612 CHAPTER 32. SOLUTIONS NEAR A REGULAR SINGULAR POINT

This reduces toxr (r (r−1)+ar+b) = 0

and so you have to solve the equation

r (r−1)+ar+b = 0

to find the values of r. If these values of r are different, say r1 ̸= r2 then the general solutionmust be

C1xr1 +C2xr2

because the Wronskian of the two functions will be nonzero. I know this because the ratioof the two functions is not a constant so Proposition 30.4.4 implies this gives the generalsolution. The reason for this is that the quotient rule gives the numerator as ±1 times theWronskian.

Example 32.1.2 Find the general solution to x2y′′−2xy′+2y = 0.

You plug in xr and look for r. Then as above this yields

r (r−1)−2r+2 = r2−3r+2 = 0

and so the two values of r are 1,2. Therefore, the general solution to this equation is

C1x+C2x2.

Of course there are three cases for solutions to the so called indicial equation

r (r−1)+ar+b = 0

Either the zeros are distinct and real, distinct and complex or repeated. Consider the casewhere they are distinct and complex next.

Example 32.1.3 Find the general solution to x2y′′+3xy′+2y = 0.

This time you haver2 +2r+2 = 0

and the solutions are r =−1± i. How do we interpret

x−1+i,x−1−i?

It is real easy. You assume always that x > 0 since otherwise the leading coefficient couldvanish. Then

x−1+i = eln(x)(−1+i) = e− ln(x)+i ln(x)

and by Euler’s formula this equals

x−1+i = eln(x−1) (cos(ln(x))+ isin(ln(x)))

=1x(cos(ln(x))+ isin(ln(x)))

Corresponding to x−1−i we get something similar.

x−1−i =1x((cos(ln(x))− isin(ln(x))))