y' + y/x = xy^2 Solve the Bernoulli differential equation.

Given equation is y'+y/x=xy^2
 
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'+y/x=xy^2
and
y'+Py=Qy^n
on comparing we get
P=(1/x) , Q=x , n=2
so the linear form of first order of the equation y'+y/x=xy^2 is given as
 
=> (u')/(1-n) +P u =Q where u= y^(1-n) =y^(1-2)=1/y
=> (u')/(1-2) +(1/x) u =x
 
=> -u' +(1/x) u =x
 
=>u' -(1/x) u = -x so this linear equation is of the form
u' + pu=q
p=-(1/x) , q=-x
so I.F (integrating factor )
= e^(int p dx) = e^(int -(1/x) dx) = e^(-lnx)=1/x
 
and the general solution is given as
u (I.F)=int q * (I.F) dx +c
=>u (1/x)=int (-x) * (1/x) dx +c
=>u (1/x)=int (-1) dx +c
=>u (1/x)= -x+c
=>u= (c-x)/(1/x) = x(c-x)
but u=1/y
1/y = x(c-x)
y=1/(x(c-x))
 
is the general solution.