From Boyce & DiPrima's "Elementary Differential Equations and Boundary Value Problems" book:
pp. 171, 172. Example 3:
Given that y1(t) = t^(-1) is a solution of 2t^2y'' + 3ty' - y = 0, t > 0, (31)
find a second linarly independent solution.
We set y = v(t)t^(-1); then
y' = v't^(-1) - vt^(-2), y'' = v''t^(-1) - 2v't^(-2) + 2vt^(-3)
Substituting for y, y', and y'' in Eq. (31) and collecting terms, we obtain
2t^2(v''t^(-1) - 2v't^(-2) + 2vt^(-3)) + 3t(v't^(-1) - vt^(-2)) - vt^(-1)
= 2tv'' + (-4+3)v' + (4t^(-1) - 3t^(-1) - t^(-1))v
= 2tv'' - v' = 0, (32)
Note that the coefficient of v is zero, as it should be, this provides a useful check on our algebra.
Separating the variables in Eq. (32) and solving for v'(t), we find that
v'(t) = ct^(1/2), (33)
...
I have difficulty following the step from 32 to 33.
Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?