Growth and Decay

Growth and Decay Web Assign Objectives: • Use separation of variaЬles to solve а simple differential equation. • Use exponentia\ functions to model growth and decay in applied proЫems. Differential Equations We have learned to analyze the solutions visually of differential equations using slope fields and to approximate solutions numerically using Euler's Method. Analytically, you have learned to solve only two types of differential equations-those of the forms у'= f (х) and у"= f (х). ln this section, you will learn how to solve а more general type of differential equation. The strategy is to rewrite the equation so that each variaЫe occurs on only one side of the equation. This strategy is called separation of variaЫes. 2 Example 1: Solve the differential equation у'= ;. - �х dj � ) ¾ =)·2,)(d� ci "'2- '/.. -t -=-) � "--'f:>€. t • с.. Notice that when you integrate both sides of the equation in Example 1, you do not need to add а constant of integration to both sides. When you do, you still obtain the same result. l \' 2 2. f+ ydy = С = 2 ½ /� = ½ у2 = f+ 2xdx х2 + 2 х + х 2 СЗ (С3 - С2) С1 Some people prefer to use Leibniz notation and differentials when applying separation of variaЬles. The solution to Example 1 is shown below using this notation. cly == 2х dx у f у с/у == 2х dx ydy- f 2xdx• .!.у, 2 == x z 2 у 2 - 2х2 == С + С 1 . Growth and Decay Models })\�ТL'{ ln many applications, the rate of change of а variaЫe у is proportional to the value of у. lf у is а function of time t, the portion can Ье written as follows. • CZ.LE.5 D�C-""'(L. Rate 01· ctш11gc of' у dy is I ' pюpol'lional LO у. = k,y dt The general solution of this differential equation is given in the following theorem. THEOREM 5.1 Exponential Growth and Decay Model If у is а dift·ereпtiaЫe fuпction of t such that у > О and у' = ky for sоше constant k, theп у= Ce k' where С is the initial value of у, апd k is the proportioпality constant. Exponential growtl1 occuгs when k > О. алd expoпential decay оссшs wheп k < О. Example 2: The rate of change of is pro12ortio а to у. When t = О, у = 2, and when t = 2, у = 4. What is the value of у when t= З? k 4 l!2! 3 �; f �щt.№- • Sct..urtotJ·. Ат -L=3 � j - 2е ""1,,- "'v б, ft;S+ 7 1 (3, 5.657) l ro. 2! lf the rate of change of у is proportional to у, then у follov.'S an exponentlal mooel. Fioure 5.7 Radioactive decay is measured in terms of ha/f-life-the number of years required for half of the atoms in а sample of radioactive material to decay. The rate of decay is proportional to the amount resent. The half-lives of some common radioactive isotopes are shown below. 8':Ь Се Uranium( 238U) 4,470,000,000 years Carbon( 14C) 5715 years Radium( 226 Ra) 276 days Radon( 222 Rn) Einsteinium( 254Es) Nobelium( 257 No) 24,100 years 1599 years 3.82 days • 25 seconds Example З: Ten grams of the plutonium isotope 239 Pu were released in а nuclear accident. How long will it take for the 10 grams to decay to 1 gram? =') dt=c,�=-\D i == \De�t. L,бе. I-IALf-цF6" --го J)ВГ�J � Е> "'- •�Q.б � .i 6,Q...\"" =) \ ::. \ D е..,, 10 _ _j_ к('2А-, 100) t,; 'Щ-, tСО \ �t -_,... eL-� � /.J� n \JO) - �Jbl) t \e,�to\ 5.::-1 е .:==) l \1'\ .':) (.,\) � �•t.L tD -= Ce' ::::=) � -;::::i' '"С= 24',IO(') \1'} 1 -t- � �О, Об<6--44>-=1' YГLS � � 'Ъ'b1 C)5 &.1lG,1. lr1 L '� (,,5') Example 4: An experimental population of fruit flies increases according to the law of exponential growth. There were 100 flies after the second day of the experiment and 300 flies after the fourth day. Approximately how many flies were in the original population? �') С. S МТ 01=- 02..\6 Ut.дTtO/J 012. :,- -t_ Q = .\Ь = 't Се. k.-t L-:.2 � � � lt>o -=) lt:>O= С,�� - - ) 1:.,=-t., , ::::.зоо "3оо ::= 60D '3,:: Зсо 4-1< Се -2� tooe � e7...'l" \У)�== 2.t. Q =) -=- [У\3 2- ] sv& lkl - с- ----- - /CJO LDC €�� е'2(�) G1'..Ъ3-366 .-� � � Yl,\ -e_t;; - - юо - � /tJO 3 •