32.5. FINDING THE SOLUTION 619

From the product rule,(y1y′2− y2y′1

)′= y′′2y1 + y′1y′2−

(y′′1y2 + y′1y′2

)=(y1y′′2− y2y′′1

)but y1y′2− y2y′1 =W (y1,y2) . Hence 32.7 is of the form

W ′+ p(x)W = 0

and so, from the theory of linear equations,

W (x) =Ce−P(x) where P′ (x) = p(x) .

This proves Abel’s formula.

Proposition 32.4.1 Let y1,y2 be two solutions to y′′+ p(x)y′+ q(x)y = 0 for x in someinterval on which p(x) ,q(x) are continuous. Then

W (y1,y2)(x) =Ce−P(x), P′ (x) = p(x) .

Note how this shows directly that the Wronskian either vanishes identically or not atall. This also motivates the following procedure.

PROCEDURE 32.4.2 Suppose y is a known solution to

y′′+ p(x)y′+q(x)y = 0 (32.8)

To find another solution z which solves

z′′+ p(x)z′+q(x)z = 0

do the following: For

W =

∣∣∣∣∣ y zy′ z′

∣∣∣∣∣Find a nonzero solution W (x) of

W ′ (x)+ p(x)W (x) = 0

Then solve for z in the equationz′y− zy′ =W

BE SURE THAT THE EQUATION IS IN THE FORM DESCRIBED IN THEABOVE PROCEDURE. THIS MEANS THE COEFFICIENT OF y′′ IS 1!

32.5 Finding the SolutionSuppose you have reduced the equation to

x2y′′+ xp(x)y′+q(x)y = 0 (32.9)

where each of p,q is analytic near 0. Then letting

p(x) = b0 +b1x+ · · ·