xy' + y = xy^3 Solve the Bernoulli differential equation.

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.