`y' + y/x = xsqrt(y)` Solve the Bernoulli differential equation.

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Bernoulli equation has form `y'+P(x)y=Q(x)y^alpha.` We can convert any such equation into linear equation differential equation by using substitution `z=y^(1-alpha).` To learn more about this method of solving Bernoulli equation check out the links below.

We will show a somewhat different method which can also be used for solving linear equations.

`y'+y/x=x sqrt(y)`

Make substitution `y=uv` `=>` `y'=u'v+uv'`

`u'v+uv'+(uv)/x=x sqrt(uv)` 

`v(u'+u/x)+uv'=x sqrt(uv)`                                                                (1)

Let us now assume that the expression in brackets equals zero (we can choose almost anything instead of zero, but this makes the calculation easier) in order to calculate `u.`

`u'+u/x=0`

`u'=-u/x`

`(du)/u=-dx/x`

Integrating both sides yields

`ln u=-ln x`

`ln u=ln x^-1`

`u=x^-1`

Plugging that into (1) gives

`x^-1 v'=x sqrt(x^-1 v)`

Now we multiply by `x` to get only `v'` on the left hand side. 

`v'=x^2x^(-1/2)sqrt v`

`v'=x^(3/2)sqrt v`

`(dv)/sqrt v=x^(3/2)dx`

Integrating the above equation gives us

`2sqrt v=(2x^(5/2)+C)/5`

Divide by 2 and square the whole equation in order to get `v.`

`v=(x^5+2Cx^(5/2)+C^2)/25`

Now we just plug the obtained `u` and `v` into the substitution to get the final result.

`y=x^-1 cdot (x^5+2Cx^(5/2)+C^2)/25`  

`y=(x^5+2Cx^(5/2)+C^2)/(25x)`  `lArr` The general solution. 

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