By Michel Willem

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8) v0 sinh ωt , ω v(t ) = v 0 cosh ωt + ωr0 sinh ωt. r (t ) = r0 cosh ωt + (d) Figure 1. 8. The motion on a hyperbola under the action of repulsive forces Mechanical interpretation. With respect to an oblique co-ordinate system Ox’y’, determined by the conjugate diameters corresponding to the vectors r0 and v 0 , we get the parametric equations of the trajectory x ′ = r0 cosh ωt , y′ = v0 sinh ωt , ω (e) which is an arc of hyperbola, of equation x′2 r02 − y′2 (v 0 / ω)2 =1. (f) 1. Linear ODEs of First and Second Order 55 It is seen that the centre O is a labile position of equilibrium, as the orbit cannot be contained inside an arbitrarily small circle and the velocity of the particle may increase indefinitely.

It is with constant coefficients and its associated characteristic equation allows only the purely imaginary roots ±iν . So, its general solution is the linear combination y (x ) = C1 cos νx + C 2 sin νx . Introducing the boundary conditions, we obtain for C1 , C 2 the linear algebraic system ⎧C1 ⋅1 + C 2 ⋅ 0 = 0, ⎨ ⎩C1 cos νl + C 2 sin νl = 0. ODEs WITH APPLICATIONS TO MECHANICS 36 This leads to C1 = 0, C 2 sin νl = 0 . The only option in order to get non-zero solutions is that sin νl = 0 , which yields νk = kπ , k∈N .

The Taylor’s formula is currently used especially to solve Cauchy problems. Yet, the obtained approximation has a local character and for this reason it serves to set up one-step methods in the frame of the numerical analysis. 52). 86) starting from y1 . 87) is the desired solution. 11) for which q(x ) ≠ 0 . The equation may be written in the form y = p 0 (x ) y ′ + q 0 (x ) y . 90) if the denominators do not vanish. 89) involve y y ′′ = p 0 (x ) + q 0 ( x ) . 89) by y ′′ , we deduce y = p 0 (x ) + y′ q 0 (x ) .