Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS15.2.4 Variation of parametersThe method of variation of parameters proves useful in finding particular integralsfor linear ODEs with variable (and constant) coefficients. However, it requiresknowledge of the entire complementary function, not just of one part of it as inthe previous subsection.Suppose we wish to find a particular integral of the equationa n (x) dn ydx n + ···+ a 1(x) dydx + a 0(x)y = f(x), (15.53)and the complementary function y c (x) (the general solution of (15.53) withf(x) = 0) isy c (x) =c 1 y 1 (x)+c 2 y 2 (x)+···+ c n y n (x),where the functions y m (x) are known. We now assume that a particular integral of(15.53) can be expressed in a form similar to that of the complementary function,but with the constants c m replaced by functions of x, i.e.weassumeaparticularintegral of the formy p (x) =k 1 (x)y 1 (x)+k 2 (x)y 2 (x)+···+ k n (x)y n (x). (15.54)This will no longer satisfy the complementary equation (i.e. (15.53) with the RHSset to zero) but might, with suitable choices of the functions k i (x), be made equalto f(x), thus producing not a complementary function but a particular integral.Since we have n arbitrary functions k 1 (x),k 2 (x),...,k n (x), but only one restrictionon them (namely the ODE), we may impose a further n − 1 constraints. Wecan choose these constraints to be as convenient as possible, and the simplestchoice is given byk 1(x)y ′ 1 (x)+k 2(x)y ′ 2 (x)+···+ k n(x)y ′ n (x) =0k 1(x)y ′ 1(x)+k ′ 2(x)y ′ 2(x)+···+ ′ k n(x)y ′ n(x) ′ =0.. (15.55)k 1(x)y ′ (n−2)1(x)+k 2(x)y ′ (n−2)2(x)+···+ k n(x)y ′ n (n−2) (x) =0k 1(x)y ′ (n−1)1(x)+k 2(x)y ′ (n−1)2(x)+···+ k n(x)y ′ n(n−1) (x) = f(x)a n (x) ,where the primes denote differentiation with respect to x. The last of theseequations is not a freely chosen constraint; given the previous n − 1 constraintsand the original ODE, it must be satisfied.This choice of constraints is easily justified (although the algebra is quitemessy). Differentiating (15.54) with respect to x, we obtainy ′ p = k 1 y ′ 1 + k 2 y ′ 2 + ···+ k n y ′ n +[k ′ 1y 1 + k ′ 2y 2 + ···+ k ′ ny n ],where, for the moment, we drop the explicit x-dependence of these functions. Since508