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
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?
From v dv =3y^2dy you will get 1/2v^2 = y^3 +c
From initial conditions, ...