dy dt and z t y b a It follows that dz dt a z with z t y b a Joseph M Mahaffy h

# Dy dt and z t y b a it follows that dz dt a z with z

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= dy dt and z ( t 0 ) = y 0 + b a It follows that dz dt = a z with z ( t 0 ) = y 0 + b a Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Por — (13/50)

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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 2 The linear growth model given by dz dt = a z with z ( t 0 ) = y 0 + b a , has been solved by our previous method. The solution is: z ( t ) = y 0 + b a e a ( t - t 0 ) = y ( t ) + b a . It follows that the solution, y ( t ) is y ( t ) = y 0 + b a e a ( t - t 0 ) - b a . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Por — (14/50)
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 Solution of General Linear Model 3 The linear differential equation satisfies: dy dt = a y + b = a y + b a Method (Solution of General Linear Differential Equation) Consider the linear differential equation dy dt = a y + b a with y ( t 0 ) = y 0 . With the substitution z ( t ) = y ( t ) + b a , we obtain the solution: y ( t ) = y 0 + b a e a ( t - t 0 ) - b a . This method produces a vertical shift of the solution. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Por — (15/50)

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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 Example of Linear Model 1 Example of Linear Model Consider the Linear Model dy dt = 5 - 0 . 2 y with y (3) = 7 Rewrite equation as dy dt = - 0 . 2( y - 25) Make the substitution z ( t ) = y ( t ) - 25, so dz dt = dy dt and z (3) = - 18 dz dt = - 0 . 2 z with z (3) = - 18 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Por — (16/50)
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 Example of Linear Model 2 Example of Linear Model The substituted model is dz dt = - 0 . 2 z with z (3) = - 18 Thus, z ( t ) = - 18 e - 0 . 2( t - 3) = y ( t ) - 25 The solution is y ( t ) = 25 - 18 e - 0 . 2( t - 3) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Direction Fields and Phase Por — (17/50)

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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 Example of Linear Model 3 The linear differential equation was transformed into the IVP: dy dt = - 0 . 2( y - 25) , with y (3) = 7 The graph is given by 0 10 20 30 40 50 -10 0 10 20 30 t y ( t ) Joseph M. Mahaffy,
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