Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 62

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.