Partial Differential Equations with MATLAB

470 An Introduction to Partial Differential Equations with MATLAB©R

Now perform variation of parameters, as in the example. We get

yp=u 1 y 1 +u 2 y 2 ,

where

u′ 1 =

f(x)y 2 (x)

r(x)W(x)

,u′ 2 =−

f(x)y 1 (x)

r(x)W(x)

,

andWis the Wronskian

W(x)=y 1 (x)y′ 2 (x)−y 2 (x)y′ 1 (x)

(see Exercise 9a).¶(Oh, and how do we know thatWis never zero?) Then,

u 1 =

∫x

c 1

y 2 (ξ)

r(ξ)W(ξ)

f(ξ)dξ,

u 2 =−

∫x

c 2

y 1 (ξ)

r(ξ)W(ξ)

f(ξ)dξ

and

yp=y 1 (x)

∫x

c 1

y 2 (ξ)

r(ξ)W(ξ)

f(ξ)dξ

−y 2 (x)

∫x

c 2

y 1 (ξ)

r(ξ)W(ξ)

f(ξ)dξ.

As in the example, we’d likeypto be our solution, so we’d likec 1 andc 2
to be such thatypsatisfies the boundary conditions. First, our life is made
much easier by the fact thatr(x)W(x) is constant ona≤x≤b.‖ Further,
led by the example, our selection ofy 1 andy 2 suggests that we letc 1 =band
c 2 =a. Then,

yp=−y 1 (x)

∫b

x

y 2 (ξ)

rW

dξ−y 2 (x)

∫x

a

y 1 (ξ)

rW

dξ,

so

a 1 yp(a)+a 2 y′p(a)=−[a 1 y 1 (a)+a 2 y′ 1 (a)]

∫b

a

y 2 (ξ)

rW

dξ

−[a 1 y 2 (a)+a 2 y′ 2 (a)]

∫a

a

y 1 (ξ)

rW

dξ=0

(see Exercise 9b). Similarly, atx=b.

¶Why are we not concerned with the possibility thatrW=0atsomepoint?
‖See Exercise 3, Section 8.1.