Mathematical Methods for Physics and Engineering - Matematica.NET

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSUsing table 13.1,q 1 (t) = 1 V 2 0C(cos ω 1 t − cos ω 2 t),where ω1 2(L + M) =G and ω2 2 (L − M) =G. Thus the current is given byi 1 (t) = 1 V 2 0C(ω 2 sin ω 2 t − ω 1 sin ω 1 t). ◭Solution method. Perform a Laplace transform, as defined in (15.31), on the entireequation, using (15.32) to calculate the transform of the derivatives. Then solve theresulting algebraic equation for ȳ(s), the Laplace transform of the required solutionto the ODE. By using the method of partial fractions and consulting a table ofLaplace transforms of standard functions, calculate the inverse Laplace transform.The resulting function y(x) is the solution of the ODE that obeys the given boundaryconditions.15.2 Linear equations with variable coefficientsThere is no generally applicable method of solving equations with coefficientsthat are functions of x. Nevertheless, there are certain cases in which a solution ispossible. Some of the methods discussed in this section are also useful in findingthe general solution or particular integral for equations with constant coefficientsthat have proved impenetrable by the techniques discussed above.15.2.1 The Legendre and Euler linear equationsLegendre’s linear equation has the forma n (αx + β) n dn ydx n + ···+ a 1(αx + β) dydx + a 0y = f(x), (15.36)where α, β and the a n are constants and may be solved by making the substitutionαx + β = e t . We then havedydx = dt dydx dt = α dyαx + β dtd 2 ydx 2 = ddxdydx = α 2 ( d 2 y(αx + β) 2 dt 2 − dy )dtand so on for higher derivatives. Therefore we can write the terms of (15.36) as(αx + β) dydx = αdy dt ,(αx + β) 2 d2 ydx 2 = α2 d dt.(αx + β) n dn ydx n = αn d dt( ddt − 1 )y,( ) ( )d ddt − 1 ···dt − n +1 y.503(15.37)