MATH
chapter4

# Solution by the data given in this example we have pt

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Solution: By the data given in this example we have P(t ) = N(t) n=4, p=1, m=300 in the model (1.23) 75 4 4 . 300 4 ) ( P P dt t dP - = - = or 4 75 P dt ) t ( dP = + This is a linear differential of first order in P whose integrating factor is t e dt e 75 1 75 1 = (See Section 2.3) Solution is given by c dt t e t e t P + = 75 1 4 75 1 ). ( P(t) = 300 + c t e 75 1 . - Since P(0) = 20 is given we get 20=P(0) = 300+Ce 0 , that is c= -280 Thus P(t) = 300 - 280 t e 75 1 - 93

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4.5 Series Circuits Let a series circuit contain only a resistor and an inductor as shown in Figure 4.2 Figure 4.2 LR Series circuit By Kirchhoff’s second law the sum of the voltage drop across the inductor Ι dt di and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. Current at time t, i(t), is the solution of the differential equation. ) ( t E Ri dt di = + Ι (4.10) where Ι and R are constants known as the inductance and the resistance respectively. The voltage drop across a capacitor with capacitance C is given by C ) t ( q , where q is the charge on the capacitor. Hence, for the series circuit shown in Figure 4.3 we get the following equation by applying Kirchhoff’s second law ) ( 1 t E q C Ri = + (4.11) 94
Figure 4.3 RC Series Circuit Since dt dq i = , (4.11) can be written as ) t ( E q C 1 dt dq R = + (4.12) Example 4.9 Find the current in a series RL circuit in which the resistance, inductance, and voltage are constant. Assume that i(o)=0; that is initial current is zero. Solution: It is modeled by (4.10) ) t ( E Ri dt di = + Ι or Ι = Ι + ) t ( E i R dt di (4.13) Since Ι , R and E are constant (4.13) is linear equation of first-order in i with integrating factor Ι = Ι t R e dt R e The solution of (4.13) is c R t R e E t R e ) t ( i or dt t R e E t R e ) t ( i + Ι Ι Ι = Ι Ι Ι = Ι or i(t) = t R ce R E Ι - + (4.14) 95

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Since i(0) = 0, c = - R E Putting this value of c in (4.14) we get ) 1 ( ) ( t R e R E t i Ι - - = Example 4.10 A 100-volt electromotive force is applied to an RC series circuit in which the resistance is 200 ohms and the capacitance is 10 -4 farads. Find the charge q(t) on the capacitor if q (0)=0. Find the current i(t). Solution: The phenomenon is modeled by (4.12): ) ( 1 t E q C dt dq R = + , where R=200, C=10 -4 , E(t) = 100 Thus 2 1 4 10 200 1 = + q dt dq (4.15) This is a linear differential equation of first-order The integrating factor is t 50 e dt 50 e = and so the solution of (4.15) is q(t)e 50t = c dt t 50 e 2 1 + or q(t)= t ce 50 100 1 - + q(0)=0= 0 . 50 ce 100 1 - + 96
or 100 1 c - = and so t 50 e 100 1 100 1 ) t ( q - - = t 50 e 2 1 dt ) t ( dq - = But dt ) t ( dq i = and so i= t 50 e 2 1 - 4.6 Survivability with AIDS Equation (1.31) provides survival fraction S(t). It is a separable equation and its solution is S(t) =S i +(1-S i )e -kt : Given equation is ) ) ( ( ) ( i S t S k dt t dS - - = kdt i S t S dS - = - ) ( Integrating both sides, we get ln|S(t)-S i |=-kt+lnc kt c | i S ) t ( S | ln - = - or kt e c i S ) t ( S - = - S(t)=S i +ce -kt Let S(0)=1 then c=1-Si. Therefore 97

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S(t) =S i +(1-S i )e -kt We can rewrite this equation in the equivalent form. S(t)=S i +(1-S i )e -t/T where, in analogy to radioactive nuclear decay, T is the time required for half of the mortal part of the cohort to die-that is, the survival half life.
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