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

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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