Riccati equationRiccatiEquation22009-05-29 15:59:332009-05-31 01:46:13Topicseparation of variablesextensions of Bernoulli equationhomogeneous linear differential equation% this is the default PlanetPhysics preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands hereThe nonlinear differential equation
\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 is a question of a linear differential equation; if\, $f(x) \equiv 0$,\, of 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 \PMlinkescapetext{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 function \,$w =w(x)$.\\
\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 \PMlinkname{first order equation}{LinearDifferentialEquationOfFirstOrder} 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.