A First Course in Differential Equations, Modeling, and by Carlos A. Smith

By Carlos A. Smith

IntroductionAn Introductory ExampleModelingDifferential EquationsForcing FunctionsBook ObjectivesObjects in a Gravitational FieldAn instance Antidifferentiation: procedure for fixing First-Order usual Differential EquationsBack to part 2-1Another ExampleSeparation of Variables: strategy for fixing First-Order usual Differential Equations again to part 2-5Equations, Unknowns, and levels of Read more...

summary: IntroductionAn Introductory ExampleModelingDifferential EquationsForcing FunctionsBook ObjectivesObjects in a Gravitational FieldAn instance Antidifferentiation: procedure for fixing First-Order traditional Differential EquationsBack to part 2-1Another ExampleSeparation of Variables: process for fixing First-Order traditional Differential Equations again to part 2-5Equations, Unknowns, and levels of FreedomClassical strategies of standard Linear Differential EquationsExamples of Differential EquationsDefinition of a Linear Differential EquationIntegrating issue MethodCharacteristic Equation

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Extra resources for A First Course in Differential Equations, Modeling, and Simulation

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14b) and can also be written as where r = P/m. 4. 7. 14a cannot be solved using antidifferentiation because the dependent variable vy appears on the right side of the equation. We now offer another method that can be used when antidifferentiation cannot be used. 15) The position y can be obtained by antidifferentiation. 16) A couple of comments before concluding this section are in order. ” Experiments suggest that a better model is for the drag force to vary with the square of the velocity. Think about the solution using this new model; it is the topic of a problem at the end of the chapter.

10) dx where P(x) = a0(x)/a1(x) and Q(x) = r(x)/a1(x). 11) y=e ∫  Q( x) e  The integration constant C is obtained using the initial condition. 4, dv = − g with v (0) = v i dt dv + (0)v = − g dt dv + P(t )v = Q(t ) dt where P(t) = 0 and Q(t) = –g. 14a dv = − g − rv dt with v (0) = v i dv + rv = − g dt dv + P(t )v = Q(t ) dt where P(t) = r and Q(t) = –g. Then, − P (t ) dt − r dt P (t ) dt e ∫ = e ∫ = e − rt e ∫ = e rt − P (t ) dt  ∫ P (t ) dt dt + C  = e − rt  − g e rt dt + C  = e − rt  − g e rt + C  = − g + C e − rt v=e ∫  r   Q(t ) e r      ∫ ∫ Applying the initial condition, results in C = vi + (g/r),  g g v =  v i +  e − rt −  r r The previous two differential equations could have been solved as well using separation of variables.

12. 13) The model for the velocity is composed of four equations, and using simple algebra we obtain a single equation. 14b) and can also be written as where r = P/m. 4. 7. 14a cannot be solved using antidifferentiation because the dependent variable vy appears on the right side of the equation. We now offer another method that can be used when antidifferentiation cannot be used. 15) The position y can be obtained by antidifferentiation. 16) A couple of comments before concluding this section are in order.

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