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I have been tasked with solving y'' - 3y^2 =0 using the technique used substituting v for y', therefore substituting v dv/dy for y''.

I have been tasked with solving y'' - 3y^2 =0 using the technique used substituting v for y', therefore substituting v dv/dy for y''.  (Equation with "x" missing)
I broke it down as follows
Y'' -3y^2 =0
Y'' =3y^2
Substituting I get v dv/dy = 3y^2
Separating variables, I get v dv =3y^2dy
Integrating I get 1/2v^2 +c = y^3 +c
Solving for v, I get v = (2y^3-2c)^1/2 which  back substituting, = dy/dx
From here, it is a blur!!  Initial conditions are  y(0) =2, y'(0) =4
Can anybody help me from here?

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