1996.1.superyu 1. Find the solution satisfying the given condition (10%)

1996.1.superyu
1. Find the solution satisfying the given condition (10%)
y =
1
x2 y2 .
1
x
y + 1 ; y(1) = 3
j. . Ñ Riccati j˙ , .øøÔjÑ yp = x, ]I y =
1
v
+ x, ＋ y = .
v
v2 + 1, H
. ODE 2ª)
v +
1
x
v = .
1
x2
}ÄäÑ
I = e 1
x dx = eln x = x
]
Iv = x(.
1
x2 )dx = .ln |x| + c
1
v =
c . ln |x|
x
Ĥ
y(x) =
1
v
+ x =
x
c . ln |x|
+ x
â y(1) = 3, ª) 3 =
1
c
+ 1, ] c =
1
2
, 1
y(x) =
x
1
2 . ln |x|
+ x
2. Given that x and xex are solutions of the homogeneous equation corresponding to
x2y . x(x + 2)y + (x + 2)y = 2x3 (x > 0)
Find the general solution by using the Method of Variation of Parameter. (10%)

2
j. I y1 = x, y2 = xex, yI yp = y1φ1 + y2φ2, Ä
W(y1 , y2) =
x xex
1 ex + xex
= x2ex
]
φ 1 = .
2xy2
W(y1 , y2)
= .
2x2ex
x2ex = .2
＋ φ1 = .2x, /
φ 2 =
2xy1
W(y1 , y2)
=
2x2
x2ex = 2e.x
＋ φ2 = .2e.x, ]
yp = y1φ1 + y2φ2 = .2x2 . 2x
Ĥ
y(x) = c1x + c2xex . 2x2 . 2x
3. Let field .→G(x, y, z) = .→a x(x . yz) + .→a y(y2 . zx) + .→a z(z2 . xy). (10%)
(1) Find the line integral C1
.→G(x, y, x) · d.→ , where C1 is a segment of the curve
y = x2, z = x from (0,0,0) to (1,1,1).
(2) Compute the sum of line integrals C2
.→G(x, y, x) · d.→ + C3
.→G(x, y, z) · d.→
where C2 is the straight line from (1,1,1) to (0,0,0) and C3 is along the z.axis
from (0,0,1) to (0,0,0).
j.
(1)
C1
.→G · d.→ = C1
[(x . yz)dx + (y2 . zx)dy + (z2 . xy)dz]
= 1
x=0
[(x . x2 · x)dx + (x4 . x2)(2xdx) + (x2 . x3)dx]
=
1
6
(2) Ä ▽×.→G = 0, / .→G ƒb￡ø¼Rûbc/, ]
C2
.→G · d.→ + C3
.→G · d.→ = . C1
.→G · d.→ = .
1
6

3
4. Evaluate the integrals along the path C that is the counterclockwise circle with
|z| = 3.
(a) C
z2 . 1
z2 + 1
ezdz (b) C
sinh 3z
(z2 + 1)2 dz (10%)
j.
(a) I f(z) = (
z2 . 1
z2 + 1
)ez, ] f(z) x z = ±i íø¼ poles, /
Resf(i) = (
z2 . 1
2z
ez)
z=i
= iei , Resf(.i) = (
z2 . 1
2z
ez) z=.i
= .iei
]
C
f(z)dz = 2πi[Resf(i) + Resf(.i)] = 2πi[iei . ie.i] = .2πi sin(1)
(b) I f(z) =
sinh 3z
(z2 + 1)2 , ] f(z) x z = ±i íù¼ poles, /
Resf(i) = lim
z→i
d
dz
[(z . i)2 sinh 3z
(z2 + 1)2 ]
= lim
z→i
d
dz
[
sinh 3z
(z + i)2 ]
= lim
z→i
3 cosh3z · (z + i)2 . sinh 3z · 2(z + i)
(z + i)4
=
3 cosh(3i)(2i)2 . sinh (3i) · 2(2i)
(2i)4
= .3 cos 3+sin3
4
°Ü
Resf(.i) = lim
z→.i
d
dz
[
sinh 3z
(z . i)2] = .3 cos 3+sin3
4
]
C
f(z)dz = 2πi[Resf(i) + Resf(.i)] = πi(sin 3 . 3 cos 3)

4
5. The Fourier transform of x(t) is defined as (15%)
X(jw) = ∞
.∞
x(t)e.jwtdt
Suppose
x(t) = 1 |t| < 1 2 0 otherwise (a) Find X(jπ) (b) Calculate the following convolution integral ∞ .∞ sin(πτ) sin[2π(t . τ )] π2τ (t . τ ) dτ j. (a) X(jw) = ∞ .∞ x(t)e.jwtdt = 2 1 2 0 cos wt dt = 2 w sin w 2 ] X(jπ) = 2 π sin π 2 = 2 π (b) âú.4”ªø F{ sin t 2 t 2 } = 2πx(w) yâ ìܪø F{ sin(πt) πt } = 2π 2π x( w 2π ) = x( w 2π ) = 1 |w| < π 0 otherwise F{ sin(2πt) 2πt } = 2π 4π x( w 4π ) = 1 2 x( w 4π ) = 1 2 |w| < 2π 0 otherwise ] F{ ∞ .∞ sin(πτ) sin[2π(t . τ )] π2τ (t . τ ) dτ} = F{2 sin πt πt . sin(2πt) 2πt } = 2x( w 2π ) 1 2 x( w 4π ) 5 = 1 |w| < π 0 otherwise = G(w) ] ∞ .∞ sin(πτ) sin[2π(t . τ )] π2τ (t . τ ) dτ = F.1{G(w)} = 1 2π ∞ .∞ G(w)ejwt dw = 1 2π π .π ejwtdw = 1 π π 0 cos wt dw = sin πt πt 6. Solve the initial value problem (10%) y + 9y = f(t) ; y(0) = y (0) = 1 f(t) = 0 if0. t < π cos t if t . π j. Ä f(t) = 0 if0. t < π cos t if t . π = cost u(t . π) / L{f(t)} = e.πsL{cos(t + π)} = . s s2 + 1 e.πs ú ODE s.| L–T ª) s2ˆy(s) . sy(0) . y (0) + 9ˆy(s) = L{f(t)} = . s s2 + 1 e.πs w2 L{y(t)} = ˆy(s), cܪ) ˆy(s) = s + 1 s2 + 9 . s (s2 + 1)(s2 + 9 e.πs = s + 1 s2 + 9 + 1 8 ( s s2 + 9 . s s2 + 1 )e.πs 6 ] y(t) = L.1{ˆy(s)} = cos3t + 1 3 sin 3t + 1 8 [cos 3(t . π) . cos(t . π)]u(t . π) 7. Find A100, if (a) A = .. 3 1 1 2 4 2 .1 .1 1 .. . (10%) (b) A50 = .. 3 1 1 2 4 2 .1 .1 1 .. . (5%) j. (a) â det (A . λI) = 0 ª)Ô MÑ λ = 4 2 2 , / rank (A . 2I) = 1, ] A í| üÖá Ñ f(x) = (x . 4)(x . 2) 1 f(A) = (A . 4I)(A . 2I) = A2 . 6A + 8I = 0, I g(x) = x100 = Q(x)f(x) + ax + b = Q(x)(x . 4)(x . 2) + ax + b Ä 4100 = 4a + b 2100 = 2a + b . a = 1 2 (4100 . 2100) , b = (2 · 2100 . 4100) ] g(A) = A100 = aA + bI = 1 2 (4100 . 2100)A + (2 · 2100 . 4100)I (b) I B = A50, / B Å— B2 . 6B + 8I = 0, ] A100 = B2 = 6B . 8I = .. 10 6 6 12 16 12 .6 .6 .2 .. 7 8. By employing the power series method for the equation of .x2y + y + xy . y = 0 the solution can be expressed as the form of y = a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + · · · a0 = 0 Please calculate the value of a4a5 a3 + a4 + a5 . (10%) j. Ä x = 0 Ñ O.D.E. í õ, ]I y = ∞ n=0 y(n) n! xn = a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + · · · ＋ y(0) = a0, y (0) = a1, â (1 . x2)y (x) = .xy(x) + y(x) ] y (0) = y(0) = a0 = a22!, , s. }ª) (1 . x2)y (x) . 2xy (x) = .xy (x) . y (x) + y (x) . (1 . x2)y (x) = xy (x) ] y (0) = 0 = a33!, , s. }ª) (1 . x2)y(4)(x) . 2xy (x) = xy (x) + y (x) ] y(4)(0) = y (0) = a0 = a44!, , s. }ª) (1 . x2)y(5)(x) . 2xy(4)(x) = 3xy(4)(x) + 3y (x) + y (x) ] y(5)(0) = 0 = a55!, Ĥ a4a5 a3 + a4 + a5 = 0 (∵ a5 = 0) 9. (μ2æ) The steady-state temperature u(x, y) in s semi-infinity plate (x > 0,
y > 0) is determined from the boundary conditions of u(0, y) = e.y, u(x, 0) = e.x.
Find P, and Q and S of the solution u(x, y) =
2
π ∞
0
α
P
(Qsin αy + S sin αx)dα
(A) P = 1+α2, (B)Q = e.αx, (C)S = e.αx, (D)Q = e.αy,
(E) P = 1+α. (10%)

8
j. I u = u1 + u2, /
▽u1 =
∂2u1
∂x2 +
∂2u1
∂y2 = 0 u1(0, y) = 0 , u1(x, 0) = e.x (x > 0 , y>0) (1)
▽u2 =
∂2u2
∂x2 +
∂2u2
∂y2 = 0 u2(0, y) = e.y , u2(x, 0) = 0 (x > 0 , y >0) (2)
ú (1) | Fourier sine 2ª)
.α2U1 + αu1(0, y) +
d2U1
dy2 = 0
w2 Fs{u1} = U1(α , y), cܪ)
d2U1
dy2 . α2U1 = 0
j, ª)
U1(α , y) = c1e.αy + c2eαy
â lim
y→∞
U1(α , y) = bounded, ª) c2 = 0, ]
U1(α , 0) = Fs{u(x, 0)} = ∞
0
e.x sin αx dx =
α
1 + α2 = c1
]
U1(α , y) =
α
1 + α2 e.αy
Ĥ
u1(x, y) = F.1
s {U1(α , y)} =
2
π ∞
0
α
1 + α2 e.αy sin αx dα
°Ü
u2(x, y) =
2
π ∞
0
α
1 + α2 e.αx sin αy dα
]
u(x, y) = u1(x, y) + u2(x, y) =
2
π ∞
0
α
1 + α2 (e.αy sin αx + e.αx sin αy)dα
2 (A) ￡ (B)