./ MAP 2302 Exam #3
Name:
Prnw Klj
HONOR CODE: On my honor , I have neither given nor received any aid on this
examination.
Signature: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
Instructions: Do all scratch work on the test itself. Make sure your final answers
are clearly labelled. Be sure to simplify all answer~ whenever possible. SHOW ALL
WORK ON THIS EXAM IN ORDER TO RECEIVE FULL CREDIT!!!
No.
Score
1
2
3
4
I
Total
I
/30
/10
/20
/40
/100 I
(1) Apply the definition to find the Laplace transform of the function. State which
values of s the Laplace transform exists for. (15 points each)
(a) f(t)
=
te t
t~tt~J = fOO e-st . te~Jt
o
"-.
:::.
~
i.f
s-\ >0
~L'Ho~f;tJ'5 ~\t .. -= 0 it ~
't:> >1
(b)
2345678
x
-1
....-
-
0
-:) t~+(t)1~r~e-St.l dt +-r~e-5~(-llolt rJ.~iSt.oot{:
J2
o
'i
'---r---­
'::0
-;: 12. i~tJb -f~e-5'\lt
o
:c -
1
~ e-~t\: + *I~t\:
I
- 2S
:: - S e
~
-2S
- -f:e
v-JAJ
.1. - 1.\ ~
,
+s ~ s e
\
\
+--S+-Se
~
\ -lS
- S;e
- ~~
edt rtJ r\1ffl\~~
(2) Use Laplace transforms to transform the initjal value problem in y(t) into an
algebraic equation in Y (s). Solve the algebraic equation for Y (s) but do not do
partial fractions or solve the initial value problem. (10 points)
y(4) _ 3y" + 5y = t 2 e- 3t
+ 4e 2t cos V5t + 4 sin 3t y(O) = 1, y'(O) = 0, y"(O) = -2, y"'(O) = -8, y(4) (0) = -6 -.
-
~V
\
r:
a~
l.H
l5 +~}
3
~ ((~) -~
-r ~
~ -;t
f.
"2.
l~ -J.)
f'S"
+~
z
+-Is +-~ - 35 ,((c.,') -+ s'.) -t- S
~
~ z:
I
v
.,...
'2.
'
~ ;J..
'((~') ~ (s ..:~)\
+ II
1
S -;2.
(s-:lf+~
+- u
-(
f3 s L.+
~
5'1y(~) -~S2.y~)+-S-Y(~):: (~~)' +~ (~:;)~+~ t-~ sfl~ "+S3_S-S-~ . . (~~)"\ ~ 'i {~:~~+-~
Sl{ -
of-
~ s~~
~Sz. ~S"
+- 6
3
-
~~ -g
(3) Find the inverse Laplace transform. (10 points each)
2S
(a) 2- 1 {s3_
+6 }
3
8
(S-3)
_
.....
~2 (S-i) +-SS(~-!.) +- ((~-3)+ 1x3
-------
-& ~(s-3)
,As?. (&-3) +- B.s(~-J) ~ C(b -~) +-DS~ :: ~ 3 -
~S +Co
~~ _ ~~2. +-ttt-1.'BS +-CS _3C + 'D~~ -= <S~-2S+"
---
.....".."....
A+'D=t
~ -3~
-~C=
-:=-
--
~ ~
--­
~O+1)=-l ~b:::1
=0
L-~~~
~
-;}
~ 0 -),\==-0
=:> A---o
~-~-5B::: -J ~ -~B'::O ~g=O
lo =i> C
~-;;
­
t~' 5
G,S
~
7 £,Z -zs+-n - S'-2Str.} -
(~~~)'i"'") :: (.,(etu»~t ~~ - 3ti
= ~/C0~t-?'H-~tl
1t
(4) Use Laplace transforms to solve the initial value problem. (20 points each)
+ y' = 0;
(a) yl!!
t 1:rJ
l
y(O) = -4, y'(O) = -2, yl/(O) = 1
t~7'} "- t{O}
)"(0\ +-S ({\) - J(b) ~O
J'
S3 Y(s) _f,2J(~)- 5 (0\ .................-..-.
-~
---
-2
....-­
\
-'(
<~?y(~) +-~ Y(~) = _~~2 -2S - ~
Y(!>l (S} ~S)::-
~'-t~Z -2.S- ~
yt)) ~ -'{{,~ - 2$- ~
2~ - "3
5(~:t. "-1) -:.
- l{S'l -
~ ~ t-~
-Lh"l. -2.s-3.-=
_ 4 (SL+~ ~ (&.K) s
:;. S6 ¥1')
1
A(s~+-;').f (6s+-C)S -l{r,1. - 1S-"3 -=. ~7.+A +-f>s~ tD :. (~"\l. +C~ ~A A-;-S-=-
-tt
C
-:J.
=:.
~ -~+e,
= - l{
~
l3 =. - 1
A-- =- ~3
Jl~\ ~ ...P" - \ S1 -5"'3 -
S
s~+t -
;;-.1
¥H' J
:: -Yr'HI - r~s£tl-j-.:2 ('~ &1~,
-=- \-:, - ~ I:. - .::l·5\"'~
}
(b) y" - 6y'
+ 13y =
0; y(O) = 4, y'(O) = 8
h{J" 3- ~ -.(~~' Jt 13 ;(~:JJ ~ i fo}
~/"Yb) -5j(0)- J'(b) -1o(sM-~1.)H3nS):: 0
v--
~
~')..y(~)
..;.y­
~
'1
"\
-- ~~ -<6 -(os yes)
2"1
t-
+ f3r(~) =- v
'>
S~ ((~) -~s ((S) 4---{3 yes) = Ys - (~
y( ~') (~"Z.-~s 'H~\ ~ \{ (5- 4)
y(~):: . ~!>-4)
_
~ _\.(_(~_-4_)_ _ _
5 -G,st-r~
(s~-<¢..f-'\)}(l3"~)
-
'{(s-'t)
C~_3)'2. +- Y
'1(t)-=: ~ ,,~t5 ~t ~\{...p-'5 (S-~)+-(-\iH) 1
J
1 6-~Y~ J
"-) (':J-1")1 + \( )
_ \ { [ . . , ..
--
A
t~ s-3
1 6-j)1. f~
1
~"'5- '
- " ) (~-~y +-4
,1J J