XMPLS An Initial-Value Problem + Solve y' y f(t), yO 5, where =
f(t)
=
{ 0•
=
0 : t < 7 3 cost, t ; 7. 3 costU(t tU(t- 7 d so by neariy, e
SLUTN The ncion!can be wen asf(t) resus o Exampe 8, and he sa s a paral par al acons we h ave =
!{y} + !y} s - y Y s sY s y((O)
+
3{costU(ttU(t-
s Y s
=
- 3
s
2 s+ e-1 1
3+ s 1 s + Ys 5 s 1
Now proceedng as we did n Example 7 i llows om o he es n e bracke e
e-1 s
}
=
- 1o U(t - 7), e-< t t-
!1
d
e-1 s } s + 1
1 =
5 wh a 7 ha e inverses =
e- s
}
=
sin(t sin (t - 7)o U(t - 7,
co c os(t - 7) U U( ( t t - 7.
Ths e nverse o (17) is
3 2 1 -1
sn(t-)U(t -)U(t- 7 l cos(t -)U(t- 7 y 5 + l (t !U(t 7 - l sn(t cos(t-)U(t 5 + % 1 + snt + cost U(t- 7 +igonomeic idenies _
=
[
2n
3n
FIGURE 4..7 Gaph of functon (18) in
Example9
=
{ 5 + � 5e-
]
e-< t- 1)
+�
sint sint
+
� c cost ost ,
0 t < t
; 7.
Wih he ad o a graphng uiiy we ge he graph o (18) shown n
FIGURE 4..7 4..7
beam o lengh L D B Secion 3.9 we saw ha he saic deecion y o a n beam
cang oad per n engh s nd om he near rhorder dierena eqaion
d4 y
I
=
9
where s Young's modlus o easciy d I is a momen o nera o a cross secon o e beam The aple s is pcly use when is piecewise dened, dened, b n n order o use he ans we mus acy assume a y d e dened on 0 raer h on 0 L) Noe oo, a e nex expe s a boundvaue pbem raer nve problem.
XMPL 0 A onar-Vale Poblem A beam o lengh L s embedded a boh ends as shown n FIGURE 48 48 Fnd he deecon o
w(x)
he beam when he load s gven by
wal
x
FIGURE 4..8 Embedded beam wi a
varabe oad in Example 10
228
{( o
1-
)
�L
0
where s a consan.
0 < < < L
SLUTN Recal ha snce he beam is embedded a boh ends he bondary condiions e yO 0 yO 0 y 0 y 0 Now by 0 we can express n es o
he uni sep ncon: ncon:
CHER h pc f
( i)
( i ) ( -�)
w( x ) = wo 1 - x - wo 1 - x a x
( �) a( x - �)].
=
- x + x -
Transfrming (9) wit spec o t e varable x gives
] [ 2w0 /2 1 s4Y(s) - sy"(O) - y"'(O) = [ - + e un ] . EI s s s
2w0 L2 1 1 E l (s 4 Y (s) - s3 y(O) - s2 '(O) - sy"(O) - y" '(O)) = - + e un s s s or I we e CJ
= y"(O) and c = y"'(O), ten CJ c 2w0 /2 1 1 Y(s) + + - + eun s3 s4 EI s s6 s6
[
J
s -
ad consequenly
{ } { } 2w0 L/2 � -J 4! _ � -J 5! } _ � -J 5! e -un } } + { { s6 + 5! { s6 ] [ EI ! s 5!
C J ! � J 2! ( x ) 2! s3 + 3! s4 _
Applying te condions y() fr CJ and c :
= and y () = o he las resl yelds a syse o eqaions
3 9w0 4 c+c+ O J 92£ 2
85w0 3 CJ + C z + %OE Sovng, we nd CJ = 2wo 9EI and c = -9woEI Thus he deecon s y( x ) =
[
Exe re is es C
) ( L)]
2wo wo Wo 5 4 - x + x - 2 a x - 2 . x 3 + x X T SO E / 60 E I L 1920 E /
(
Answers to selected odd-numbeed problems begin on page ANS-9.
T h x
In Probes 2 nd eher F(s) or() as ndcaed. 1.
� {te 10}
2
� te - 6}
3.
� {t 3e -2 }
4
� t 10 e -
5.
� {t( e +e 2 )2 }
6
7 9
{e sn t} 8 {( - e +e 4 ) cos 5t
e (t - 1) } e cos t}
J 15. �
{ s + s + 5 } s
43
22
L
0
c I Problems 21-30, use the Lapace transf to solve the
given
niavae probem. 21. y'+ 4y = y(O) = 2 y' y 1+ t , y(O) 0 y"+ 2y+ y 0, y(O) 1, y'(O) 1 24. y" 4y+ 4y y(O) 0, y'(O) 0 25. y" 6y+ 9y t, y(O) 0, y'(O) 1 26. y" 4y+ 4y , y(O) 1, y(O) 0 27. y" 6y+ 13y 0 y(O) 0 y'(O) -3 28. 2y+ 20y+ 5y 0 y(O) 2, y(O) 0 y" y' cos y(O) 0, y'(O) 0 3 y" 2y+ 5y 1+ t y(O) 0 y(0) 4
I obes 37-48 ind eiher F() orf(t)
!{(t- l)U(t-1)} !{tU(t- 2) 41. !cos t U(t- ) - -1 37. 3 9.
se the Lapace ransfrm d the procedure outined in Exampe 0 to solve e given bondarvalue probem 1. y"+ 2y+ y 0, y(O) 2 y() 2 y"+ Sy+ 2 0 y(O) 0 y') 0 A b weight stretches a sprng 2 The weigh is released m s 18 in above e eqbm postion d e reslt ing moton tes pace in a medim oeng a damping frce numercal eqa to� imes the instteos veoci Use the Laplace trsfrm o nd the eqaion o moion x(t). 4. Recal ha he dierenta eqaon fr the instantaneos charge q(t) on e capactor in Cseres crct s (20)
ee ecion 3.8. Use he Laplace trasfrm o nd q(t) when =1 h=20, C= 0.005 ()=150 V t> 0, q(O)=0 and i(O) = 0. Wha is the cuent i(t)? 5. onsider e bae o const volageat chrges e cap or shown n FGURE 4.9 Dvde equaon (20) b d deine 2 I d /C Use the Laplace trsf o show that e souon q(t) o q"+ 2 + q / sbjec to q(O) 0 i(O) 0 is
[ [ ] [ {
1 -
A q(t) 1 - ( + At) +
_
230
45.
47.
1
s} } } e �·
+ 1 •
( + 1)
- w2t
a
)]
)]
!- 1
} s•n
+ 4 1 ( -1)
FGURE 4 aph fr Pobems -
() b c d f 49.
f(t) -f(t) U(t- f(t- U(t- f(t) U(t- f(t) -f(t) U(t- f(t) U(t- -f(t) U(t - f(t- U(t- -f(t- U(t-
51.
� II II
A =
I a
, <
44.
FGURE 41 aph fr Poblem
A >
as indcaed. !{ U(t- 2)} (3t+ )U(t-1) sntU(t - ?/2) (1 + -1 +2
n Probems 49-54 match the gven graph with one o the gven nctons n (a)-(). The graph of(t) s given in FGURE 4
a
1 - cos - 2t A sn - A2t + _
Use the Lapace ransrm o nd he charge q(t) n an Cseres when (O) = 0 and () =k k > 0. onsder two cases: k * /C ad k= /C.
C Translation on the -axis
I Pbes 31 d 32,
d2q dq 1 + + q = (). dt dt2
FGURE 49 Ciut n Poblem
I
FGURE 4 aph fr Poblem
CHAPER 4 Lp m
( II
II I
a
FGURE 4..2 aph fr Poblem
