`y' - y = y^3` Solve the Bernoulli differential equation.

Given equation is `y' -y =y^3`

An equation of the form `y'+Py=Qy^n`

is called as the Bernoullis equation .

so, to proceed to solve this equation we have to transform the equation into a linear equation form of first order as follows

=>` y' (y^-n) +P y^(1-n)=Q`

let `u= y^(1-n)`

=> `(1-n)y^(-n)y'=u'`

=> `y^(-n)y' = (u')/(1-n)`

so ,

`y' (y^-n) +P y^(1-n)=Q`

=> `(u')/(1-n) +P u =Q `

so this equation is now of the linear form of first order

Now,

From this equation ,

`y' -y =y^3`

and

`y'+Py=Qy^n`

on comparing we get

`P=-1 , Q=1 , n=3`

so the linear form of first order of the equation `y' -y =y^3 ` is given as

=> `(u')/(1-n) +P u =Q ` where` u= y^(1-n) =y^-2 `

=> `(u')/(1-3) +(-1) u =1`

=> `(-u')/2 -u=1`

=> `u'+2u = -2`

so this linear equation is of the form

`y' + py=q`

`p=2 , q=-2`

so I.F (integrating factor ) = `e^(int p dx) = e^(int 2dx) = e^(2x)`

and the general solution is given as

`u (I.F)=int q * (I.F) dx +c `

=> `u(e^(2x))= int (-2) *(e^(2x)) dx+c`

=> `u (e^(2x))= (-2) int (e^(2x)) dx+c`

let us solve `int (e^(2x)) dx `

=>let` t= 2x`

`dt = 2dx`

=> `int (e^(t)) dt/2`

=>`1/2 (int (e^(t)) dt) = 1/2 e^t = (e^(2x))/2`

so, `int (e^(2x)) dx =(e^(2x))/2`

so  ,now

`u (e^(2x))= (-2) ((e^(2x))/2)+c`

=>`u (e^(2x))= -(e^(2x))+c`

=> `u = ((-(e^(2x)))+c)/(e^(2x))`

but `u=y^-2`

so,

`y^-2=((-(e^(2x)))+c)/(e^(2x))`

=> `y^2 = (e^(2x))/((-(e^(2x)))+c)`

=> `y = sqrt((e^(2x))/((-(e^(2x)))+c))`

=> `y = e^x/(sqrt(c-e^(2x)))`

the general solution.