# 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.