I have noticed a certain similarity between Riccati 1st Order ODEs and linear 1st Order ODEs. Specifically, the general solution for each is given by any particular solution plus some function of the independent variable.

In the linear case, this function is a scalar multiple of any particular solution to the associated homogeneous ODE, so the answer takes the form $y = y_p + Cy_h$, where $y$ is the general solution, $y_p$ is any particular solution, $C$ is an arbitrary constant, and $y_h$ is any particular solution to the homogeneous ODE. This is much the same as the observation that the solution to a linear system of equations is given by a particular solution plus the nullspace.

The answer to a Riccati ODE takes the form $y = y_p + u$, where $u$ is some function of the independent variable. Given this similarity, can we think of $u$ as the zero set (the generalization of the nullspace into nonlinear contexts), or as in some way highly related to $Cy_h$?

If so, is there an analog to the linear ODE fact that $y_p + Cy_h = y_h \int \frac{f}{y_h} dx$, where $f$ is the non-homogenous term of the linear ODE, which could possibly give rise to another solution method for the Riccati?

  • $\begingroup$ I don't believe that every solution of a Riccati equation is given by $y= y_p + u$. In general, it is much harder to find the general solutions of a non-linear ODE. For a $n^{th}$ order linear ODEs, we know when we have a basis because there is one particular solution and $n$ linearly independent homogenous solution. But we have no such theory for non-linear ODE; there can be solutions not of the form $y= y_p + u$ and I don't see why not. I think it's just a nice trick to convert the non-linear ODE into a Bernoulli Equation, which we know how to solve. $\endgroup$ – Story123 Dec 27 '18 at 6:00

The case you are considering is not the general case for solving the Riccati ODEs. In fact, it is a very simple case when a particular solution can be guessed and found easily. Of course that is often the case in academic exercises. Depending on the level of the students, it is more or less difficult to find a particular solution, then to apply the textbook method of solving which after some well-known transformations reduces to a first order linear ODE. This makes understand the similarity of solving method that you noticed.

But the majority of Riccati ODEs have not evident or simple particular solution. When one cannot find a particular solution, what to do ?

Contrary to preconceived ideas, one have to transform the Riccati ODE to a linear second order linear ODE. The Riccati ODE is a non-linear first order ODE. The linearity is more important that the order. A second order linear ODE is generally easier to solve than a first order non-linear ODE.

The method to transform the general Riccati ODE $$y'(x)=p(x)+q(x)y(x)+r(x)(y(x))^2$$ consists in the change of function : $$y(x)=-\frac{u'(x)}{r(x)u(x)}$$ This leads to a linear second order homogeneous ODE $$r(x)u''(x)-\left(r'(x)+q(x)r(x)\right)u'(x)+p(x)(r(x))^2u(x)=0$$

Of course, all linear second order ODEs cannot be solved in terms of a combination of a finite number of elementary and/or standard special functions. Thus all Riccati ODEs as well. Nevertheless many can be solved thanks to this approach when no particular solution can be found.

For example : $$y'=x-y^2$$ At first sight this equation seems simple. But you will fail to find a particular solution.

Thanks to the above method, the equation is transformed to a linear second order ODE of the Bessel kind. Finally the solution is : $$y(x)=\frac{\text{Ai}'(x)+c\:\text{Bi}'(x) }{\text{Ai}(x)+c\:\text{Bi}(x)}$$ Ai$(x)$ and Bi$(x)$ are the Airy functions which are particular forms of Bessel functions.


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