MATH 308 Week in Review 9 (Sections 6.5 and 6.6) October 21, 2020

1. A 2-kg mass is attached to a spring with the Hooke’s constant of 3 N/m, and it is subject to moving in a

medium with a damping constant of 5 N-s/m. The mass is initially displaced +0.5 m from its equilibrium

position and is released. Then after 2 seconds, a hammer hits the mass in such a way that its velocity

suddenly changes by 5m/s in the positive direction (i.e., �v = +5 m/s).

(a) Find the impulse of the hammer and its units.

(b) Write the hammer force using an appropriate Dirac delta function. Note that your force must result

in the same impulse value you found in part (a).

(c) Set up an IVP and find the position x(t) of the mass for t � 0.

1

r k

om r OO

-

ca) I = DP = m Do = 2 (5) = to N - s

(b) Fits = I get - c) = to SH - 2 ) N

=

⇒ I? Feldt = to N- s = I -

cc , 2 a 5k + 3 x = ( o S Ct- 27.

{xco ) = Byz x' Col - O

# s ' =L Exit , y ⇒LEX} - sites > - z

LEE}= sites , - Is -o

2 ( s'# css - { s ) + 5 (SIG ) - f ) +3 Ics ) = to e-Is

(25+55+3) Ics ) - s - Iz = Loe

-2 '

s + 512 -2 ? I

# Cs) = --+ lo e -

C-252555+5 ,254543

'i- e -

-'

£D= V

"- 4mk = 25-4127131=25-24--1 70

over clamped .

25+55+3 = (2s +3) (Stl )

i:÷. . . :÷i÷*÷÷÷i.

{D=. -1+512=32

- 21-3

= Is.is?fc= IT ,

= - z

D= - I

- I

s +312 ¥312--- --

.

r

-- -

-

y l l

* is . -- i it ÷,

+ we" f ÷.

.

:- ÷

. )'

/i-- s

'

\.-

' &delay⇒ x its = - e-

%Zz e-t

+ to uzH, [- e-Z" -I e- '

t- Z )

)

2. Find the solution of the initial value problem

(a) y00 � y = 4�(t� 2) + t2, y(0) = 0, y0(0) = 2.

2

THS ) =LSgt ⇒ LEY 't =s Yes , -y lo) = STIS )

L { y" 1=5 Test - Syco ) - y

-

cos = 5Ycs , - 2

Eyes ) - z - Tess = 4 e-"

t Fs delay

--

-

--

-

r n E --Yes , = ÷,

+ it -4¥,¥5 '

= # +FistSKS- 1) (still

D=2 ,

E-

- o,C = - 2

253+2 =A5(5- 1) + Bs (S2- 1) + C Cs'

- 1) + D 5%+17+ E S3( s - t )

I 0 = A tDtEf°⇒AI 2 = B + D -Ef°⇒B=2o253+2 - 2

5¥=

J- Et I,Hittite,

It= Ii

- I,✓

(b) y00 + y = �(t� 2⇡) cos t, y(0) = 0, y0(0) = 1.

3

= 8 (t - kit ) - Cos (2h)

MJT

= 8 Ct - zit )

lT w u

- ZITS --21T S

S-

Yes ) - I + Tls ) = e . Cos ( zit ) = e

✓ -

2T

? IIls ) = 1- + e -

SZ + I s 't l

Yet) = Sin Ct) t U#(t ) .

Sin (t - 21T)

-Sin Ct ) .

yet ) = Sin Ctl ( i t dat Iti ) .

3. Find the following convolutions using the definition only.

(a) et ⇤ e3t

(b) t ⇤ tn, where n = 0, 1, 2, . . . .

4. Using the Laplace transform (instead of the definition), compute the following convolutions.

(a) ua(t) ⇤ ub(t)

(b) tn ⇤ tm, n,m = 0, 1, 2, . . .

5. Show, both by the definition and by the Laplace transform, that f(t) ⇤ �(t� a) = ua(t)f(t� a) for a � 0.

In particular,

Z t

0�(⌧ � a) d⌧ = ua(t).

4

#* g) its -_ fotfct- E) gridTLIfxygt-Fcsi.GG ,

tote? !"

de = etfoetd-c-et.IE "o

= et [ tzet- I ] = test - Eet .

lot et - riende = gotten- rn" den -12

* I - ÷:*. -

I'

(htt ) (n -12)

→ a

L { nahh * Ublt) } =L { halts ) - L { ubct, }

a-fun. = =e÷""

Haiti * anti =L"fe"

I .ua#tiCt- cats ,"

atb

LETT -Html --s:÷i÷,

= msIz .--

.

'"

I =÷÷÷÷¥±÷HE't -1¥'

tkti I-N(k - manti )

T T=

convolution delayqS Lt -a ,

fit , * Set -a) = fotfct - T ) SIT-a) DT II.um a-

off La ⇒ fctix Sit-a) = o -

t>a ⇒ fct , * SH-a) = lot fit - ⇒ S CT- al de = fct - a ){= -

⇒ fcti*set-ai=uaAifH-T ""'II: '?'

6. In each of the following cases find a function (or a generalized function) g(t) that satisfies the equality for

t � 0.

(a) t ⇤ g(t) = t4

(b) 1 ⇤ 1 ⇤ g(t) = t2

(c) 1 ⇤ g(t) = 1

7. Write the inverse Laplace transform of the function F (s) =s

(s+ 1)2(s+ 4)3in terms of a convolution

integral.

5

In the tonne of distributions-

-4 stz . Gcs ) = 4¥ ⇒ G- Is ) = ¥3

⇒ gets = 2¥.

I = 12 t'

.

t 70

-7 Is . Is . Gcs ) = ¥3⇒ G-Csi = Is =, gu 470

Is GLS ) = Is = , G- G) =/I

9Hs--

.

-

F""up

= ÷p . ¥4,3

=. ¥4,3 =L'T,

- esta) - ¥4,3tf TL

"

e-t- e- t.te-4.tt

2

⇒ fits = e-th - t ) * Iz te- 4T

=#ote- ta - e) et -e5 e-""" '

de

7'

) Find the Laplace transform of

lot e-" Lt -et de .

←

LET '* est ) = 5¥ . ÷, = IIIs,

8. Solve the initial value problem y00 � 2y0 � 3y = g(t), y(0) = 1, y0(0) = �3.

6

-S

"

Yes , - s -13.

- 2 (Hess - l ) - 3 Yess = Gcs )

( s'- Zs -3) Yes , = s - 5 t Els )-

( stills -3 )- s - 5

lls ) = +

(stil C- 3 )

S - 5312 -213

¥5,=

I,+ Is

1¥,iii. is

yet , = zit - zest # [ I, est- I, e-t

) # 9¥=

= Zeit - Eze't+tegCt - E) de- -

Yen Yp

= Zz et - Zz esta f ! ¥ ( e3"-

e-'t- t ') gee , de

TFIE : G Lt , t ) is called the

Green function of the equation

'ji, jijY"to response )#7#