MAP 2302 Exam #3

Name: At.wtr...ouL K~=--.l....Uo

(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)

e-3t (a) f (t) =

s~ - !Jl -~~ J - h~ rC -(~\-~t d o e e (J...t. - C.....,Oo J e lo

\ - (~))c -- e +St~ '5+3 ~

~o if ~~'>O

(b)

(1,1)

y

oL-------------------x

::: JI u.-st dt + h", re-St cit. o (~oo I

(2) Use Laplace transforms to transform the initial 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)

et2ylll + 3y" - 3y' - 2y = t 4 - 3e-2t sin 37ft - 4 cos./3t y(O) = I , y'(O) = 0, y"(O) = -2

2t{ YeS) ::.(~:!> t"'>S' - y, - ;))(S_I) S"

+

(3) Find the inverse Laplace transform. (10 points each)

(a) XO- 1 {CS+2 )(s2+4)}

_ A(~~ t\{)+ (f>stc.) (Sf;l)- ------:-..:.....--(~;1)(S2 t \( )

s -= A(s"tt\() t (f15tC)(S-t-;l) S::.-~~ -~ -=~tl ~A=-~

_1. -t ~ ::: 0 -=;> ~)::: J. \{

JotC ~ , ;tt - \ ~O => c:: i

1 '-15 - J.s-! ~ - ~ i ~I ~ s~~ ~ +t 1 :1 ~ ~ J - ~ t" ~ ~ J~ ~;(' J/~~ ) ~ i ~ -'{ S:~~ ~

s s s

-

-

(4) Use Laplace transforms to solve the initial value problem. (20 points each) (a) y" + 8y' + 15y = 0; y(O) = 2, y'(O) = -3

trr+-~ tfy'}l-IS tlr l" 0 ~~Y(Sh)~(b) -:1'(0) +~-(s'I'(S) -~(ol) tIS y(~) =0

'-"-'"'" ----- ~ ~ - ) do.

yeS) (S' ti'~ tIS) ::: ~ ~ +13 A B _ A-(Sf3) +s(~A-s")Y(s)::. C:Zs+ (3 -= ~~ ~ l ~

- ;:s + '5+"3 -- (S~5)(S+~)S2.~stIS (5tS)(~+~)

Jb+ \3:: A(S4-3)\- S(srs) ~~- 3: 1-:: 2~ ~S= -; S:: - '5 : 3 = - 2A ~ A- =- - "23

V{r.\ - _ ..2. -.J-.. + 3:.. .-L. HJJ - 2 '5-+5 .;2 5-.3

_ ~ - s{: :L - 3t - - -sf:, + t e

(b) y' + 4y = e-4t ; y(O) = 2

t fJ' )+~ tfJ);: ;( ~ e- ~t ~ S'(0,)- iO' t If 't'CS) ~ St~

~

U(\ I_+.-:L.. 1 ~ J 7:: ( St\fY s+~

j(t)~ -Cr (~t~Y ) I ~r'fsAd