Ecuación diferencial ordinaria de Riccati

${\displaystyle {\frac {dy}{dx}}=-p(x)y-q(x)y^{2}+f(x)={\frac {dy_{1}}{dx}}-{\frac {1}{z^{2}}}{\frac {dz}{dx}}}$

${\displaystyle \Rightarrow -p(x)y-q(x)y^{2}+f(x)=-{\frac {1}{z^{2}}}{\frac {dz}{dx}}-p(x)y_{1}-q(x)y_{1}^{2}+f(x)}$

${\displaystyle \Rightarrow {\frac {1}{z^{2}}}{\frac {dz}{dx}}=p(x)(y-y_{1})+q(x)(y^{2}-y_{1}^{2})}$

teniendo en cuenta que ${\displaystyle y=y_{1}+{\frac {1}{z}}}$

${\displaystyle \Rightarrow {\frac {1}{z^{2}}}{\frac {dz}{dx}}={\frac {p(x)}{z}}+q(x)({\frac {1}{z^{2}}}+{\frac {2y_{1}}{z}})}$

${\displaystyle \Rightarrow {\frac {dz}{dx}}-(p(x)+2q(x)y_{1})z=q(x)}$ ∎

${\displaystyle {\frac {dy}{dx}}=-2xy+y^{2}+x^{2}}$

una solución particular es ${\displaystyle y_{1}(x)=x+1}$

luego:

${\displaystyle {\frac {dy}{dx}}=-\underbrace {2x} _{p(x)}y+\underbrace {1} _{-q(x)}y^{2}+\underbrace {x^{2}} _{f(x)}}$

${\displaystyle \Rightarrow {\frac {dz}{dx}}-(\underbrace {2x} _{p(x)}+2(\underbrace {-1} _{q(x)})(\underbrace {x+1} _{y_{1}}))z=\underbrace {-1} _{q(x)}}$

${\displaystyle \Rightarrow {\frac {dz}{dx}}-(2x-2x-2)z=-1}$

${\displaystyle \Rightarrow {\frac {dz}{dx}}=-2z-1}$

${\displaystyle \Rightarrow \int {\frac {dz}{-2z-1}}=\int {dx}}$

${\displaystyle \Rightarrow -{\frac {1}{2}}ln(2z+1)=x+C}$

${\displaystyle \Rightarrow 2z+1=Ce^{-2x}}$

${\displaystyle \Rightarrow z={\frac {Ce^{-2x}-1}{2}}}$

puesto que ${\displaystyle y=y_{1}+{\frac {1}{z}}=x+1+{\frac {1}{z}}}$

${\displaystyle \Rightarrow z={\frac {1}{y-x-1}}}$

Luego la solución general es:

${\displaystyle y=x+1+{\frac {2}{Ce^{-2x}-1}}}$