Riccati方程

先同乘${\displaystyle q_{2}(x)}$ ，使得${\displaystyle q_{2}y'=q_{0}q_{2}+q_{1}q_{2}y+q_{2}^{2}y^{2}}$ 

再以${\displaystyle v=yq_{2}}$ 代入：

${\displaystyle v'=v^{2}+P(x)v+Q(x)}$ ；其中令 ${\displaystyle Q(x)=q_{0}q_{2};P(x)=q_{1}+{\frac {q_{2}'}{q_{2}}}}$  。

再以${\displaystyle v=-{\frac {u'}{u}}}$ 代入上式。

1. ${\displaystyle v'=-\left({\frac {u'}{u}}\right)'=-{\frac {u''}{u}}+\left({\frac {u'}{u}}\right)^{2}=-{\frac {u''}{u}}+v^{2}\!}$ 

则

${\displaystyle {\frac {u''}{u}}=v^{2}-v'=-Q-Pv=-Q+P{\frac {u'}{u}}}$ 

因此

${\displaystyle u''-Pu'+Qu=u''-(q_{1}+{\frac {q_{2}'}{q_{2}}})u'+q_{0}q_{2}u=0}$ 

最终 ${\displaystyle y=-{\frac {u'}{q_{2}u}}}$ .

${\displaystyle S(w)\equiv \left({\frac {w''}{w'}}\right)'-{\frac {\left({\frac {w''}{w'}}\right)^{2}}{2}}=f}$ 

显然可设${\displaystyle y={\frac {w''}{w'}}}$ ：

${\displaystyle y'-{\frac {y^{2}}{2}}=f}$ 

再代入 ${\displaystyle -{\frac {2u'}{u}}=y}$  ，得线性微分方程：

${\displaystyle u''-{\frac {1}{2}}fu=0}$ 

因为 ${\displaystyle {\frac {w''}{w'}}=-{\frac {2u'}{u}}}$  ，积分得${\displaystyle w'={\frac {C}{u^{2}}}}$ 。另一方面，若线性微分方程有其他线性独立解U，则有：

${\displaystyle w'={\frac {U'u-Uu'}{u^{2}}}}$ 

${\displaystyle w={\frac {U}{u}}}$ 

已知 ${\displaystyle y=y_{1}}$  是一特定解，可设通解${\displaystyle y=y_{1}+{\frac {1}{z}}}$ ，代入整理得一阶线性常微分方程：

${\displaystyle z'+(q_{1}+2q_{2}y_{1})z=-q_{2}}$ 

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