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

Expert Answers
kseddy123 eNotes educator| Certified Educator

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.