Mathematical Methods for Physics and Engineering : A Comprehensive Guide

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS

we are free to choose our constraints as we wish, let us define the expression in

parentheses to be zero, giving the first equation in (15.55). Differentiating again

we find

y′′p=k 1 y′′ 1 +k 2 y′′ 2 +···+kny′′n+[k′ 1 y′ 1 +k′ 2 y′ 2 +···+k′ny′n].

Once more we can choose the expression in brackets to be zero, giving the second

equation in (15.55). We can repeat this procedure, choosing the corresponding

expression in each case to be zero. This yields the firstn−1 equations in (15.55).

Themth derivative ofypform<nis then given by

y(pm)=k 1 y

(m)

1 +k^2 y

(m)

2 +···+kny

(m)

n.

Differentiatingyponce more we find that itsnth derivative is given by

y(pn)=k 1 y 1 (n)+k 2 y 2 (n)+···+knyn(n)+[k 1 ′y( 1 n−1)+k 2 ′y 2 (n−1)+···+kn′y(nn−1)].

Substituting the expressions foryp(m),m=0ton, into the original ODE (15.53),

we obtain

∑n

m=0

am[k 1 y 1 (m)+k 2 y 2 (m)+···+knyn(m)]+an[k′ 1 y( 1 n−1)+k′ 2 y 2 (n−1)+···+k′ny(nn−1)]=f(x),

i.e.

∑n

m=0

am

∑n

j=1

kjy(jm)+an[k 1 ′y( 1 n−1)+k 2 ′y( 2 n−1)+···+kn′y(nn−1)]=f(x).

Rearranging the order of summation on the LHS, we find

∑n

j=1

kj[anyj(n)+···+a 1 yj′+a 0 yj]+an[k′ 1 y( 1 n−1)+k 2 ′y( 2 n−1)+···+kn′y(nn−1)]=f(x).

(15.56)

But since the functionsyjare solutions of the complementary equation of (15.53)

we have (for allj)

any(jn)+···+a 1 y′j+a 0 yj=0.

Therefore (15.56) becomes

an[k 1 ′y 1 (n−1)+k 2 ′y( 2 n−1)+···+k′ny(nn−1)]=f(x),

which is the final equation given in (15.55).

Considering (15.55) to be a set of simultaneous equations in the set of unknowns

k′ 1 (x),k 2 ′,...,k′n(x), we see that the determinant of the coefficients of these functions

is equal to the WronskianW(y 1 ,y 2 ,...,yn), which is non-zero since the solutions

ym(x) are linearly independent; see equation (15.6). Therefore (15.55) can be solved

for the functionsk′m(x), which in turn can be integrated, setting all constants of