I am currently studying analysis of Algorithms and have come across a paper about Median of 3 partition for quick select. The authors have solved the recurrence using the generating functions and came up to a partial differential equation (eq 6 in the paper):
$\frac{1}{6} \frac{\partial^2Q}{\partial z^2} - \left [ \frac{1}{(1-z)^2} + \frac{u^2}{(1-uz)^2} \right]Q = \frac{u}{(1-u)} \left [ \frac{1}{(1-z)^4} - \frac{u^3}{(1-uz)^4} \right].$
Expressing $Q(z,u)$ as (eq 7 in paper)
$Q(z,u) = \frac{1}{(1-z)^2(1-uz)^2}E(z,u)$
and substitute $z = 1+t(1-u)/u$.
$G(t,u)=E\left(1+\frac{1-u}{u}t,u\right),$
(eq 8 in paper) the original equation will transforms to:
$t(1-t)\frac{\partial^2 G}{\partial t^2} + (-4+8t)\frac{\partial G}{\partial t} - 8G = 6(1-u)\frac{u(1-t)^4 - t^4}{t(1-t)}$
(eq 9 in paper). Now, I am not able to understand how the authors get this equation because when I tried to solve using the same substitutions I get the below one:
$t^2\frac{\partial^2 G}{\partial t^2} - 8t\frac{\partial G}{\partial t} + 2G \left [ 10 + \frac{u^2}{(1-u)^2} \right] = 6u^2t^2.$
Are the both equations same or am I doing some mistake?