Riccati equation

This article will gives out a brief introduction about Riccati euqation¹.

Mathematical form of Riccati equation

Consider a first order ( $\frac{d y}{d x}$ ) quadratic ( $y^2$ ) equation:

$\begin{equation} \frac{d y}{d x}=A(x) y^{2}+B(x) y+C(x) \end{equation}$

we are going to look at a solution which form is:

$\begin{equation} y=y_1+\frac{1}{v(x)} \end{equation} \label{solution11}$

where $y_1$ can be represent a particular solution and $v(x)$ is a function of $x$ which we don’t know what that function is.

Now differentiate the above formula, we get:

$\begin{equation} \frac{d y}{d x}=\frac{d y_1}{d x}-\frac{1}{v^2}\frac{d v}{d x} \end{equation}$

thus we get

$\begin{equation} A(x) y^{2}+B(x) y+C(x)=A(x) y_1^{2}+B(x) y_1+C(x)-\frac{1}{v^2}\frac{d v}{d x} \end{equation}$

bring $\eqref{solution11}$ to above equation:

$\begin{equation} A(y+\frac{1}{v})^2 +B(y+\frac{1}{v}) +C=A y_1^{2}+B y_1+C-\frac{1}{v^2}\frac{d v}{d x} \end{equation}$

expend we get:

$\begin{equation} Ay_1^2 +2\frac{Ay_1}{v} +\frac{A}{v^2} +By_1+\frac{B}{v} +C=A y_1^{2}+B y_1+C-\frac{1}{v^2}\frac{d v}{d x} \end{equation}$

simplification

$\begin{equation} \frac{dv}{dx}+(2Ay_1+B)v=-A \end{equation}$

From the equation above we find that we write the problem to a linear first-order problem

Now we solve the rewritten problem

$\begin{equation} \frac{dv}{dx}+P(x)v=Q(x) \end{equation}$

First introduce a $\rho=e^{\int P(x)dx}$ and multiply the above equation:

$\begin{equation} e^{\int P(x)dx}\frac{dv}{dx}+P(x)e^{\int P(x)dx}v=Q(x)e^{\int P(x)dx} \end{equation}$

thus (wow it’s really amazing)

$\begin{equation} D_x(ve^{\int P(x)dx})=Q(x)e^{\int P(x)dx} \end{equation}$

integrate both sides with respect to $x$ and solve for $v$

$\begin{equation} v=\frac{\int{Q(x)e^{\int P(x)}}dx}{e^{\int P(x)dx}} \end{equation}$

Now we get our v.

Solve $\begin{equation} y^{\prime}+2 x y=1+x^{2}+y^{2} \end{equation}$ where $y_1=x$ is a solution.