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`=> `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.