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.
Make substitution `y=uv` `=>` `y'=u'v+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.`
Integrating both sides yields
`ln u=-ln x`
`ln u=ln 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.
Integrating the above equation gives us
Divide by 2 and square the whole equation in order to get `v.`
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.