# y' + xy = xy^-1 Solve the Bernoulli differential equation.

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)

=>...

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(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.

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