Differential Equations
The purpose of this chapter is to present several methods to solve differential equations, this chapter begins with a motivation based on physical phenomena that can be represented with differential equations, after that we introduce definitions and metho
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Differential Equations
Abstract The purpose of this chapter is to present several methods to solve differential equations, this chapter begins with a motivation based on physical phenomena that can be represented with differential equations, after that we introduce definitions and methods for first order and second order linear differential equations, various exercises are given to the reader to better illustrate the contents of this chapter.
9.1 Motivation: Some Physical Origins of Differential Equations 9.1.1 Free Fall Free fall is any motion of a body where gravity is the only force acting upon it. Consider the free fall of a body with mass m and y(0) = 0, y (0) = 0 (Fig. 9.1). By second law of Newton we have: +↓ Fy = ma mg = ma d2 y g=a= 2 dt d2 y dv then dv = gdt hence: since 2 = dt dt v = gt + k1 Integrating the above equation, and taking into account that v = y=
vdt =
(gt + k1 )dt =
dy : dt
1 2 gt + k1 t + k2 2
© Springer Nature Switzerland AG 2019 R. Martínez-Guerra et al., Algebraic and Differential Methods for Nonlinear Control Theory, Mathematical and Analytical Techniques with Applications to Engineering, https://doi.org/10.1007/978-3-030-12025-2_9
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9 Differential Equations
Fig. 9.1 Free fall
mg
Finally, by initial conditions of the problem: k1 = y (0) − g(0) = 0 1 k2 = y(0) − g(0)2 + k1 (0) = 0 2 therefore y=
1 2 gt 2
9.1.2 Simple Pendulum Problem Now, according to Fig. 9.2, it is not difficult to obtain the following analysis based in kinetic and potential energy for the simple pendulum that consists in a mass m attached to a rope with length L. From points 1 and 2 in the figure: 1 2 mv + mgL (1 − cos θ ) = mgL (1 − cos α) 2 1 2 v + gL (1 − cos θ ) = gL (1 − cos α) 2 v2 + 2gL (1 − cos θ ) = 2gL (1 − cos α) since s = θ L then v =
ds dθ = L , we have: dt dt L
dθ dt
2 + 2g (cos α − cos θ ) = 0
9.1 Motivation: Some Physical Origins of Differential Equations
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Fig. 9.2 Simple pendulum
9.1.3 Friction Problem According to Fig. 9.3, we have: mg − F f = m
d2 y dt 2
then d2 y dt 2 d2 y dy =m 2 mg − K¯ dt dt mg − μK v = m
where K¯ = μK . Finally we have: d2 y K¯ dy −g=0 + 2 dt m dt
Fig. 9.3 Friction problem Ff
mg
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9 Differential Equations
9.2 Definitions Definition 9.1 A differential equation of order n is a relationship of the form: F = y (n) , y (n−1) , . . . , y (1) , y, x = 0 Remark 9.1 Note that y (1) =
dy = y dx
y (2) =
d2 y = y dx 2
and
(9.1)
y = y˙ = y (1) , y = y¨ = y (2) are commonly used to represent the first and second derivatives with respect to the time respectively. Definition 9.2 It is said to be f : (a, b) → R, f ∈ C n , is a solution of (9.1), if for all x ∈ (a, b): F = f (n) (x), f (n−1) (x), . . . , f (x), f (x), x = 0
(9.2)
Example 9.1 1. The relation F = y − y = 0 is a differential equation of second order. Note that f 1 (x) = ex f 2 (x) = k1 ex , k1 ∈ R f 3 (x) = ex + e−x f 4 (x) = k2 ex + k3 e−x , k2 , k3 ∈ R are solutions of F. 2. The solutions of the differential equation of second order y −
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