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: process for fixing First-Order traditional Differential Equations again to part 2-5Equations, Unknowns, and levels of FreedomClassical recommendations of normal Linear Differential EquationsExamples of Differential EquationsDefinition of a Linear Differential EquationIntegrating issue MethodCharacteristic Equation. Read more...
summary: IntroductionAn Introductory ExampleModelingDifferential EquationsForcing FunctionsBook ObjectivesObjects in a Gravitational FieldAn instance Antidifferentiation: method for fixing First-Order usual Differential EquationsBack to part 2-1Another ExampleSeparation of Variables: procedure for fixing First-Order usual Differential Equations again to part 2-5Equations, Unknowns, and levels of FreedomClassical options of normal Linear Differential EquationsExamples of Differential EquationsDefinition of a Linear Differential EquationIntegrating issue MethodCharacteristic Equation
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Extra info for A First Course in Differential Equations, Modeling, and Simulation
4 to show the procedure. 17) Note that in this equation we know the value of m but not that of vy or ΣFy; thus, there is one equation with two unknowns. As we well know, there must be the same number of equations as unknowns before reaching a solution. 17) dt 1 equations, 2 unknowns [vy, ΣFy] So, we need one more equation. The solution of the differential equation will give us the velocity vy ; thus, we need an equation for ΣFy. 18) 2 equations, 4 unknowns [Fg, Fd] This new equation adds two new unknowns; we write the new unknowns, increase the number of equations by 1 and the numbers of unknowns by 2.
Also, because the power of the exponential terms is negative, as the independent variable t becomes very large, the value of the term decays down to zero, and the response reaches a final value; in this case, y|t=∞ = y(∞) = 0. 7 Obtain the solution of y″ + 8y′ + 20y = 0 with y(0) = 1 y′(0) = 0 Assume y = ert, then r2 ert + 8r ert + 20 ert = 0 ⇒ r2 + 8r + 20 = 0 Applying the quadratic solution to obtain the roots, r1, r2 = –4 ± i2. Both roots have negative real parts indicating a stable response; they are also complex conjugates indicating an oscillatory response.
This rare instance may only happen when there are multiple roots. Note also that because for homogeneous differential equations there is no forcing function, the qualitative behavior of the system does not depend on the type of forcing function, f(t), only on the characteristics of the system itself. We can also express these last statements as Root α ± iβ Response is stable or unstable solely depending on the sign of α. If negative, the response is stable; if positive, the response is unstable Response is monotonic or oscillatory solely depending on the numerical value of β.
A First Course in Differential Equations, Modeling, and Simulation by Carlos A. Smith