微分方程式の解き方

$$\begin{eqnarray} \dv{y}{x}+xy&=&\frac{1}{2}xy^5\\ xy(1-\frac{1}{2}y^4)\mathrm{d}x+\mathrm{d}y&=&0\\ x\mathrm{d}x+\frac{1}{y(1-\frac{1}{2}y^4)}\mathrm{d}y&=&0\\ \int x\mathrm{d}x+\int\Big(\frac{1}{y}+\frac{\frac{1}{2}y^3}{1-\frac{1}{2}y^4}\Big)\mathrm{d}y&=&C\quad(\because\text{補題2})\\ \frac{1}{2}x^2+\ln|y|-\frac{1}{4}\ln\Big|1-\frac{1}{2}y^4\Big|&=&C\\ \frac{y^4}{1-\frac{1}{2}y^4}&=&C'e^{-2x^2} \end{eqnarray}$$