MATH 308 Week in Review 2 (Sections 2.1-2.3) September 2, 2020

1. Find the general solution of the given di↵erential equation.

(a) y0 + 2ty = 2te�t2

(b) 2px y0 =

p1� y2

1

y't play = get '

µ = esp =eS2tdt= et

'

standard

-

et'

( y' tzty) = fate- t ' ) et

'

[ etyy'

,

#2T I> e%=fztdt=t2tC

⇒ ?y=Ee-tIce-

nonlinear . but separable

zrxdI×=FT⇒fd!FyE⇐r×-

Note that y -- II are equilibrium solutions

-

arc sin ly) = Tx t C : thegeneralimplicit

general explicit )↳ =± , solution.

(c) ty0 + y = 3t cos t, t > 0.

2. Find the solution to the initial value problem and the interval of validity in each case.

(a) 2pxdy

dx= cos

2 y, y(4) =⇡

4.

2

O linear , non-standard→ y't¥y=#ot

standard linear ODB

Ptt ) = tfIpltsdt -but as µ=eht=#t ( y't f y) =@ cost) t

[ try]-

=3 t cost ⇒ ty = fztucosddttc

try = ur - Sunda - c(dF=3dt "visit

= 3T Sint - 3) sintdt + C=3t sit +Kost t C

Y=3snt_3n-ET--

C DIC - Yoji'm,

.

=.

{ d÷y=fdzfT, equilibrium-

Solis CaFy=o.

)seEydy=fx t C y -- Ia -1kt KIIta| ¥amof Ether ,tan CE

,) -- Ft t C =, C = - I #

"'z

F T -

'⇒ EEEE.EE#an.cr.... ,

"

±

-7¥a menIf yo = 5%

,= IT +

'I,

needy XI for Vx⇒ y = IT + arctancrx - t ) I . V . = ( 0,0 )

(b)dy

dt+

2y

t=

cos t

t2y(1) =

1

2, t > 0.

3

1. v. = the largest interval containing xo , in which y is

differentiable

linear r standard V ⇒ pal = z ⇒ µ=es¥dtt' ( Edt + I y ) = cost

= e-ht.I

-

Iffy ) -- cost ⇒ thy = Sint + C

l CE ) = Sin Cn t C

sings * ¥ ,€Cy =sinltl-I.si#

radacutenoon. Ext -

5^7 "

I. V . = ( o, x)th=/←

Interval of validity

c-÷

t=o to=L

3. Consider the initial value problem

y0 + 2y = 5� t, y(0) = y0

Find the value y0 for which the solution touches, but does not cross the t-axis.

4

Ptt ) - 2 ⇒ µ it , =efzdt, eat

*B•aLre

e't( y 't zy ) -- ( s -t )

it ↳[etty )

'

= Cst ) eat Itmintegration u y

'

e' try = fes- t) e'tdt tf C bypartsd=

= ⇐the"

+ titter '

{tyg. ⇒ e°Yo=C5-ol 'ze° -if it C + obtect⇒ Yo = Iz tf t C ⇒C=2o

Y=tzC=ti-I-(yo-E)eZtTt; ?

-Yeti -- o ⇒ Hq - Iz te Cyo - Hz ) e- Zhao-

atned y'Ctp=o⇒-zxz(yo-¥)e-2t¥ - It , = - ( yo - ¥) e-

Zt

= - C -41=4It , =

'I ⇒t

Kyo - ⇒ e-Zt's -E ⇒ yo - ¥ = - felt '

⇒ yo = ¥ - I, e"

=L,( H- e" ) to

4. A 120-gallon tank initially contains 90 lb of salt dissolved in 90 gallons of water. Brine containing 2 lb/gal

of salt flows into the tank at a rate of 4 gal/min, and the well-stirred mixture flows out of the tank at a

rate of 3 gal/min. How much salt does the tank contain when it is full?

5

salt-water

--

mo = 90I b Vo = 90 gallons

Cin -- 2K rout =3 gal Im ingal{ rain = 401 rin > rout ⇒ Tots increasingmin

✓,= 120 Stl

Lyft = nine - rout = 4-3=1 SET{Col = go ⇒fV*t-9#the taek gets fall at

= Cette t1=3omiTUHM #

Fma; !!!-Douthat = 2141 -Fox

lot → t t a o

E :*,

t.am ⇒ ¥

µ ..es#eodt=e3htt+" ' = ( t + go ,

'

[ Ct -19073 m ]'

= 8 ( t +9013,u ,du = It

⇒ Ct -19033 m = 8 f Pdt t C- 4

= 8 . I,Lt +90 ) t C

=2(tt9o{ ten go

⇒ 90 GOT = 2 ( go ) " t C ⇒ C = - 90

"

#=

⇒m=2(t-9o)-90(ttso Ks

#t1=30

" "⇒ (m,

= z (1207 - 9%4-3 =202 1¥

5. Consider a large lake formed by damming a river that initially holds 200 million liters of water. Because

a nearby chemical plant uses the lake’s water to clean its reservoirs, 1,000 liters of brine, each containing

100(1 + cos t) kilograms of dissolved pollution, run into the lake every hour. Let’s make the simplifying

assumption that the mixture, kept uniform by stirring, runs out at the rate of 1,000 liters per hour and

no additional spraying causes the lake to become even more contaminated. Find the amount of pollution

p(t) in the lake at any time t, and determine its variations in the long run.

6

=

✓0=200 . million liters =2x#-

Vin = I ooo liter

cinch = Kotka Kfz = o. , Gt cost )1000#

Q-2

rout = tooo high Cin Xo

dv

§,= Kin - rout = too . - too . =o

Vo = 2×108 litas =)Vct) = Vo .

,

I 0

µL, = rin Cin - tout . Coat = too oxo . I (Hurst ) - to -2¥PCO ) -- Po dff + 5×15

"

p = too ( It cost )

= £pcoµ = et '"Tdt es " 't!; ( e5

"out. p y'

= too ( i + cost )E"'t !

e°tp= too f e"""tCie cost > It + C-

÷,oo-f es"-6¥de + too fe

""- Yost It t C

tbTaabtbb{IT#¥long-ruexio-6.it#bti='sIio.ut'aEbIaaosbt-ibebtI

6. A cake is removed from an oven at 210�F and left to cool at room temperature, which is 70

�F. After 30

min the temperature of the cake is 140�F. When will it be 100

�F?

7

O-

{u¥uo du

-

a

⇐Eof- hat↳ near

diff + ku =k -1=7ok ,

acts =ebh=ekt

Lek tu )'=To keht⇒ehtu-zoe.tt#f=u=oeht2io--70-C=sC=14o-

ta ! ?4! ⇒ 140=70+140 e-"h

÷:÷÷:::÷÷÷÷÷.1- 140 e

- k t

1405kt so ⇒ e

- kt= ¥,

⇒ -Kt -- In )

+=÷÷÷÷T

7. A ball with mass 1kg is thrown upward with initial velocity 20 m/s from the roof of a building 50 m high.

A force due to the resistance of the air of v/10, where the velocity is measured in m/s, acts on the ball.

Find the maximum height above the ground that the ball reaches

8

-G=No = 20 Mlsh o = 50 m

/ m

!, .

- tug - Fo - m diething.

- no - Yo -- Et{ Ufo) = 20 Ms⇒ gV=-l0g-Czo-iog)e#hma tma

,

such that~

⇒-E -

- hlI÷!⇒ tmax = 10h (204%09)=1.86

see

←

hm 50¥08 ! log + Got log ) e -% It⇒hna×e68mJ

8. In a typical Swedish summer day, temperature fluctuates sinusoidally between 10 (at midnight) and 30

(at noon) degrees Celsius. Assume at a party someone forgets a beer with a temperature of 5 degrees

at 10 pm, but finds it again at 2 am. What will be the temperature of the beer then, assuming a heat

transfer coe�cient of 0.1 per hour.

9

--¥t=-ku -Eff k=alpmhry

C- - noon =ofhumj#

tmdwj -- 24W' TCtl-20-locosLZ.LIz4h = 20+10 Cos (Ft )

io# it -- ions"

midnj 2dm i'f- 14 hrsnot

* yzhow2

Can be solved as a linear equator .

If + ku =L -1=0.1 Ezo -110 as CFI )]--