Solving a given Differential Equation I have tried quiet a lot of methods to solve the given ODE. $$xdy-(3y+x^5y^{1/3})dx = 0$$
Can anybody share any clue on solving this?
 A: We have 
$$xy'=3y+x^5y^{1/3}.$$
Of course $y=0$ ia a solution. 
Now let $y\ne 0$. Consider $y=u^3$, then $y'=3u^2u'$. Substitute:
$$3xu^2u'=3u^3+x^5u$$
or
$$u'=\frac{u}{x}+\frac{x^4}{3u}$$
Now suppose $u=vx$, so $u'=v+xv'$ and we have
$$v+xv'=v+\frac{x^3}{3v}$$
or $$xv'=\frac{x^3}{3v}$$
and you can separate now.
A: $$xdy-(3y+x^5y^{1/3})dx = 0$$
Multiply by $x^2$ :
$$x^3dy-(3yx^2+x^7y^{1/3})dx = 0$$
Rearrange terms:
$$(x^3dy-3yx^2dx)-x^7y^{1/3}dx = 0$$
Rewrite it as:
$$d \left (\frac {y}{x^3} \right )-xy^{1/3}dx = 0$$
Multiply by $\frac {x}{y^{1/3}}$:
$$\left (\frac {x}{y^{1/3}} \right )d \left (\frac {y}{x^3} \right )-x^2dx = 0$$
Integrate.
A: We are given
$$xdy-(3y+x^5y^{1/3})dx = 0$$
Rearrange the differential equation to
$$y'-\frac{3y}{x}=x^4y^{1/3}\tag{1}$$
where $y'=dy/dx$. As suggested in the comments, this is a Bernoulli differential equation of the type
$$y'+a(x)y=b(x)y^m$$
To convert the Bernoulli equation into a linear equation we use the change of variable
$$z=y^{1−m}$$
As $m=1/3$, substitute 
$$z=y^{2/3}, \quad z'=\frac{2}{3}y^{-1/3}y'=\frac{2}{3y^{1/3}}y'$$
and then multiply every term in $(1)$ by $$\frac{2}{3y^{1/3}}$$
so that $(1)$ becomes
$$\frac{2}{3y^{1/3}}y'-\frac{2y^{2/3}}{x}=\frac{2}{3}x^4\tag{2}$$
or
$$z'-\frac{2}{x}z=\frac{2}{3}x^4\tag{3}$$
which is a linear first order differential equation of the form
$$z' + P(x)z = Q(x)$$
therefore you can find the integrating factor and then the general solution for $z(x)$. You then need to make the substitution $y=z^{3/2}$ to obtain an implicit solution.
