Qualitative Behavior of Differential Equations More Examples Maple Direction

Qualitative behavior of differential equations more

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Qualitative Behavior of Differential Equations More Examples Maple - Direction Fields Solution of Linear Growth and Decay Models Mathematical Modeling Newton’s Law of Cooling Murder Investigation Linear Differential Equation Murder Example 2 Solution: From the model for Newton’s Law of Cooling and the information that is given, if we set t = 0 to be 8:30 am, then we solve the initial value problem dT dt = - k ( T ( t ) - 22) with T (0) = 30 Make a change of variables z ( t ) = T ( t ) - 22 Then z 0 ( t ) = T 0 ( t ), so the differential equation above becomes dz dt = - kz ( t ) , with z (0) = T (0) - 22 = 8 This is the radioactive decay problem that we solved The solution is z ( t ) = 8 e - kt Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Portraits - 1D — (9/50) Mathematical Modeling Introduction to MatLab Qualitative Behavior of Differential Equations More Examples Maple - Direction Fields Solution of Linear Growth and Decay Models Mathematical Modeling Newton’s Law of Cooling Murder Investigation Linear Differential Equation Murder Example 3 Solution (cont): From the solution z ( t ) = 8 e - kt , we have z ( t ) = T ( t ) - 22 , so T ( t ) = z ( t ) + 22 T ( t ) = 22 + 8 e - kt One hour later the body temperature is 28 C T (1) = 28 = 22 + 8 e - k Solving 6 = 8 e - k or e k = 4 3 Thus, k = ln ( 4 3 ) = 0 . 2877 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Port — (10/50) Mathematical Modeling Introduction to MatLab Qualitative Behavior of Differential Equations More Examples Maple - Direction Fields Solution of Linear Growth and Decay Models Mathematical Modeling Newton’s Law of Cooling Murder Investigation Linear Differential Equation Murder Example 4 Solution (cont): It only remains to find out when the murder occurred At the time of death, t d , the body temperature is 37 C T ( t d ) = 37 = 22 + 8 e - kt d Thus, 8 e - kt d = 37 - 22 = 15 or e - kt d = 15 8 = 1 . 875 This gives - kt d = ln(1 . 875) or t d = - ln(1 . 875) k = - 2 . 19 The murder occurred about 2 hours 11 minutes before the body was found, which places the time of death around 6:19 am Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Portraits - 1D — (11/50) Mathematical Modeling Introduction to MatLab Qualitative Behavior of Differential Equations More Examples Maple - Direction Fields Solution of Linear Growth and Decay Models Mathematical Modeling Newton’s Law of Cooling Murder Investigation Linear Differential Equation Murder Example 5 Graph of Body Temperature over time -8 -6 -4 -2 0 2 4 6 8 10 12 20 22 24 26 28 30 32 34 36 38 40 Time of death, t d Room Temperature t T ( t ) Body Temperature Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Port — (12/50)

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Mathematical Modeling Introduction to MatLab Qualitative Behavior of Differential Equations More Examples Maple - Direction Fields Solution of Linear Growth and Decay Models Mathematical Modeling Newton’s Law of Cooling Murder Investigation Linear Differential Equation Solution of General Linear Model 1 Solution of General Linear Model : Consider the Linear Model dy dt = a y + b with y ( t 0 ) = y 0 Rewrite equation as dy dt = a y + b a Make the substitution
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