`y' + y/x = xy^2` Solve the Bernoulli differential equation.

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Given equation is `y'+y/x=xy^2`

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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Given equation is `y'+y/x=xy^2`

 

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)`

=> `(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'+y/x=xy^2`

and

`y'+Py=Qy^n`

on comparing we get

`P=(1/x) , Q=x , n=2`

so the linear form of first order of the equation `y'+y/x=xy^2 ` is given as

 

=> `(u')/(1-n) +P u =Q ` where` u= y^(1-n) =y^(1-2)=1/y `

=> `(u')/(1-2) +(1/x) u =x `
 
=> `-u' +(1/x) u =x `
 
=>`u' -(1/x) u = -x `

so this linear equation is of the form

`u' + pu=q`

`p=-(1/x) , q=-x`

so I.F (integrating factor )

= `e^(int p dx) = e^(int -(1/x) dx) = e^(-lnx)=1/x`

 

and the general solution is given as

`u (I.F)=int q * (I.F) dx +c `

=>`u (1/x)=int (-x) * (1/x) dx +c `

=>`u (1/x)=int (-1) dx +c `

=>`u (1/x)= -x+c `

=>`u= (c-x)/(1/x) = x(c-x)`

but `u=1/y`

`1/y = x(c-x)`

`y=1/(x(c-x))`

 

is the general solution.

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