Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 63

Given equation is xy' + y = xy^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 ,
xy' + y = xy^3
=> y'+(1/x)y =y^3
and
y'+Py=Qy^n
on comparing we get
P=(1/x) Q=1 , n=3
so the linear form of first order of the equation xy' + y = xy^3 is given as

=> (u')/(1-n) +P u =Q where u= y^(1-n) =y^-2
=> (u')/(1-3) +(1/x) u =1
=> -(u')/2 +u/x=1
=> (u')-2u/x = -2

so this linear equation is of the form
u' + pu=q
p=-2/x , q=-2
so I.F (integrating factor ) = e^(int p dx) = e^(int (-2/x)dx) = e^(ln(1/x^2))=1/x^2

and the general solution is given as
u (I.F)=int q * (I.F) dx +c
=> u(1/x^2)= int (-2) *(1/x^2) dx+c
=> u (1/x^2)= (-2) int (1/x^2) dx+c

so now,
u (1/x^2)= (-2) (-x^-1 /1)+c
=> u (1/x^2)=(2) (x^-1) +c
=>u (1/x^2)= (2/x) +c
=> u = ((2/x) +c)*x^2
but u=y^-2
so,
y^-2=((2/x) +c)*x^2
=> y^2 = 1/((2x) +c*x^2)
=> y = sqrt(1/((2x) +c*x^2))

=> y = sqrt(1/(2x +cx^2))
the general solution.