Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 64

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.