常微分方程

利用首次积分法(First Integral)求解对称形式的常微分方程组：
[frac{{ m\,d}x}{-x+y+z}=frac{{ m\,d}y}{x-y+z}=frac{{ m\,d}z}{x+y-z}]

[frac{{ m\,d}x}{-x^2+y^2+z^2}=frac{{ m\,d}y}{x^2-y^2+z^2}=frac{{ m\,d}z}{x^2+y^2-z^2}]

egin{align*}
&&frac{{ m\,d}x}{-x+y+z}&=frac{{ m\,d}y}{x-y+z}\
&Rightarrow&frac{{ m\,d}left(x-y ight)}{-2left(x-y ight)}&=frac{{ m\,d}left(x+y+z ight)}{x+y+z}\
&Rightarrow&frac{{ m\,d}left(x-y ight)}{left(x-y ight)}&=-frac{2{ m\,d}left(x+y+z ight)}{left(x+y+z ight)}\
&Rightarrow&lnleft|x-y ight|&=-2lnleft|x+y+z ight|+ln|C_1|\
&Rightarrow&left(x-y ight)left(x+y+z ight)^2&=C_1\
end{align*}

同理
In like manner
egin{align*}
&&frac{{ m\,d}y}{x-y+z}&=frac{{ m\,d}z}{x+y-z}\
&Rightarrow&frac{{ m\,d}left(y-z ight)}{-2left(y-z ight)}&=frac{{ m\,d}left(x+y+z ight)}{x+y+z}\
&Rightarrow&frac{{ m\,d}left(y-z ight)}{left(y-z ight)}&=-frac{2{ m\,d}left(x+y+z ight)}{left(x+y+z ight)}\
&Rightarrow&lnleft|y-z ight|&=-2lnleft|x+y+z ight|+ln|C_2|\
&Rightarrow&left(y-z ight)left(x+y+z ight)^2&=C_2\
end{align*}

egin{align*}
left{ egin{aligned}
frac{ ext{d}x}{-x+y+z}&=frac{ ext{d}y}{x-y+z}\
frac{ ext{d}y}{x-y+z}&=frac{ ext{d}z}{x+y-z}\
end{aligned} ight. Rightarrow left{ egin{aligned}
left( x-y ight) left( x+y+z ight) ^2&=C_1\
left( y-z ight) left( x+y+z ight) ^2&=C_2\
end{aligned} ight.
end{align*}

egin{align*}
left{ egin{aligned}
frac{ ext{d}x}{-x+y+a}&=frac{ ext{d}y}{x-y+a}\
frac{ ext{d}y}{b-y+z}&=frac{ ext{d}z}{b+y-z}\
end{aligned} ight. Rightarrow left{ egin{aligned}
x-y&=C_1e^{-frac{x+y}{z}}\
y-z&=C_2e^{-frac{y+z}{x}}\
end{aligned} ight.
end{align*}

egin{align*}
left{ egin{aligned}
frac{ ext{d}x}{-x^2+y^2+a^2}&=frac{ ext{d}y}{x^2-y^2+a^2}\
frac{ ext{d}y}{b^2-y^2+z^2}&=frac{ ext{d}z}{b^2+y^2-z^2}\
end{aligned} ight. Rightarrow left{ egin{aligned}
x-y&=C_1e^{-frac{left( x+y ight) ^2}{2a^2}}\
y-z&=C_2e^{-frac{left( y+z ight) ^2}{2b^2}}\
end{aligned} ight.
end{align*}
另外：
egin{align*} &&frac{{ m\,d}x}{-x^3+y^3+a^3}&=frac{{ m\,d}y}{x^3-y^3+a^3}\ &Rightarrow&frac{{ m\,d}left(x-y ight)}{-2left(x^3-y^3 ight)}&=frac{{ m\,d}left(x+y ight)}{2a^3}\ &Rightarrow&frac{{ m\,d}left(x-y ight)}{{ m\,d}left(x+y ight)}&=frac{-2left(x^3-y^3 ight)}{2a^3}\ &Rightarrow&frac{{ m\,d}left(x-y ight)}{{ m\,d}left(x+y ight)}&=frac{-left(x-y ight)left(x^2+xy+y^2 ight)}{a^2}\ &Rightarrow&frac{left(x-y ight){ m\,d}left(x-y ight)}{{ m\,d}left(x+y ight)}&=frac{-left(x-y ight)^2left(3left(x+y ight)^2+left(x-y ight)^2 ight)}{a^2}\ &Rightarrow&frac{{ m\,d}left(left(x-y ight)^2 ight)}{{ m\,d}left(x+y ight)}&=-frac{3left(x+y ight)^2}{a^2}left(x-y ight)^2-frac{1}{a^2}left(x-y ight)^4\ &Rightarrow&frac{{ m\,d}u}{{ m\,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{3left(y'' ight)^2+x!cdot!left(y' ight)^5}{y'}]
egin{align*}
frac{{ m\,d}x}{{ m\,d}y}&=frac{1}{y'}\
frac{{ m\,d}^2x}{{ m\,d}y^2}&=frac{{ m\,d}}{{ m\,d}y}left(frac{{ m\,d}x}{{ m\,d}y} ight)
=frac{{ m\,d}}{{ m\,d}y}left(frac{1}{y'} ight)
=frac{{ m\,d}}{{ m\,d}x}left(frac{1}{y'} ight)frac{{ m\,d}x}{{ m\,d}y}\
&=-frac{y''}{left(y' ight)^2}frac{{ m\,d}x}{{ m\,d}y}=-frac{y''}{left(y' ight)^3}\
frac{{ m\,d}^3x}{{ m\,d}y^3}&=frac{{ m\,d}}{{ m\,d}y}left(frac{{ m\,d}^2x}{{ m\,d}y^2} ight)
=frac{{ m\,d}}{{ m\,d}y}left(-frac{y''}{left(y' ight)^3} ight)
=frac{{ m\,d}}{{ m\,d}x}left(-frac{y''}{left(y' ight)^3} ight)frac{{ m\,d}x}{{ m\,d}y}\
&=-frac{y'''!cdot!left(y' ight)^3-3left(y' ight)^2y''!cdot!y''}{left(y' ight)^6}frac{{ m\,d}x}{{ m\,d}y}=-frac{y'''!cdot!left(y' ight)^3-3left(y' ight)^2left(y'' ight)^2}{left(y' ight)^7}\
&=frac{3left(y' ight)^2left(y'' ight)^2-y'''!cdot!left(y' ight)^3}{left(y' ight)^7}=frac{3left(y'' ight)^2-y'y'''}{left(y' ight)^5}
end{align*}

egin{align*}
&&frac{{ m\,d}^3x}{{ m\,d}y^3}&=frac{color{red}{3left(y'' ight)^2-y'y'''}}{left(y' ight)^5}=frac{color{red}{-x!cdot!left(y' ight)^5}}{left(y' ight)^5}=-x\
&Rightarrow&frac{{ m\,d}^3x}{{ m\,d}y^3}+x&=0\
&Rightarrow&x'''+x&=0\
end{align*}

egin{align*}
x'''+x&=0\
x&=C_1e^{-y}+C_2e^{frac{y}{2}}cosleft(frac{sqrt{3}}{2}y ight)+C_3e^{frac{y}{2}}sinleft(frac{sqrt{3}}{2}y ight)
end{align*}

更一般：

[
y ^ { prime } y ^ {prime prime prime } - 3 left( y ^ { prime prime } ight) ^ { 2 } + a left( y ^ { prime } ight) ^ { 2 } y ^ { prime prime } - b left( y ^ { prime } ight) ^ { 4 } - c x cdot left( y ^ { prime } ight) ^ { 5 } = 0
]

[xrightarrow[frac{{ m\,d}^3x}{{ m\,d}y^3}=frac{3left(y'' ight)^2-y'y'''}{left(y' ight)^5}]
{frac{{ m\,d}x}{{ m\,d}y}=frac{1}{y'}quadfrac{{ m\,d}^2x}{{ m\,d}y^2}=-frac{y''}{left(y' ight)^3}}]

[f ^ { prime } ( x ) y ^ { prime } y ^ { prime prime prime } - 3 f ^ { prime } ( x ) left( y ^ { prime prime } ight) ^ { 2 } + 3 f ^ { prime prime } ( x ) y ^ { prime } y ^ { prime prime } + a f ^ { prime } ( x ) left( y ^ { prime } ight) ^ { 2 } y ^ { prime prime } = a f ^ { prime prime } ( x ) left( y ^ { prime } ight) ^ { 3 } + b f ^ { prime } ( x ) left( y ^ { prime } ight) ^ { 4 } + f ^ { prime prime prime } ( x ) left( y ^ { prime } ight) ^ { 2 } + c f ( x ) left( y ^ { prime } ight) ^ { 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}

egin{align*}
min_xquad f(x)=sum_{i=1}^{n} f_i(x),\
s.t.quad g_i(x)leq 0, A_ix=b_i,xincap_{i=1}^nOmega_i.
end{align*}

egin{align*}
frac{d}{dt}left( egin{array}{c}
y_i\
lambda _i\
mu _i\
x_i\
end{array} ight) &=left( egin{array}{c}
-y_i+x_i- abla f_ileft( x_i ight) -left( abla g_ileft( x_i ight) ight) ^Tleft( lambda _i+g_ileft( x_i ight) ight) ^+-A_{i}^{T}left( mu _i+A_ix_i-b_i ight) +u_i\
-lambda _i+left( lambda _i+g_ileft( x_i ight) ight) ^+\
A_ix_i-b_i\
x=P_{Omega _i}left( y_i ight)\
end{array} ight)
\
u_i &=mathcal{k}_Psum_{j=1}^n{a_{ij}left( x_j-x_i ight)}+mathcal{k}_Iint_0^t{sum_{j=1}^n{a_{ij}left( x_jleft( s ight) -x_ileft( s ight) ight) ds}}
end{align*}