The solution of $y^{5}x+y-x\frac{\mathrm{d}y}{\mathrm{d}x}=0$ is 

The given differential equation can be written

as $y^{5}x\mathrm{d}x+y\mathrm{d}x-x\mathrm{d}y=0$ Multiplying by $x^3/y^5$, we have

                       $x^4\mathrm{d}x+\frac{x^3}{y^3}\left(\frac{y\mathrm{d}x-x\mathrm{d}y}{y^2}\right)=0$

Integrating, we get $x^5/5+(1/4)(x/y{)}^4=C,$

since $\frac{x^3}{y^3}\left(\frac{y\mathrm{d}x-x\mathrm{d}y}{y^2}\right)=u^3\mathrm{d}u\text{ where }u=\frac{x}{y}$