PlanetPhysics/Riccati Equation 2

The nonlinear differential equation

${\begin{matrix}{\frac {dy}{dx}}=f(x)+g(x)y+h(x)y^{2}\end{matrix}}$

is called the Riccati equation .\, If\, $h(x)\equiv 0$,\, it becomes a linear differential equation; if\, $f(x)\equiv 0$,\, then it becomes a Bernoulli equation.\, There is no general method for integrating explicitely the equation (1), but via the substitution $y\,:=\,-{\frac {w'(x)}{h(x)w(x)}}$ one can convert it to a second order homogeneous linear differential equation with non-constant coefficients.\\

If one can find a particular solution \,$y_{0}(x)$,\, then one can easily verify that the substitution

${\begin{matrix}y\,:=\,y_{0}(x)+{\frac {1}{w(x)}}\end{matrix}}$

converts (1) to

${\begin{matrix}{\frac {dw}{dx}}+[g(x)\!+\!2h(x)y_{0}(x)]\,w+h(x)=0,\end{matrix}}$

which is a linear differential equation of first order with respect to the function \,$w=w(x)$.\\

Example. \, The Riccati equation

${\begin{matrix}{\frac {dy}{x}}=3+3x^{2}y-xy^{2}\end{matrix}}$

has the particular solution\, $y:=3x$.\, Solve the equation.

We substitute\, $y:=3x+{\frac {1}{w(x)}}$\, to (4), getting ${\frac {dw}{dx}}-3x^{2}w-x=0.$ For solving this first order equation we can put\, $w=uv$,\, $w'=uv'+u'v$,\, writing the equation as

${\begin{matrix}u\cdot (v'-3x^{3}v)+u'v=x,\end{matrix}}$

where we choose the value of the expression in parentheses equal to 0: ${\frac {dv}{dx}}-3x^{2}v=0$ After separation of variables and integrating, we obtain from here a solution\, $v=e^{x^{3}}$,\, which is set to the equation (5): ${\frac {du}{dx}}e^{x^{3}}=x$ Separating the variables yields $du={\frac {x}{e^{x^{3}}}}\,dx$ and integrating: $u=C+\int xe^{-x^{3}}\,dx.$ Thus we have $w=w(x)=uv=e^{x^{3}}\left[C+\int xe^{-x^{3}}\,dx\right],$ whence the general solution of the Riccati equation (4) is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle y \,:=\, 3x+\frac{e^{-x^3}}{C+\int xe^{-x^3}\,dx}.\\}

It can be proved that if one knows three different solutions of Riccati equation (1), then any other solution may be expressed as a rational function of the three known solutions.