# 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)...

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)

### 1 Answer | Add Yours

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 sides yields:

`int vdv = int 3y^2 dy =gt (v^2)/2 = 3y^3/3 + c`

`v^2 = 2y^3 + c =gt v = sqrt(2y^3+c)`

At `x = 0 =gt y(0) = 2 and v = y'(0)=4=gt16 = 2*2^3 + c =gt c = 0`

`` `v^2 = 2y^3 =gt v = ysqrt2*sqrty`

`v = dy/dx =gt dy/dx = ysqrt2*sqrty =gt dy/(y^(3/2)) = sqrt2 dx`

Integrating both sides yields:

`int dy/(y^(3/2)) = int sqrt2 dx =gt y^(-3/2 + 1)/(-3/2 + 1) = sqrt2* x + c`

`` `-2/sqrt y = sqrt2*x + c =gt sqrt y = -2/(sqrt2*x + c)`

`y = 4/((sqrt2*x + c)^2)`

The problem provides that y=2 at `x=0 =gt 2 = 4/(c^2) =gt c^2 = 2 =gt c = +-sqrt 2`

Taking `c = -sqrt 2 =gt y = 2/((x-1)^2)`

Taking `c = sqrt 2 =gt y = 2/((x+1)^2)`

**Hence, evaluating solutions to differential equation yields:`y = 2/((x-+1)^2).` **