Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSFor example, if we consider the second-order case with boundary conditionsy(a) =α, y(b) =β then a suitable change of variable isu = y − (mx + c),where y = mx + c is the straight line through the points (a, α) and(b, β), for whichm =(α − β)/(a − b) andc =(βa − αb)/(a − b). Alternatively, if the boundaryconditions for our second-order equation are y(0) = y ′ (0) = γ then we wouldmake the same change of variable, but this time y = mx + c would be the straightline through (0,γ) with slope γ, i.e.m = c = γ.Solution method. Require that the Green’s function G(x, z) obeys the original ODE,but with the RHS set to a delta function δ(x − z). This is equivalent to assumingthat G(x, z) is given by the complementary function of the original ODE, with theconstants replaced by functions of z; these functions are different for xz. Now require also that G(x, z) obeys the given homogeneous boundary conditionsand impose the continuity conditions given in (15.64) and (15.65). The generalsolution to the original ODE is then given by (15.60). For inhomogeneous boundaryconditions, make the change of dependent variable u = y − h(x), whereh(x) is apolynomial obeying the given boundary conditions.15.2.6 Canonical form for second-order equationsIn this section we specialise from nth-order linear ODEs with variable coefficientsto those of order 2. In particular we consider the equationd 2 ydx 2 + a 1(x) dydx + a 0(x)y = f(x), (15.70)which has been rearranged so that the coefficient of d 2 y/dx 2 is unity. By makingthe substitution y(x) =u(x)v(x) we obtain( ) ( 2uv ′′ ′u+u + a 1 v ′ ′′ + a 1 u ′ )+ a 0 u+v = f uu , (15.71)where the prime denotes differentiation with respect to x. Since (15.71) would bemuch simplified if there were no term in v ′ ,letuschooseu(x) such that the firstfactor in parentheses on the LHS of (15.71) is zero, i.e.2u ′{ ∫u + a 1 =0 ⇒ u(x) =exp − 1 2We then obtain an equation of the form}a 1 (z) dz . (15.72)d 2 v+ g(x)v = h(x), (15.73)dx2 516