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

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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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=> `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


From this equation ,




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`


 = `1 +ce^(-x^2)`


`u=y^2` ,so

`y^2=(1 +ce^(-x^2))`

`y= sqrt (1 +ce^(-x^2))`

is the general solution.

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