0

B′′n(r) +

1

rB′

n(r) − n2

r2Bn(r) = 0

即 ⎧⎪⎪⎪⎨⎪⎪⎪⎩r2A′′

0(r) + rA′0(r) = 0

r2A′′n(r) + rA′

n(r) − n2An(r) = 0

r2B′′n(r) + rB′

n(r) − n2Bn(r) = 0

(1)

(1) 式為 Euler–Cauchy 方程式, 其解為⎧⎨⎩A0 = a0 + b0 ln r

An = anr−n + bnr

n

Bn = cnr−n + dnr

n

故

T (r , θ) = a0 + b0 ln r +∞∑

n=1

{(anr−n + bnr

n) cosnθ + (cnr−n + dnr

n) sinnθ}

因

limr→∞

T (r , θ) = bounded ⇒ b0 = bn = dn = 0

且

T (b , θ) = a0 +∞∑

n=1

{an bn cosnθ + cn b

n sin nθ}

其中

a0 =1

2π

∫ 2π

0

T (b , θ)dθ =1

2π{∫ π

0

T0 dθ +

∫ 2π

π

(−T0) dθ} = 0


an bn =

1

π

∫ 2π

0

T (b , θ) cosnθ dθ =1

π{∫ π

0

T0 cos nθ dθ +

∫ 2π

π

(−T0) cosnθ dθ} = 0

cn bn =

1

π

∫ 2π

0

T (b , θ) sinnθ dθ

=1

π{∫ π

0

T0 sinnθ dθ +

∫ 2π

π

(−T0) sinnθ dθ}

=2T0(1 − cosnπ)

n

故

T (r , θ) =∞∑

n=1

2T0(1 − cosnπ)

nbnr−n sin nθ

10. (Complex analysis) Prove the following identity (8%)

sin−1 z + cos−1 z =1

2(4n+ 1)π , n = 0 , ±1 , ±2 , · · ·


《解》 � 令 w1 = sin−1 z 、 w2 = cos−1 z, 故

z = sinw1 = cosw2

因此 w1 = 2nπ +π

2− w2 (n = 0 , ±1 , ±2 · · ·) , 故

sin−1 z + cos−1 z = w1 + w2 = 2nπ +π

2− w2 + w2 = 2nπ +

π

2(n = 0 , ±1 , ±2 · · ·)


14. 100中山大學材料工程研究所

1. Find the general solution of 2yd3y

dx3+ 2(y + 3

dy

dx)d2y

dx2+ 2(

dy

dx)2 = sin x.

《喻超凡 100中山材料》

《提示》 � 本題為高階非線性正合 ODE 。

《解》 � 原式可改寫成

d

dx(2y

d2y

dx2) − 2

dy

dx

d2y

dx2+ 2(y + 3

dy

dx)d2y

dx2+ 2(

dy

dx)2 = sin x

即d

dx(2y

d2y

dx2) + 2y

d2y

dx2+ 2(

dy

dx)2 + 4

dy

dx

dy2

dx2= sin x

故d

dx(2y

d2y

dx2) +

d

dx(2y

dy

dx) + 2

d

dx(dy

dx)2 = sin x

兩端積分可得

2yd2y

dx2+ 2y

dy

dx+ 2(

dy

dx)2 = − cosx+ c1

又d

dx(2y

dy

dx) +

d

dx(y2) = − cosx+ c1

兩端積分可得

2ydy

dx+ y2 = − sin x+ c1x+ c2

令 u = y2, 故du

dx= 2y

dy

dx, 代回上式可得

du

dx+ u = − sin x+ c1x+ c2

積分因子為 I = ex, 且

Iu(x) =

∫ex(− sin x+ c1x+ c2) dx

=1

2(−ex sin x+ ex cos x) + c1(xe

x − ex) + c2ex + c3

故

u(x) = y2(x) =1

2(− sin x+ cosx) + c1(x− 1) + c2 + c3e

−x

14. 中山大學材料工程研究所 99–101

2. Use Frobenius method to find the general solution of x2y′′ +6xy′ +(6−4x2)y = 0.

(20%) 《喻超凡 100中山材料》

《解》 � 因 x = 0 為規則奇點, 令

y =∞∑

n=0

anxn+r (0 < |x| <∞ , a0 �= 0)

代回 ODE 中可得

x2∞∑

n=0

(n+ r)(n+ r − 1)anxn+r−2 + 6x ·

∞∑n=0

(n + r)anxn+r−1

+6∞∑

n=0

anxn+r − 4x2

∞∑n=0

anxn+r = 0

即∞∑

n=0

[(n+ r)2 − (n+ r) + 6(n+ r) + 6]anxn+r −

∞∑n=0

4anxn+r+2 = 0

∞∑n=0

(n+ r + 2)(n+ r + 3)anxn+r −

∞∑n=2

4an−2xn+r = 0

故

(r + 2)(r + 3)a0xr + (r + 3)(r + 4)a1x

1+r

+∞∑

n=2

[(n + r + 2)(n+ r + 3)an − 4an−2]xn+r = 0

可得

(r + 2)(r + 3)a0 = 0 (1)

(r + 3)(r + 4)a1 = 0 (2)

(n+ r + 2)(n+ r + 3)an − 4an−2 = 0 (n = 2 , 3 , · · ·) (3)

因 a0 �= 0, 由 (1) 式可得 r = −2 、 −3, 由 (2) 式可得 a1 = 0, 且

an =4

(n+ r + 2)(n+ r + 3)an−2 (n = 2 , 3 , · · ·)


且

n = 2 ⇒ a2 =4

(r + 4)(r + 5)a0

n = 3 ⇒ a3 = 0

n = 4 ⇒ a4 =4

(r + 6)(r + 7)a2 =

42

(r + 4)(r + 5)(r + 6)(r + 7)a0

n = 5 ⇒ a5 = 0

· · · · · · · · ·故 a2n+1(r) = 0, 且

a2n(r) =4n

(r + 4)(r + 5) · · · (r + 2n+ 3)a0

(1) 當 r = −3, 時

a2n(r = −3) =4n

1 · 2 · 3 · · · (2n)a0 =

4n

(2n)!a0 ; a2n+1(−3) = 0

故

y1(x) =∞∑

n=0

anxn+r∣∣∣r=−3

=∞∑

n=0

a2nx2n−3 +

∞∑n=0

a2n+1x2n−2

= a0x−3

∞∑n=0

4n

(2n)!x2n = a0x

−3∞∑

n=0

(2x)2n

(2n)!

= a0x−3 cosh 2x

(2) 當 r = −2, 時

a2n(r = −2) =4n

2 · 3 · 4 · · · (2n+ 1)a0 =

4n

(2n + 1)!a0 ; a2n+1(−2) = 0

故

y2(x) =∞∑

n=0

anxn+r∣∣∣r=−2

=∞∑

n=0

a2nx2n−2 +

∞∑n=0

a2n+1x2n−1

= a0x−2

∞∑n=0

4n

(2n+ 1)!x2n =

a0

2x−3

∞∑n=0

(2x)2n+1

(2n+ 1)!

=a0

2x−3 sinh 2x

故

y(x) = c1y1(x) + c2y2(x) = c∗1x−3 cosh 2x+ c∗2x

−3 sinh 2x


3. (15%) Find the eigenvalues, eigenfunctions and verify the orthogonality of the

solution by direct calculation.

y′′ + λy = 0 , y(π) = y(−π) , y′(π) = −y′(−π)


《解》 �(a) λ < 0 ; 令 λ = −p2 (0 < p <∞) , 故 ODE 可改寫成

y′′(x) − p2y(x) = 0

ODE 的通解為

y(x) = c1epx + c2e

−px (1)

且

y′(x) = p c1epx − p c2e

−px (2)

將邊界條件 y(−π) = y(π) 代回 (1) 式, 可得

c1e−pπ + c2e

pπ = c1epπ + c2e

−pπ

即

c1(e−pπ − epπ) + c2(e

pπ − e−pπ) = 0

因 (e−pπ − epπ) �= 0, 故

c1 − c2 = 0 (3)

再將邊界條件 y′(−π) = y′(π) 代回 (2) 式, 可得

p c1e−pπ − p c2e

pπ = p c1epπ − p c2e

−pπ

即

c1(e−pπ − epπ) − c2(e

pπ − e−pπ) = 0

因 (e−pπ − epπ) �= 0, 故

c1 + c2 = 0 (4)

聯立 (3) 式、 (4) 式可解得 c1 = c2 = 0, 將 c1 = c2 = 0 代回 (1) 式 , 可得 y(x) = 0 ,

故 ODE 無非零解。


(b) λ = 0 , 故 y′′(x) = 0 , 則 ODE 的通解為

y(x) = c1x+ c2 且 y′(x) = c1 (5)


c1(−π) + c2 = c1π + c2

故 c1 = 0 , 再將邊界條件 y′(−π) = y′(π) 代回 (5) 式, c1 = c1 , 自動滿足 , 因

此 c2 為任何數, 故令 c2 �= 0 , 則 λ = 0 為 ODE 的特徵值 , 且對應的特徵函數為

y(x) = c2 (c2 �= 0) 。

(c) λ > 0 ; 令 λ = p2 (0 < p <∞) , 故 ODE 可改寫成

y′′(x) + p2y(x) = 0

ODE 的通解為

y(x) = c1 cos px+ c2 sin px (6)

且

y′(x) = −p c1 sin px+ p c2 cos px (7)


c1 cos(−pπ) + c2 sin(−pπ) = c1 cos pπ + c2 sin pπ

即

c2 sin pπ = 0 (8)

再將邊界條件 y′(−π) = y′(π) 代回 (7) 式, 可得

−p c1 sin(−pπ) + p c2 cos(−pπ) = −p c1 sin pπ + p c2 cos pπ

即

c1 sin pπ = 0 (9)

若 (8) 式、 (9) 式中的 c1 、 c2 要具有非全為 0 的解時 , 可得 sin pπ = 0, 故

p = n (n = 1 , 2 , 3 , 4 · · ·)

故 ODE 的特徵值為 λ = p2 = n2 (n = 1 , 2 , 3 , 4 · · ·)

對應的特徵函數為 y(x) = c1 cosnx+ c2 sin nx (c1 、 c2 不全為0)


4. A surface is defined by z = x2+y2, calculate the surface are defined on the domain

0 ≤ x2 + y2 ≤ 4. (15%) 《喻超凡 100中山材料》

《解》 � 令曲面 S 為 φ = x2 + y2 − z = 0, 故曲面單位法向量為

−→n =∇φ|∇φ| =

2x−→i + 2y

−→j −−→

k√4x2 + 4y2 + 1

則

dA =dx dy

|−→k · −→n | =√

4x2 + 4y2 + 1 dx dy

故曲面的面積為

A =

∫∫S

dA =

∫∫Sxy

√4x2 + 4y2 + 1 dx dy

Sxy 為曲面 S 在 xy 平面上的投影區域

=

∫ θ=2π

θ=0

∫ r=2

r=0

√4r2 + 1 r dr dθ

=1

6(−1 + 17

√17 )π

5. f(x) = x, −π ≤ x ≤ π, f(x) = f(x+ 2π), solve

(1) Fourier series, (10%)

(2) complex Fourier series. (10%) 《喻超凡 100中山材料》

《解》 �(1) 因 f(x) 為奇函數, 故

f(x) =∞∑

n=1

bn sin nx

其中

bn =2

π

∫ π

0

x sin nx dx = −2 cosnπ

n


(2) 令

f(x) =∞∑

n=−∞cne

i nx

其中

cn =1

2π

∫ π

−π

f(x)e−i nx dx =1

π

∫ π

0

x sinnx dx = −cos nπ

n(n �= 0)

c0 =1

2π

∫ π

−π

f(x) dx =1

2π

∫ π

−π

x dx = 0

6. Solve nonhomogeneous boundary condition of hear equation

∂u(x , t)

∂t= c2

∂2u(x , t)

∂x2

u(0 , t) = A1, u(L , t) = A2, u(x , 0) = f(x), where 0 < x < L, t > 0.


《解》 � 令 u(x , t) = φ(x , t) +A2 − A1

Lx+ A1, 代回 PDE 中可得

∂φ

∂t= c2

∂2φ

∂x2(1)

由

u(0 , t) = φ(0 , t) + A1 = A1 ⇒ φ(0 , t) = 0

u(L , t) = φ(L , t) + A2 − A1 + A1 = A2 ⇒ φ(L , t) = 0

u(x , 0) = φ(x , 0) +A2 − A1

Lx+ A1 = f(x) ⇒ φ(x , 0) = f(x) − A2 − A1

Lx− A1

令

φ(x , t) =∞∑

n=1

an(t) sinnπx

L

代回 (1) 式中可得

∞∑n=1

a′n(t) sinnπx

L= c2

∞∑n=1

an(t)(−n2π2

L2) sin

nπx

L

整理可得∞∑

n=1

{a′n(t) + (nπc

L)2an(t)} sin

nπx

L= 0


即

a′n(t) + (nπc

L)2an(t) = 0

可得

an(t) = Bne−(nπc

L)2t

因此

φ(x , t) =∞∑

n=1

Bne−(nπc

L)2t sin

nπx

L

再中

φ(0 , t) = f(x) − A2 − A1

Lx−A1 =

∞∑n=1

Bn sinnπx

L

可得

Bn =2

L

∫ L

0

{f(x) − A2 − A1

Lx− A1} sin

nπx

Ldx

故原 PDE 的解為

u(x , t) = φ(x , t) +A2 −A1

Lx+ A1


15. 100台聯系統 A 卷

1. (Total 15%) Let Mn×n(C) be the vector space consisting of all n×n matrices with

complex entries. Two matrices A, B ∈Mn×n(C) are said to be unitarily equivalent

if there exists a unitary matrix P ∈ Mn×n(C) such that A = P ∗BP , where P ∗ is

the conjugate transpose of P

(a) (10%) Let A, B ∈Mn×n(C) be unitarily equivalent. Show that

tr(A∗A) = tr(B∗B).

(b) (5%) Determine whether the matrices A =

[1 2

2 + i 3

]and B =

[1 + 2i 1

4i 2

]are unitarily equivalent (you need to justify your answer ) ?

《喻超凡 100台聯 A》

《解》 �

(a) 因

A∗A = (P ∗BP )∗(P ∗BP ) = P ∗B∗PP ∗BP = P ∗B∗BP

故

tr(A∗A) = tr(P ∗B∗BP ) = tr(PP ∗B∗B) = tr(B∗B)

(b) tr(A∗A) = 12 + 22 + |2 + i|2 + 32 = 19 , tr(B∗B) = |1 + 2i|2 + 12 + |4i|2 + 22 = 26 ,

因 tr(A∗A) �= tr(B∗B), 故 A 與 B 不為 unitarily equivalent 。

2. (Total 10%) Let P2(R) be a vector space that consists of all polynomials with real

coefficients and with degree less or equal to 2. Let T be a linear operator on P2(R)

defined by

Tf(x) = f(2x− 1) − 2xf ′(x)

for all f(x) ∈ P2(R)

(a) (6%) Suppose that β = {1 + x2 , x + x2 , 1 + x + x2} is an ordered basis for

P2(R). Find the matrix representation of T in β, i.e., [T ]β

(b) (4%) Let A = [T ]β and let U be a linear operator on R3 defined by U(x) = Ax

for all x ∈ R3. Find a basis for the range space of U .


15. 台聯 A 卷 99–109

《解》 �(a) 令 P2(R) 的標準基底為 γ = {1 , x , x2} , 由⎧⎪⎨

⎪⎩T (1) = 1 − 2x(0) = 1

T (x) = 2x− 1 − 2x(1) = −1

T (x2) = (2x− 1)2 − 2x(2x) = 1 − 4x

可得

[T ]γ =

⎡⎢⎣

1 −1 1

0 0 −4

0 0 0

⎤⎥⎦

故

[T ]β = [I]βγ [T ]γγ[I]γβ

=

⎡⎢⎣

1 0 1

0 1 1

1 1 1

⎤⎥⎦−1 ⎡⎢⎣

1 −1 1

0 0 −4

0 0 0

⎤⎥⎦⎡⎢⎣

1 0 1

0 1 1

1 1 1

⎤⎥⎦

=

⎡⎢⎣

4 4 4

−2 0 −1

−2 −4 −3

⎤⎥⎦

(b) 因 R (U) = CS (A), 又 rank (A) = 2, 故 U 的值域的基底為⎧⎪⎨⎪⎩⎡⎢⎣

4

−2

−2

⎤⎥⎦ ,

⎡⎢⎣

4

0

−4

⎤⎥⎦⎫⎪⎬⎪⎭

3. (a) (6%) Solve y′ +1

x· y = 3x2

(b) (14%) Given the differential : x(t) + a · x(t) + b · x(t) = u(t) where u(t) is a

unit-step function, the response x(t) is expressed as :

x(t) = 0.01 · {1 − c · e−5√

2·t · sin(5√

2t+ θ)}

All initial conditions are zero. Please calculate the constants a, b, c, and the

angle θ (the unit is radian). 《喻超凡 100台聯 A》

《解》 �


(a) ODE 的積分因子為 I(x) = e∫

1x

dx = eln |x| = x, 故

I(x)y =

∫x · 3x2dx =

3

4x4 + c

故

y =3

4x3 +

c

x

(b) 由

x(t) = 0.01 · {1 − c · e−5√

2·t · sin(5√

2t+ θ)}可知 ODE 的特解為 xp(t) = 0.01, 齊次解為

xh(t) = −0.01c · e−5√

2 t · sin(5√

2t+ θ)} (1)

令 x(t) = emt 代回

x(t) + a · x(t) + b · x(t) = 0

中可得

m2 + am+ b = 0 (2)

由 (1) 式可知, m = −5√

2 ± 5√

2 i, 因此

a = −{(−5√

2 + 5√

2 i) + (−5√

2 − 5√

2 i)} = 10√

2

b = (−5√

2 + 5√

2 i)(−5√

2 − 5√

2 i) = 100


x(t) = 0.01 + e−5√

2 t(c1 cos 5√

2 t+ c2 sin 5√

2 t)

由 {x(0) = 0 = 0.01 + c1

x(0) = 0 = −5√

2 c1 + 5√

2 c2

故 c1 = c2 = −0.01 , 因此

x(t) = 0.01 − 0.01e−5√

2 t(cos 5√

2 t+ sin 5√

2 t)

= 0.01{1 −√

2 e−5√

2 t(1√2

cos 5√

2 t+1√2

sin 5√

2 t)}

= 0.001{1 −√

2 e−5√

2 t(sinπ

4cos 5

√2 t+ cos

π

4sin 5

√2 t)}

= 0.001{1 −√

2 e−5√

2 t sin(5√

2 t+π

4)}

故 c =√

2 、 θ =π

4。

15. 台聯 A 卷 99–111

4. If a periodic function whose Laplace transform is shown as :

F (s) =s

(s2 + 1)(1 − e−πs)

(a) (10%) Using power series to expand1

1 − e−πsand find the corresponding pe-

riodic function f(t).

(b) (5%) Please plot this function f(t) in t–domain. 《喻超凡 100台聯 A》

《解》 �(a) 因 L

−1{ s

(s2 + 1)} = cos t , 故

f(t) = L−1{F (s)} = L

−1{ s

(s2 + 1)

1

(1 − e−πs)}

= L−1{ s

s2 + 1(1 + e−πs + e−2πs + · · ·)}

= cos t+ cos(t− π)H(t− π) + cos(t− 2π)H(t− 2π) + · · ·

(b)

2 4 6 8 10 12t

�2

�1

1

2f


5. For a function : f(x) = | sin x|, where −π < x < π. If we wish to use a function

g(x), which is in a finite-dimensional vector space V spanned by trigonometric

functions; sinnx and cosnx for n = 0 to 5, to approximate f(x).

(a) (6%) Does the set

{sin x, sin 2x, sin 3x, sin 4x, sin 5x, cos 0x, cos x, cos 2x, cos 3x, cos 4x, cos 5x}

form an orthogonal basis in V for −π < x < π ?

(b) (9%) Please find g(x) in V that is ”closest” to f(x), i.e., g(x) is with minimum

square error from f(x) in V .

You may need the following formulas :

sin x cos y =1

2[sin(x+ y)+ sin(x− y)] , cosx sin y =

1

2[sin(x+ y)− sin(x− y)]

cosx cos y =1

2[cos(x+ y)+ cos(x− y)] , sin x sin y =

1

2[cos(x− y)− cos(x+ y)]


《解》 �

(a) 因

< sin nx , cosmx >=

∫ π

−π

sin(nx) cos(mx) dx = 0

< sin nx , sinmx > =

∫ π

−π

sin(nx) sin(mx) dx

=1

n2 −m2{−n cos(nx) sin(mx) +m sin(nx) cos(mx)}

∣∣∣π−π

= 0 (n �= m)

< cosnx , cosmx > =

∫ π

−π

cos(nx) cos(mx) dx

=1

n2 −m2{n sin(nx) cos(mx) −m cos(nx) sin(mx)}

∣∣∣π−π

= 0 (n �= m)

< sin nx , sinnx >=

∫ π

−π

sin2(nx) dx = π

15. 台聯 A 卷 99–113

< cosnx , cosnx >=

∫ π

−π

cos2(nx) dx = π

故集合為 V 的一組正交基底。

(b) 因 f(x) = | sin x| 週期為 π 之偶函數 , 故令

f(x) = a0 +∞∑

n=1

an cos(2nx)

其中

a0 =2

π

∫ π2

0

sin xdx =2

π

an =4

π

∫ π2

0

sin x cos(2nx)dx =4

π(1 − 4n2)

則

f(x) =2

π+

∞∑n=1

4

π(1 − 4n2)cos(2nx)

故

g(x) =2

π− 4

3πcos 2x− 4

15πcos 4x

6. Evaluate the integrals (counterclockwise) 《喻超凡 100台聯 A》

(a) (10%)

∮C

ez

cos zdz , C : |z| = 4.5

(b) (15%)1

2πi

∫ c+i∞

c−i∞F (z) dz , where F (z) =

2kz

(z2 + k2)2and k is a constant.

《解》 �

(a) 令 f(z) =ez

cos z, 由 cos z = 0, 可得 z =

2n + 1

2π (n = 0 , ±1 , ±2 · · ·) , 為 f(z) 的

一階 poles , 只有 z1 =π

2、 z2 = −π

2在 C 內 , 且

Resf(π

2) =

ez

− sin z

∣∣∣z= π

2

= −eπ2

Resf(−π2) =

ez

− sin z

∣∣∣z=−π

2

= e−π2


故 ∮C

f(z)dz = 2πi {Resf(π

2) + Resf(−π

2)} = 2πi{−eπ

2 + e−π2 }

(b) 設 c > 0 , 故

1

2πi

∫ c+i∞

c−i∞F (z)dz =

1

2πi

∫ c+i∞

c−i∞

2kz

(z2 + k2)2eztdz

∣∣∣t=0

=1

2πi

∫ c+i∞

c−i∞

2ks

(s2 + k2)2estds

∣∣∣t=0

= L−1{ 2ks

(s2 + k2)2}∣∣∣t=0

= 2 sin(kt) ∗ cos(kt)∣∣∣t=0

= 0

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