Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSSubstituting equations (15.37) into the original equation (15.36), the latter becomesa linear ODE with constant coefficients, i.e.a n α n d ( ) ( )d ddt dt − 1 ···dt − n +1 y + ···+ a 1 α dy ( e t )dt + a − β0y = f ,αwhich can be solved by the methods of section 15.1.A special case of Legendre’s linear equation, for which α = 1 and β =0,isEuler’s equation,a n x n dn ydx n + ···+ a 1x dydx + a 0y = f(x); (15.38)it may be solved in a similar manner to the above by substituting x = e t .Iff(x) = 0 in (15.38) then substituting y = x λ leads to a simple algebraic equationin λ, which can be solved to yield the solution to (15.38). In the event that thealgebraic equation for λ has repeated roots, extra care is needed. If λ 1 is a k-foldroot (k >1) then the k linearly independent solutions corresponding to this rootare x λ1 ,x λ1 ln x,...,x λ1 (ln x) k−1 .◮Solvex 2 d2 ydx + x dy − 4y = 0 (15.39)2 dxby both of the methods discussed above.First we make the substitution x = e t , which, after cancelling e t , gives an equation withconstant coefficients, i.e.( )d ddt dt − 1 y + dydt − 4y =0 ⇒ d2 y− 4y =0. (15.40)dt2 Using the methods of section 15.1, the general solution of (15.40), and therefore of (15.39),is given byy = c 1 e 2t + c 2 e −2t = c 1 x 2 + c 2 x −2 .Since the RHS of (15.39) is zero, we can reach the same solution by substituting y = x λinto (15.39). This givesλ(λ − 1)x λ + λx λ − 4x λ =0,which reduces to(λ 2 − 4)x λ =0.This has the solutions λ = ±2, so we obtain again the general solutiony = c 1 x 2 + c 2 x −2 . ◭Solution method. If the ODE is of the Legendre form (15.36) then substitute αx +β = e t . This results in an equation of the same order but with constant coefficients,which can be solved by the methods of section 15.1. If the ODE is of the Eulerform (15.38) with a non-zero RHS then substitute x = e t ; this again leads to anequation of the same order but with constant coefficients. If, however, f(x) =0inthe Euler equation (15.38) then the equation may also be solved by substituting504