44 Chapter 0 Ordinary Differential Equations
can be developed by using the variation-of-parameters solution of the differ-
ential equation (1), as presented in Section 2. To begin, we need to have two
independent solutions of the homogeneous equation
d^2 u
dx^2+k(x)du
dx+p(x)u= 0 , l<x<r. (4)Let us designate these two solutions asu 1 (x)andu 2 (x). It will simplify algebra
later if we require thatu 1 satisfy the boundary condition atx=landu 2 the
condition atx=r;
αu 1 (l)−α′u′ 1 (l)= 0 , (5)
βu 2 (r)+β′u′ 2 (r)= 0. (6)According to Theorem 3 of Section 2, the general solution of the differential
equation (1) can be written as
u(x)=c 1 u 1 (x)+c 2 u 2 (x)+∫xl(
u 1 (z)u 2 (x)−u 2 (z)u 1 (x))f(z)
W(z)dz. (7)Recall that in the denominator of the integrand, we have the Wronskian ofu 1
andu 2 ,
W(z)=∣∣
∣∣
∣
u 1 (z) u 2 (z)
u′ 1 (z) u′ 2 (z)∣∣
∣∣
∣, (8)
which is nonzero becauseu 1 andu 2 are independent. We will need to know
the following derivative of the function in Eq. (7):
du
dx=c^1 u′
1 (x)+c^2 u′
2 (x)+∫xl(
u 1 (z)u′ 2 (x)−u 2 (z)u′ 1 (x))f(z)
W(z)dz.(See Leibniz’s rule in the Appendix.)
Now let us apply the boundary condition,Eq.(2),tothegeneralsolution
u(x).First,atx=lwe have
αu(l)−α′u′(l)=c 1(
αu 1 (l)−α′u′ 1 (l))
+c 2(
αu 2 (l)−α′u′ 2 (l))
= 0. (9)
Note that the integrals inuandu′are both 0 atx=l. Because of the boundary
condition (5) imposed onu 1 ,Eq.(9)reducesto
c 2(
αu 2 (l)−α′u′ 2 (l))
= 0 , (10)
and we conclude thatc 2 =0.