# 齐次方程

$u=\frac{y}{x}, y=xu, y’=\frac{dy}{dx}=u+x\frac{du}{dx}$

\begin{align*}u+x\frac{du}{dx}=u+e^{-u}&\Rightarrow\quad x\frac{du}{dx}=e^{-u}\\ &\Rightarrow\quad e^udu=\frac{dx}{x}\\&\Rightarrow\quad \int e^udu=\int \frac{dx}{x}\\ &\Rightarrow\quad e^u=\ln|x|+C\\& \Rightarrow\quad u=\ln(\ln|x|+C)\\ &\Rightarrow\quad \frac{y}{x}=\ln(\ln|x|+C)\\ &\Rightarrow\quad y=x\ln(\ln|x|+C)\end{align*}

$y’=\left(\frac{y}{x}\right)^2+\frac{y}{x}-1$

\begin{align*}u+\frac{du}{dx}=u^2+u-1 &\Rightarrow \quad x\frac{du}{dx}=u^2-1\\ &\Rightarrow \quad \frac{du}{u^2-1}=\frac{dx}{x}\\ &\Rightarrow \quad \int\frac{du}{u^2-1}=\int\frac{dx}{x}\end{align*}

$\frac{1}{u^2-1}=\frac{1}{(u-1)(u+1)}=\frac{A}{u-1}+\frac{B}{u+1}$

\begin{align*}\int\frac{du}{u^2-1}=\int\frac{dx}{x}&\Rightarrow\quad \frac{1}{2}\int\frac{du}{u-1}-\frac{1}{2}\int\frac{du}{u+1}=\frac{dx}{x}\\ &\Rightarrow\quad \frac{1}{2}\ln|u-1|-\frac{1}{2}\ln|u+1|=\ln|x|+C\\ &\Rightarrow\quad \frac{1}{2}\ln|\frac{u-1}{u+1}|=\ln|x|+C\end{align*}

\begin{align*}\frac{1}{2}\ln|\frac{u-1}{u+1}|=\ln|x|+C &\Rightarrow\quad \ln|\frac{u-1}{u+1}|=2\ln|Cx|\\ &\Rightarrow\quad \frac{u-1}{u+1}=Cx^2\\ &\Rightarrow\quad y-x=Cx^2(y+x)\\ &\Rightarrow\quad y(-1Cx^2)=Cx^3++x\\ &\Rightarrow\quad y=\frac{x(1+Cx^2)}{1-Cx^2}\end{align*}