利用首次积分法(First Integral)求解对称形式的常微分方程组：
\[\frac{ {\rm\,d}x}{-x+y+z}=\frac{ {\rm\,d}y}{x-y+z}=\frac{ {\rm\,d}z}{x+y-z}\]\[\frac{
{\rm\,d}x}{-x^2+y^2+z^2}=\frac{ {\rm\,d}y}{x^2-y^2+z^2}=\frac{ {\rm\,d}z}{x^2+y^2-z^2}\]\begin{align*}
&&\frac{ {\rm\,d}x}{-x+y+z}&=\frac{ {\rm\,d}y}{x-y+z}\\&\Rightarrow&\frac{ {\rm\,d}\left(x-y\right)}{-2\left(x-y\right)}&=\frac{ {\rm\,d}\left(x+y+z\right)}{x+y+z}\\&\Rightarrow&\frac{ {\rm\,d}\left(x-y\right)}{\left(x-y\right)}&=-\frac{2{\rm\,d}\left(x+y+z\right)}{\left(x+y+z\right)}\\&\Rightarrow&\ln\left|x-y\right|&=-2\ln\left|x+y+z\right|+\ln|C_1|\\&\Rightarrow&\left(x-y\right)\left(x+y+z\right)^2&=C_1\\\end{align*}同理
In like manner\begin{align*}&&\frac{ {\rm\,d}y}{x-y+z}&=\frac{ {\rm\,d}z}{x+y-z}\\&\Rightarrow&\frac{ {\rm\,d}\left(y-z\right)}{-2\left(y-z\right)}&=\frac{ {\rm\,d}\left(x+y+z\right)}{x+y+z}\\&\Rightarrow&\frac{ {\rm\,d}\left(y-z\right)}{\left(y-z\right)}&=-\frac{2{\rm\,d}\left(x+y+z\right)}{\left(x+y+z\right)}\\&\Rightarrow&\ln\left|y-z\right|&=-2\ln\left|x+y+z\right|+\ln|C_2|\\&\Rightarrow&\left(y-z\right)\left(x+y+z\right)^2&=C_2\\\end{align*} \begin{align*}\left\{ \begin{aligned} \frac{\text{d}x}{-x+y+z}&=\frac{\text{d}y}{x-y+z}\\ \frac{\text{d}y}{x-y+z}&=\frac{\text{d}z}{x+y-z}\\\end{aligned} \right. \Rightarrow \left\{ \begin{aligned} \left( x-y \right) \left( x+y+z \right) ^2&=C_1\\ \left( y-z \right) \left( x+y+z \right) ^2&=C_2\\\end{aligned} \right.\end{align*} \begin{align*}\left\{ \begin{aligned} \frac{\text{d}x}{-x+y+a}&=\frac{\text{d}y}{x-y+a}\\ \frac{\text{d}y}{b-y+z}&=\frac{\text{d}z}{b+y-z}\\\end{aligned} \right. \Rightarrow \left\{ \begin{aligned} x-y&=C_1e^{-\frac{x+y}{z}}\\ y-z&=C_2e^{-\frac{y+z}{x}}\\\end{aligned} \right.\end{align*}\begin{align*}\left\{ \begin{aligned} \frac{\text{d}x}{-x^2+y^2+a^2}&=\frac{\text{d}y}{x^2-y^2+a^2}\\ \frac{\text{d}y}{b^2-y^2+z^2}&=\frac{\text{d}z}{b^2+y^2-z^2}\\\end{aligned} \right. \Rightarrow \left\{ \begin{aligned} x-y&=C_1e^{-\frac{\left( x+y \right) ^2}{2a^2}}\\ y-z&=C_2e^{-\frac{\left( y+z \right) ^2}{2b^2}}\\\end{aligned} \right.\end{align*}另外：\begin{align*} &&\frac{ {\rm\,d}x}{-x^3+y^3+a^3}&=\frac{ {\rm\,d}y}{x^3-y^3+a^3}\\ &\Rightarrow&\frac{ {\rm\,d}\left(x-y\right)}{-2\left(x^3-y^3\right)}&=\frac{ {\rm\,d}\left(x+y\right)}{2a^3}\\ &\Rightarrow&\frac{ {\rm\,d}\left(x-y\right)}{ {\rm\,d}\left(x+y\right)}&=\frac{-2\left(x^3-y^3\right)}{2a^3}\\ &\Rightarrow&\frac{ {\rm\,d}\left(x-y\right)}{ {\rm\,d}\left(x+y\right)}&=\frac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{a^2}\\ &\Rightarrow&\frac{\left(x-y\right){\rm\,d}\left(x-y\right)}{ {\rm\,d}\left(x+y\right)}&=\frac{-\left(x-y\right)^2\left(3\left(x+y\right)^2+\left(x-y\right)^2\right)}{a^2}\\ &\Rightarrow&\frac{ {\rm\,d}\left(\left(x-y\right)^2\right)}{ {\rm\,d}\left(x+y\right)}&=-\frac{3\left(x+y\right)^2}{a^2}\left(x-y\right)^2-\frac{1}{a^2}\left(x-y\right)^4\\ &\Rightarrow&\frac{ {\rm\,d}u}{ {\rm\,d}v}&=-\frac{3v^2}{a^2}u-\frac{1}{a^2}u^2\\ &\Rightarrow&\\ &\Rightarrow&\\ \end{align*}
%http://kuing.orzweb.net/viewthread.php?tid=5752&tdsourcetag=s_pctim_aiomsg
\[y'''=\frac{3\left(y''\right)^2+x\!\cdot\!\left(y'\right)^5}{y'}\]
\begin{align*}\frac{ {\rm\,d}x}{ {\rm\,d}y}&=\frac{1}{y'}\\\frac{ {\rm\,d}^2x}{ {\rm\,d}y^2}&=\frac{ {\rm\,d}}{ {\rm\,d}y}\left(\frac{ {\rm\,d}x}{ {\rm\,d}y}\right)=\frac{ {\rm\,d}}{ {\rm\,d}y}\left(\frac{1}{y'}\right)=\frac{ {\rm\,d}}{ {\rm\,d}x}\left(\frac{1}{y'}\right)\frac{ {\rm\,d}x}{ {\rm\,d}y}\\&=-\frac{y''}{\left(y'\right)^2}\frac{ {\rm\,d}x}{ {\rm\,d}y}=-\frac{y''}{\left(y'\right)^3}\\\frac{ {\rm\,d}^3x}{ {\rm\,d}y^3}&=\frac{ {\rm\,d}}{ {\rm\,d}y}\left(\frac{ {\rm\,d}^2x}{ {\rm\,d}y^2}\right)=\frac{ {\rm\,d}}{ {\rm\,d}y}\left(-\frac{y''}{\left(y'\right)^3}\right)=\frac{ {\rm\,d}}{ {\rm\,d}x}\left(-\frac{y''}{\left(y'\right)^3}\right)\frac{ {\rm\,d}x}{ {\rm\,d}y}\\&=-\frac{y'''\!\cdot\!\left(y'\right)^3-3\left(y'\right)^2y''\!\cdot\!y''}{\left(y'\right)^6}\frac{ {\rm\,d}x}{ {\rm\,d}y}=-\frac{y'''\!\cdot\!\left(y'\right)^3-3\left(y'\right)^2\left(y''\right)^2}{\left(y'\right)^7}\\&=\frac{3\left(y'\right)^2\left(y''\right)^2-y'''\!\cdot\!\left(y'\right)^3}{\left(y'\right)^7}=\frac{3\left(y''\right)^2-y'y'''}{\left(y'\right)^5}\end{align*}\begin{align*}
&&\frac{ {\rm\,d}^3x}{ {\rm\,d}y^3}&=\frac{\color{red}{3\left(y''\right)^2-y'y'''}}{\left(y'\right)^5}=\frac{\color{red}{-x\!\cdot\!\left(y'\right)^5}}{\left(y'\right)^5}=-x\\&\Rightarrow&\frac{ {\rm\,d}^3x}{ {\rm\,d}y^3}+x&=0\\&\Rightarrow&x'''+x&=0\\\end{align*}\begin{align*}
x'''+x&=0\\x&=C_1e^{-y}+C_2e^{\frac{y}{2}}\cos\left(\frac{\sqrt{3}}{2}y\right)+C_3e^{\frac{y}{2}}\sin\left(\frac{\sqrt{3}}{2}y\right)\end{align*}更一般：
\[
y ^ { \prime } y ^ {\prime \prime \prime } - 3 \left( y ^ { \prime \prime } \right) ^ { 2 } + a \left( y ^ { \prime } \right) ^ { 2 } y ^ { \prime \prime } - b \left( y ^ { \prime } \right) ^ { 4 } - c x \cdot \left( y ^ { \prime } \right) ^ { 5 } = 0\]\[\xrightarrow[\frac{
{\rm\,d}^3x}{ {\rm\,d}y^3}=\frac{3\left(y''\right)^2-y'y'''}{\left(y'\right)^5}]{\frac{ {\rm\,d}x}{ {\rm\,d}y}=\frac{1}{y'}\quad\frac{ {\rm\,d}^2x}{ {\rm\,d}y^2}=-\frac{y''}{\left(y'\right)^3}}\]\[f ^ { \prime } ( x ) y ^ { \prime } y ^ { \prime \prime \prime } - 3 f ^ { \prime } ( x ) \left( y ^ { \prime \prime } \right) ^ { 2 } + 3 f ^ { \prime \prime } ( x ) y ^ { \prime } y ^ { \prime \prime } + a f ^ { \prime } ( x ) \left( y ^ { \prime } \right) ^ { 2 } y ^ { \prime \prime } = a f ^ { \prime \prime } ( x ) \left( y ^ { \prime } \right) ^ { 3 } + b f ^ { \prime } ( x ) \left( y ^ { \prime } \right) ^ { 4 } + f ^ { \prime \prime \prime } ( x ) \left( y ^ { \prime } \right) ^ { 2 } + c f ( x ) \left( y ^ { \prime } \right) ^ { 5 }\]
令$u(y)=f(x)$
\[\frac { \mathrm { d } ^ { 3 } u ( y ) } { \mathrm { d } y ^ { 3 } } + a \frac { \mathrm { d } ^ { 2 } u ( y ) } { \mathrm { d } y ^ { 2 } } + b \frac { \mathrm { d } u ( y ) } { \mathrm { d } y } + c u ( y ) = 0.\]\end{spacing}\begin{align*}
\min_x\quad f(x)=\sum_{i=1}^{n} f_i(x),\\s.t.\quad g_i(x)\leq 0, A_ix=b_i,x\in\cap_{i=1}^n\Omega_i.\end{align*}\begin{align*}
\frac{d}{dt}\left( \begin{array}{c} y_i\\ \lambda _i\\ \mu _i\\ x_i\\\end{array} \right) &=\left( \begin{array}{c} -y_i+x_i-\nabla f_i\left( x_i \right) -\left( \nabla g_i\left( x_i \right) \right) ^T\left( \lambda _i+g_i\left( x_i \right) \right) ^+-A_{i}^{T}\left( \mu _i+A_ix_i-b_i \right) +u_i\\ -\lambda _i+\left( \lambda _i+g_i\left( x_i \right) \right) ^+\\ A_ix_i-b_i\\ x=P_{\Omega _i}\left( y_i \right)\\\end{array} \right)\\u_i &=\mathcal{k}_P\sum_{j=1}^n{a_{ij}\left( x_j-x_i \right)}+\mathcal{k}_I\int_0^t{\sum_{j=1}^n{a_{ij}\left( x_j\left( s \right) -x_i\left( s \right) \right) ds}}\end{align*}