How to solve the following differential equation. y''-3y^2 = 0, y(0)=2, y'(0)=4   Use the following the technique to solve. y''=f(y,y') v'=f(y,v) dv/dx=(dv/dy)(dy/dx)=(dv/dy)v v(dv/dx)=f(y,v) dy/dx=v(y)  

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You need to come up with the substitution `y' = (dy)/(dx) = v=> y" = (d^2 y)/(dx^2) = (dv)/(dx) = (dv)/(dy)*(dy)/(dx) = v*(dv)/(dy)`

The problem provides the fact that: `y''-3y^2 = 0 =gt y" = 3y^2` , hence `v*(dv)/(dy) = 3y^2` .

You need to separate the variables hence, multiplying by dy both sides yields:

`vdv = 3y^2 dy`

`` Integrating both...

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