HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
integration equal to zero, to givekm(x). The general solution to (15.53) is then
given by
y(x)=yc(x)+yp(x)=∑nm=1[cm+km(x)]ym(x).Note that if the constants of integration are included in thekm(x) then, as well
as finding the particular integral, we redefine the arbitrary constantscmin the
complementary function.
Use the variation-of-parameters method to solve
d^2 y
dx^2+y=cosecx, (15.57)subject to the boundary conditionsy(0) =y(π/2) = 0.The complementary function of (15.57) is again
yc(x)=c 1 sinx+c 2 cosx.We therefore assume a particular integral of the form
yp(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 fork′ 1 (x)andk′ 2 (x)gives
k 1 ′(x)=cosxcosecx=cotx,
k 2 ′(x)=−sinxcosecx=− 1.Hence, ignoring the constants of integration,k 1 (x)andk 2 (x) are given by
k 1 (x) = ln(sinx),
k 2 (x)=−x.The general solution to the ODE (15.57) is therefore
y(x)=[c 1 +ln(sinx)]sinx+(c 2 −x)cosx,which is identical to the solution found in subsection 15.2.3. Applying the boundary
conditionsy(0) =y(π/2) = 0 we findc 1 =c 2 =0andso
y(x) = ln(sinx)sinx−xcosx.Solution method.If the complementary function of (15.53) is known then assume
a particular integral of the same form but with the constants replaced by functions
ofx. Impose the constraints in (15.55) and solve the resulting system of equations
for the unknownsk′ 1 (x),k 2 ′,...,k′n(x). Integrate these functions, setting constants of
integration equal to zero, to obtaink 1 (x),k 2 (x),...,kn(x)and hence the particular
integral.