t^2 y' + 4ty - y^3 = 0
Hint was given in the question: substitute with v = y^-2
The Attempt at a Solution
Dividing by t^2 and isolating y':
t^2 y' = y^3 - 4ty
y' = y^3 / t^2 - 4y/t
dv/dt = 0
y = v^(-1/2)
dy/dt = (-1/2)v^(-3/2) v'
so y' = dy/dt = (-1/2)v^(-3/2) v' = v^(-3/2) - 4t(v^(-1/2))
But after playing with algebra I cannot separate v and t,
I end up with;
dv/dt - 8v/t = -2/t^2
Where v' = dv/dt
How should I have approached this problem?