Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 60

Given equation is y'+xy=xy^(-1)

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

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

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

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

let us first solve
int e^(x^2) 2xdx
so , let t =x^2
dt = 2xdx
int e^(x^2) 2xdx = int e^(t) dt = e^t = e^(x^2)

so now => ue^(x^2)= e^(x^2)+c
=>u=((e^(x^2))+c)/(e^(x^2))
= 1 +ce^(-x^2)
but
u=y^2 ,so
y^2=(1 +ce^(-x^2))
y= sqrt (1 +ce^(-x^2))
is the general solution.