Example 411 consider the initial value problem 1 s 16

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Example 4.11 Consider the initial-value problem 1 ) 0 ( S ) 16 . 4 ( ) i S ) t ( S ( k dt ) t ( dS = - - = as the survivability with AIDS. (a) Show that, in general, the half-life T for the mortal part of the cohort to die is k T 2 ln = (b) (b) Show that the solution of the initial value problem can be written as S(t)=S i +(1-S i )2 -t/T (4.17) Solution: The solution of the separable differential equation in (4.16) is S(t) = (1-S i )e -kt +S i (4.18) Let S(t) = 2 1 S(0), and solving for t we obtain the half=life T = k 2 ln (b) Putting 2 ln T k = in (4.18) we obtain t T e Si Si t S 2 ln ) 1 ( ) ( - - + = 98
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4.7 Draining a Tank In Section 1.4.8 modeling of draining a tank is discussed. Equation (1.26) models the rate at which the water level is dropping. Example 4.12 A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. Find the height h of water in the tank at any time t if the initial height of the water is H. Solution: As discussed in Section 1.4.8, h(t) is the solution of the equation (1.26); that is, gh 2 A B dt dh - = (4.19) where A is the cross section area of the cylinder and B is the cross sectional area of the orifice at the base of the container. (4.19) can be written as dt g 2 A B h dh - = or Cdt h dh = where g 2 A B C - = By integrating this equation we get ' c Ct 2 1 h 2 + = For t=0 h=H and so 2 1 H 2 ' c = Therefore 99
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h(t) = 2 2 2 1 H 2 Ct + 4.8 Economics and Finance We have presented models of supply, demand and compounding interest in Section 1.4.3. We solve those models, namely equations (1.11) and (1.16). (1.11), that is equation ) S D ( k dt dP - = is a separable differential equation of first-order. We can write it as dP=k(D-S) dt. Integrating both sides, we get P(t)=k(D-S)t+A where A is a constant of integration. Solution of (1.16), which is also a separable equation, is S(t)=S(0) e rt (4.20) where S(0) is the initial money in the account Example 4.13 Find solution of the model of Example 1.21 with no initial demand (D(0)=0). Solution: The model is D t k dt dD = This can be written as D 1/2 dD=k tdt 100
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Integrating both sides we get , 2 2 1 2 3 3 2 A t k D + = where A is a constant integration. If Demand D=0 at the initial time t=0, then A=0 and demand D(t) at any time t is given by 3 2 4 2 kt 3 ) t ( D = 4.9 Mathematics Police Women The time of death of a murdered person can be determined with the help of modeling through differential equation. A police personnel discovers the body of a dead person presumably murdered and the problem is to estimate the time of death. The body is located in a room that is kept at a constant 70 degree F. For some time after the death, the body will radiate heat into the cooler room, causing the body’s temperature to decrease assuming that the victim’s temperature was normal 98.6F at the time of death. Forensic expert will try to estimate this time from body’s current temperature and calculating how long it would have had to lose heat to reach this point.
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