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Mathematical Methods for Physics and Engineering - Matematica.NET

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15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONS15.3.4 Isobaric or homogeneous equationsIt is straight<strong>for</strong>ward to generalise the discussion of first-order isobaric equationsgiven in subsection 14.2.6 to equations of general order n. Annth-order isobaricequation is one in which every term can be made dimensionally consistent upongiving y <strong>and</strong> dy each a weight m, <strong>and</strong>x <strong>and</strong> dx each a weight 1. Then the nthderivative of y with respect to x, <strong>for</strong> example, would have dimensions m in y<strong>and</strong> −n in x. In the special case m = 1, <strong>for</strong> which the equation is dimensionallyconsistent, the equation is called homogeneous (not to be confused with linearequations with a zero RHS). If an equation is isobaric or homogeneous then thechange in dependent variable y = vx m (y = vx in the homogeneous case) followedby the change in independent variable x = e t leadstoanequationinwhichthenew independent variable t is absent except in the <strong>for</strong>m d/dt.◮Solvex 3 d2 ydx 2 − (x2 + xy) dydx +(y2 + xy) =0. (15.84)Assigning y <strong>and</strong> dy the weight m, <strong>and</strong>x <strong>and</strong> dx the weight 1, the weights of the five termson the LHS of (15.84) are, from left to right: m +1, m +1, 2m, 2m, m +1. For theseweights all to be equal we require m = 1; thus (15.84) is a homogeneous equation. Since itis homogeneous we now make the substitution y = vx, which, after dividing the resultingequation through by x 3 ,givesx d2 v dv+(1− v) =0. (15.85)dx2 dxNow substituting x = e t into (15.85) we obtain (after some working)d 2 vdt − v dv =0, (15.86)2 dtwhich can be integrated directly to givedvdt = 1 2 v2 + c 1 . (15.87)Equation (15.87) is separable, <strong>and</strong> integrates to give∫1t + d dv2 2 =v 2 + d 2 1= 1 ( ) vtan −1 .d 1 d 1Rearranging <strong>and</strong> using x = e t <strong>and</strong> y = vx we finally obtain the solution to (15.84) asy = d 1 x tan ( 1d )2 1 ln x + d 1 d 2 . ◭Solution method. Assume that y <strong>and</strong> dy have weight m, <strong>and</strong> x <strong>and</strong> dx weight 1,<strong>and</strong> write down the combined weights of each term in the ODE. If these weights canbe made equal by assuming a particular value <strong>for</strong> m then the equation is isobaric(or homogeneous if m =1). Making the substitution y = vx m followed by x = e tleads to an equation in which the new independent variable t is absent except in the<strong>for</strong>m d/dt.521

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