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Problem Set 6.7 No. 1

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Advanced Engineering Mathematics, by Erwin Kreyszig 10th. Ed.

Problem Set 6.7

No. 1

(2)
(3)
(4)
(5)

No. 2

y1'y2=0, y1+y2'=2cos t , y1(0)=1, y2(0)=0

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s) sY1(s)y1(0)−Y2(s)=1

Y1(s)+sY2(s)y2(0)= 2 s

s2+1 Replace y1(0)=1, y2(0)=0 sY1(s)−Y2(s)=1

Y1(s)+sY2(s)= 2 s

s2+1

× s+ (s2+1)Y1( s)=s+ 2s

s2+1

Y1(s )= s

s2+1+ 2 s

(s2+1)2

Then from Y2(s)=sY1(s )−1= s2

s2+1+ 2s2

(s2+1)2−1=−

1

s2+1+ 2s2

(s2+1)2

Using (3) of sec.6.6 ℒ-1

{

(s2s 2)2

}

=t sin βt

y1(t )=cost +2 t2 sin t=cost+t sin t In this case β=1

And using (4) of sec.6.6 ℒ-1

{

(s2+s2β2)2

}

=2 β1 (sin βt+βt cosβt)

y2(t )=−sin t+22(sin t+t cost )=−sin t+sin t+t cost=t cost In this case β=1

(6)

No. 3

y1'−2 y1+3 y2=0, y2'y1+2 y2=0, y1(0)=1, y2(0)=0

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s) sY1(s)y1(0)−2 Y1(s)+3 Y2(s)=0 sY2(s)y2(0)−Y1(s)+2 Y2(s)=0

Replace y1(0)=1, y2(0)=0 (s−2)Y1(s)+3 Y2(s)=1

−Y1(s)+(s+2)Y2(s)=0

×( s+2)- × 3 (s2−4 +3)Y1( s)=s+ 2

Y1(s)= s+2

s2−1 Then from Y2(s)=s+21 Y1(s)= 1

s2−1

Furthermore Y1(s)=s+2

s2−1=

3 2

s−1

1 2

s+1

Y2(s)= 1

s2−1=

1 2

s−1

1 2

s+1

y1(t )=32et12e−t=12et+12e−t+2(12et12e−t)=cosht +2 sinh t

y2(t)=12et12e−t=sinht

No. 4

y1'=4 y2−8cos 4t , y2'=−3 y1−9sin 4t , y1(0)=0, y2(0)=3

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s)

sY1(s)y1(0)=4Y2(s) 8 s

s2+16

sY2(s)y2(0)=−3Y1(s) 36

s2+16 Replace y1(0)=0, y2(0)=3

sY1(s)−4Y2(s)=− 8 s

s2+16

3Y1(s)+sY2(s)=3− 36

s2+16

(7)

× s+ × 4 (s2+12)Y1( s)=− 8 s2

s2+16+12−144 s2+16

(s2+12)Y1( s)=−8s2+12 s2+192−144

s2+16 (s2+12)Y1(s)=4 s2+48

s2+16=4(s2+12)

s2+16

Y1(s)= 4

s2+16

×3- × s

(−12−s2)Y2(s)=− 24 s

s2+16−3 s+36 s

s2+16=−24 s−3 s3−48s+36s

s2+16 =−36 s−3s3

s2+16 =3s(−12−s2)

s2+16 Y2(s)= 3 s

s2+16 y1(t)=sin 4 t y2(t)=3 cos4 t No. 5

y1'=y2+2−u(t−1) , y2'=−y1+1−u(t−1), y1(0)=1, y2(0)=0

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s)

sY1(s )− y1(0 )=Y2( s)+2se−s s sY2(s )− y2(0 )=−Y1( s)+1se−s

s Replace y1(0)=1, y2(0)=0

sY1(s )−Y2(s )=1+2se−s s Y1(s )+ sY2(s )=1se−s

s

× s+ (s2+1)Y1(s)=s +2−e−s+1se−s

s =2+s2+1

s (s+1s )e−s

Y1(s)= 2

s2+1+1

s s+1 s(s2+1)e

−s

= 2

s2+1+1

s

(

1sss−12+1

)

e−s=s22+1+1

s

(

1ss2s+1+ 1 s2+1

)

e−s

- × s

(8)

(−1−s2)Y2(s)=1+2se−s

s −1+e−s=2 se−s

s +e−s

Y2(s)=− 2

s(s2+1)+

[

s21+1+s(s12+1)

]

e−s=−2s +s2s2+1+

(

s21+1+1ss2s+1

)

e−s

y1(t)=1+2 sint−[1−cos(t−1)+sin(t−1)]u(t−1)

y2(t)=−2+2 cos t+[1−sin(t −1)−cos(t−1)]u(t−1)

No. 6

y1'=5 y1+y2, y2'=y1+5 y2, y1(0)=1, y2(0)=−3

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s) sY1(s)y1(0)=5 Y1(s)+Y2(s)

sY2(s)y2(0)=Y1(s)+5 Y2(s)

Replace y1(0)=1, y2(0)=−3 sY1(s)−5 Y1(s)−Y2(s)=1 (s−5)Y1(s)−Y2(s)=1

−Y1(s)+sY2(s)−5 Y2(s)=−3 −Y1(s)+(s−5)Y2(s)=−3

×( s-5)+ [(s−5)2−1]Y1(s)=s−5−3

Y1(s)= s−8

(s2−10s+24)= 2

s−4 1 s−6

+ ×( s-5)

[−1+(s−5)2]Y2(s)=1−3s+15=−3 s+16

Y2(s)= −3 s+16

(s2−10 s+24)= −2 s−4 1

s−6 y1(t )=2e4t−e6t

y2(t )=−2e4 t−e6t

(9)

No. 7

y1'=2 y1−4 y2+u(t−1) et, y2'=y1−3 y2−u(t−1) et, y1(0)=3, y2(0)=0

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s)

sY1(s )− y1(0 )=2 Y1(s )−4 Y2( s)+e e−s s−1 sY2(s )− y2(0 )=Y1(s )−3 Y2( s )−e e−s

s−1 Replace y1(0)=3, y2(0)=0

sY1(s )−2 Y1( s)+4 Y2(s )=3+e e−s

s−1 (s−2) Y1( s)+4 Y2(s)=3+ e e−s s−1

−Y1(s )+sY2(s )+3 Y2(s )=−e e−s

s−1 −Y1(s )+( s+3) Y2( s)=−e e−s s−1

× (s+3)-×4 [(s−2) ( s+3)+ 4]Y1( s)=

(

3+es−1e−s

)

( s+3 )+ 4 e e−s s−1

(s2+s−2)Y1(s )=3 s+9+(s−1s+7)ee−s

Y1(s )= 3 s+9

(s−1) (s+2)+ (s+7)

(s−1)2(s+2)ee−s= 4

s−1 1

s+2+

[

(s−1)59s+2299 +s+259

]

ee−s

=s−14 s+21 +

[

(s−1)59 +(s−1)83 2+s+259

]

ee−s

+×( s-2) [4 +(s +3) (s−2)]Y2(s)=3+es−1e−s (s−2)e e

−s

s−1

(s2+s−2)Y2(s)=3−(s−3)e e−s s−1

Y2(s)=(s−13) (s+2) s−3

(s−1)2(s+2)ee−s= 1

s−1 1

s+2

[

( s−1)59s−1192s+259

]

ee−s

=s−11 s+21

[

s−159 ( s−1)23 2s+259

]

ees

(10)

y1(t )=4 ete−2t+[59ee(t−1)+83(t−1)ee( t−1)+59ee2(t−1)]u(t−1)

=4 et−e−2t+[(83t−299 )et+59e(−2t +3)]u (t−1)

y2(t)=et−e−2t[59eet−123(t−1)eet−159ee−2 (t−1)]u(t−1)

=et−e−2 t[(23t+119 )et59e(−2 t+3)]u (t−1)

=ete−2 t+[(23t−119 )et+59e(−2t+3)]u(t−1 )

No. 8

y1'=−2 y1+3 y2, y2'=4 y1y2, y1(0)=4 , y2(0)=3

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s) sY1(s)y1(0)=−2 Y1(s)+3 Y2(s) sY2(s)y2(0)=4 Y1(s)−Y2(s)

Replace y1(0)=4, y2(0)=3 sY1(s)+2 Y1(s)−3 Y2(s)=4 (s +2)Y1(s)−3 Y2(s)=4

−4 Y1(s)+sY2(s)+Y2(s)=3 −4 Y1(s)+(s +1)Y2(s)=3

×( s+1)+×3 [(s +2) (s+ 1)−12]Y1(s)=4 s +4 +9

Y1(s)= 4 s+13

(s2+3 s−10)= 3 s−2+ 1

s+5

×4+×( s+2)

[−12+(s +2) (s+ 1)]Y2(s)=16+3 s+6=3 s +22

Y2(s)= 3 s+22

(s2+3 s−10)= 4

s−2 1 s+5 y1(t )=3e2t+e−5t

y2(t )=4 e2 t−e−5 t

(11)

No. 9

y1'=y1+y2, y2'=−y1+3 y2, y1(0)=1, y2(0)=0

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s) sY1(s)y1(0)=Y1(s)+Y2(s)

sY2(s)y2(0)=−Y1(s)+3 Y2(s)

Replace y1(0)=1, y2(0)=0 sY1(s)−Y1(s)−Y2(s)=1 (s−1)Y1(s)−Y2(s)=1 Y1(s)+sY2(s)−3 Y2(s)=0 Y1(s)+(s−3)Y2(s)=0

×( s-3)+ [(s−1) (s−3)+1]Y1(s)=s−3

Y1(s)= s−3

(s2−4 s+4)=(s−2)−1 (s−2)2 =

1

s−2 1 (s−2)2

-×( s-1)

[−1−(s−3) (s−1)]Y2(s)=1 Y2(s )=− 1

(s2−4 s+4)=− 1 (s−2)2= y1(t )=e2t−te2 t

y2(t )=−te2t

No. 10

y1'=−y2, y2'=−y1+2[1−u(t−2π )]cost , y1(0)=1, y2(0)=0

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s)

2[1−u (t−2 π )]cost=2 cost−2 u (t−2 π ) cos(t−2 π ) sY1(s)y1(0)=−Y2(s)

(12)

sY2(s )− y2(0 )=−Y1( s)+ 2 s

s2+12 se−2 πs

s2+1 Replace y1(0)=1, y2(0)=0

sY1(s)+Y2(s)=1 Y1(s )+ sY2(s )= 2 s

s2+12 se−2 πs s2+1

×s- (s2−1)Y1(s )=s− 2 s

s2+1+2 se−2 πs s2+1

Y1(s)= s

(s2−1) 2 s

(s2−1)(s2+1)+

2se−2 πs

(s2−1) (s2+1)

=

1 2

s−1+

1 2

s+1 s

s2−1+ s

s2+1+

(

s2−1s s

s2+1

)

e−2πs

=

1 2

s−1+

1 2

s+1

1 2

s−1

1 2

s+1+ s

s2+1+

(

s−112 +s+112 s2s+1

)

e−2πs

= s

s2+1+

(

s−112 +s+112 s2s+1

)

e−2 πs

-×s

(1−s2)Y2(s)=1− 2s2

s2+1+2 s2e−2 πs

s2+1 =(s2−1)

s2+1 +2s2e−2πs s2+1

Y2(s)= 1

s2+1 2s2e−2 πs

(s2−1)(s2+1)

= 1

s2+1

(

s2−11 + 1

s2+1

)

e−2 πs

= 1

s2+1

(

s−112 s+112 +s21+1

)

e−2πs

y1(t )=cos t +[12e( t−2 π)+12e−(t −2 π )−cos(t−2 π )]u (t−2 π )

=cos t+[cosh (t−2 π )−cos t]u (t−2 π )

(13)

y2(t )=14et14e−t+12sint−[12e( t −2 π )12e−(t−2 π )+sin (t−2 π )]u (t−2 π )

=sin t−[sinh (t−2 π )+sin t]u (t−2 π )

No. 11

y1} } =y rSub { size 8{1} } +3y rSub { size 8{2} } ,y rSub { size 8{2} } rSup { size 8{

=4 y1−4et, y1(0)=2, y1'(0)=3, y2(0)=1, y2'(0)=2

Writing ℒ {y1}=Y1(s),

{y2}=Y2(s)

s2Y1(s)−sy1(0)− y1'(0)=Y1(s)+3Y2(s)

s2Y2(s)−sy2(0)y2'(0)=4Y1(s)s−14 Replace

y1(0)=2, y1'(0)=3, y2(0)=1, y2'(0)=2 (s2−1)Y1(s)−3 Y2( s)=2 s+3

−4Y1(s)+s2Y2(s)=s+2−s−14

× s2 +×3

[s2(s2−1)−12]Y1(s)=(2s+3)s2+3s+6−s−112 =2 s3+3s2+3s+6−s−112

Y1(s)=2 s3+3 s2+3 s+6 s4s2−1212

(s4−s2−12)(s−1)

= 2 s3+3 s2+3 s+6

(s24)(s2+3)

12

(s2−4)(s2+3)(s−1)

=

11 7s+187

s2−4+

3 7s+37

s2+3

(

47s2s+−447+

3 7s+37 s2+3 1

s−1

)

= s+2

s2−4+ 1 s−1= 1

s−2+ 1 s−1

×4+× (s2−1) [−12+s2(s2−1)]Y2( s )=8 s +12+(s2−1)(s+2−s−14 )

(s4s2−12)Y2(s)=8s+12+s3−s+2s2−2−4(s2−1) s−1

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