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 from32to33.

Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?