Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSintegration equal to zero, to give k m (x). The general solution to (15.53) is thengiven byn∑y(x) =y c (x)+y p (x) = [c m + k m (x)]y m (x).Note that if the constants of integration are included in the k m (x) then, as wellas finding the particular integral, we redefine the arbitrary constants c m in thecomplementary function.◮Use the variation-of-parameters method to solved 2 y+ y =cosecx, (15.57)dx2 subject to the boundary conditions y(0) = y(π/2) = 0.The complementary function of (15.57) is againy c (x) =c 1 sin x + c 2 cos x.We therefore assume a particular integral of the formy p (x) =k 1 (x)sinx + k 2 (x)cosx,and impose the additional constraints of (15.55), i.e.k 1(x)sinx ′ + k 2(x)cosx ′ =0,k 1(x)cosx ′ − k 2(x)sinx ′ =cosecx.Solving these equations for k 1 ′ (x) andk′ 2 (x) givesk 1(x) ′ =cosx cosec x =cotx,k 2(x) ′ =− sin x cosec x = −1.Hence, ignoring the constants of integration, k 1 (x) andk 2 (x) are given byk 1 (x) = ln(sin x),k 2 (x) =−x.The general solution to the ODE (15.57) is thereforey(x) =[c 1 +ln(sinx)] sin x +(c 2 − x)cosx,which is identical to the solution found in subsection 15.2.3. Applying the boundaryconditions y(0) = y(π/2) = 0 we find c 1 = c 2 =0andsoy(x) = ln(sin x)sinx − x cos x. ◭m=1Solution method. If the complementary function of (15.53) is known then assumea particular integral of the same form but with the constants replaced by functionsof x. Impose the constraints in (15.55) and solve the resulting system of equationsfor the unknowns k ′ 1 (x),k′ 2 ,...,k′ n(x). Integrate these functions, setting constants ofintegration equal to zero, to obtain k 1 (x),k 2 (x),...,k n (x) and hence the particularintegral.510