P n 1 r Δ t P n 1 1 r Δ t n P The solution of this discrete model is P n 1 r Δ

P n 1 r δ t p n 1 1 r δ t n p the solution of this

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. . P n = (1 + r Δ t ) P n - 1 = (1 + r Δ t ) n P 0 The solution of this discrete model is P n = (1 + r Δ t ) n P 0 , which is an exponential growth Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equations — (15/47) The Class — Overview The Class... Introduction Applications of Differential Equations Malthusian Growth Example Definitions - What is a Differential Equation? Classification Malthusian Growth 3 Discrete Malthusian Growth : P n +1 = (1 + 0 . t ) P n P 0 = 4 0 2∆ 4∆ 6∆ 8∆ 10∆ 0 2 4 6 8 10 12 n P n Discrete Malthusian Growth Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (16/47)
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The Class — Overview The Class... Introduction Applications of Differential Equations Malthusian Growth Example Definitions - What is a Differential Equation? Classification Malthusian Growth 4 Malthusian Growth : Let P ( t ) be the population at time t = t 0 + n t and rearrange the model above P n +1 - P n = r tP n P ( t + ∆ t ) - P ( t ) = t · rP ( t ) P ( t + ∆ t ) - P ( t ) t = rP ( t ) Let ∆ t become very small lim t 0 P ( t + ∆ t ) - P ( t ) t = dP ( t ) dt = rP ( t ) , which is a Differential Equation Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equations — (17/47) The Class — Overview The Class... Introduction Applications of Differential Equations Malthusian Growth Example Definitions - What is a Differential Equation? Classification Malthusian Growth 5 Solution of Malthusian Growth Model : The Malthusian growth model dP ( t ) dt = rP ( t ) The rate of change of a population is proportional to the population Let c be an arbitrary constant, so try a solution of the form P ( t ) = ce kt Differentiating dP ( t ) dt = cke kt , which if k = r is rP ( t ), so satisfies the differential equation Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (18/47) The Class — Overview The Class... Introduction Applications of Differential Equations Malthusian Growth Example Definitions - What is a Differential Equation? Classification Malthusian Growth 6 Solution of Malthusian Growth Model The Malthusian growth model satisfies P ( t ) = ce rt With the initial condition, P ( t 0 ) = P 0 , then the unique solution is P ( t ) = P 0 e r ( t - t 0 ) Malthusian growth is often called exponential growth Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equations — (19/47) The Class — Overview The Class... Introduction Applications of Differential Equations Malthusian Growth Example Definitions - What is a Differential Equation? Classification Example: Malthusian Growth 1 Example: Malthusian Growth Consider the Malthusian growth model dP ( t ) dt = 0 . 02 P ( t ) with P (0) = 100 Skip Example Find the solution Determine how long it takes for this population to double Joseph M. Mahaffy, h [email protected] i Lecture Notes – Introduction to Differential Equa — (20/47)
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