Talk:PlanetPhysics/Riccati Equation 2
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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: Riccati equation %%% Primary Category Code: 02.30.Hq %%% Filename: RiccatiEquation2.tex %%% Version: 6 %%% Owner: pahio %%% Author(s): bci1, pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}
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The nonlinear \htmladdnormallink{differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} \begin{align}
\frac{dy}{dx} = f(x)+g(x)y+h(x)y^2 \end{align} is called {\em 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{align} y \,:=\, y_0(x)+\frac{1}{w(x)} \end{align} converts (1) to \begin{align} \frac{dw}{dx}+[g(x)\!+\!2h(x)y_0(x)]\,w+h(x) = 0, \end{align} which is a linear differential equation of first order with respect to the \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} \,$w =w(x)$.\\
\textbf{Example.}\, The Riccati equation \begin{align} \frac{dy}{x} = 3+3x^2y-xy^2 \end{align} has the particular solution\, $y := 3x$.\, Solve the equation.
We substitute\, $y := 3x+\frac{1}{w(x)}$\, to (4), getting $$\frac{dw}{dx}-3x^2w-x = 0.$$ For solving this first order equation we can put\, $w = uv$,\, $w' = uv'+u'v$,\, writing the equation as \begin{align} u\cdot(v'-3x^3v)+u'v = x, \end{align} where we choose the value of the expression in parentheses equal to 0: $$\frac{dv}{dx}-3x^2v = 0$$ 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 $$\displaystyle y \,:=\, 3x+\frac{e^{-x^3}}{C+\int xe^{-x^3}\,dx}.\\$$
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.
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