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?

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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