Riccati Equations

Before we give the formal definition of Riccati equations, a little introduction may be helpful. Indeed, consider the first order differential equation

$$\frac{dy}{dx} = f(x,y).$$

If we approximate f(x,y), while x is kept constant, we will get

$$f(x,y) = P(x) + Q(x) y + R(x) y^2 + \cdots$$

If we stop at y, we will get a linear equation. Riccati looked at the approximation to the second degree: he considered equations of the type

$$\frac{dy}{dx} = P(x) + Q(x) y + R(x) y^2.$$

These equations bear his name, Riccati equations. They are nonlinear and do not fall under the category of any of the classical equations. In order to solve a Riccati equation, one will need a particular solution. Without knowing at least one solution, there is absolutely no chance to find any solutions to such an equation. Indeed, let y₁ be a particular solution of

$$\frac{dy}{dx} = P(x) + Q(x) y + R(x) y^2.$$

Consider the new function z defined by

$$z = \frac{1}{y - y_1}.$$

Then easy calculations give

$$\frac{dz}{dx} = -\Big(Q(x) + 2y_1 R(x)\Big) z - R(x)$$

which is a linear equation satisfied by the new function z. Once it is solved, we go back to y via the relation

$$y = y_1 + \frac{1}{z}.$$

Keep in mind that it may be harder to remember the above equation satisfied by z. Instead, try to do the calculations whenever you can.

Example. Solve the equation

$$\frac{dy}{dx} = -2 -y + y^2.$$

knowing that y₁ = 2 is a particular solution.

Answer. We recognize a Riccati equation. First of all we need to make sure that y₁ is indeed a solution. Otherwise, our calculations will be fruitless. In this particular case, it is quite easy to check that y₁ = 2 is a solution. Set

$$z = \frac{1}{y - 2}.$$

Then we have

$$y = 2 + \frac{1}{z}$$

which implies

$$y ' = - \frac{z'}{z^2}.$$

Hence, from the equation satisfied by y, we get

$$- \frac{z'}{z^2} = -2 -\left(2 + \frac{1}{z}\right)+ \left(2 + \frac{1}{z}\right)^2.$$

Easy algebraic manipulations give

$$- \frac{z'}{z^2} = \frac{3}{z} + \frac{1}{z^2}.$$

Hence

z' = -3z -1.

This is a linear equation. The general solution is given by

$$z = \frac{-1/3 e^{3x} + C}{e^{3x}} = -\frac{1}{3} + C e^{-3x}.$$

Therefore, we have

$$y = 2 + \frac{1}{\displaystyle-\frac{1}{3} + C e^{-3x}}.$$

Note: If one remembers the equation satisfied by z, then the solutions may be found a bit faster. Indeed in this example, we have P(x) = -2, Q(x) = -1, and R(x) = 1. Hence the linear equation satisfied by the new function z, is

$$\frac{dz}{dx} = -\Big(Q(x) + 2y_1 R(x)\Big) z - R(x) = -\Big(-1 + 4\Big) z - 1 = - 3 z - 1.$$

Example. Check that $y_1 = \sin(x)$ is a solution to

$$\frac{dy}{dx} = \frac{2 \cos^2(x) - \sin^2(x) + y^2}{2 \cos(x)}.$$

Then solve the IVP

$$\left\{\begin{array}{lll} \displaystyle \frac{dy}{dx} = \frac... ... - \sin^2(x) + y^2}{2 \cos(x)}\\ y(0) = -1 \end{array} \right.$$

We will let the reader check that $\sin(x)$ is indeed a particular solution of the given differential equations. We also recognize that the equation is of Riccati type. Set

$$z = \frac{1}{y - \sin(x)}$$

which gives

$$y = \sin(x) + \frac{1}{z}.$$

Hence

$$y' = \cos(x) - \frac{z'}{z^2}.$$

Substituting into the equation gives

$$\cos(x) - \frac{z'}{z^2} = \frac{2 \cos^2(x) - \sin^2(x) + \left(\displaystyle \sin(x) + \frac{1}{z} \right)^2}{2 \cos(x)}.$$

Easy algebraic manipulations give

$$- \frac{z'}{z^2} = \frac{\left(\displaystyle 2\sin(x) \frac{1... ...sin(x)}{\cos(x)} \frac{1}{z} + \frac{1}{2\cos(x)}\frac{1}{z^2}.$$

Hence

$$z' = -\frac{\sin(x)}{\cos(x)} z - \frac{1}{2\cos(x)}.$$

This is the linear equation satisfied by z. The integrating factor is

$$u(x) = e^{\displaystyle \int \frac{\sin(x)}{\cos(x)}} = e^{-\ln(\cos(x))} = \frac{1}{\cos(x)} = \sec(x).$$

The general solution is

$$z = \frac{-1/2 \int \sec^2(x) dx + C}{u(x)} = \cos(x) \left(-\frac{1}{2} \tan(x) + C\right) = -\frac{1}{2} \sin(x) + C \cos(x).$$

Now it is time to go back to the original function y. We have

$$y = \sin(x) + \frac{1}{ \displaystyle -\frac{1}{2} \sin(x) + C \cos(x)}.$$

The initial condition y(0) = -1 implies 1/C = -1, or C = -1. Therefore the solution to the IVP is

$$y = \sin(x) + \frac{1}{ \displaystyle -\frac{1}{2} \sin(x) - \cos(x)}.$$

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Last Update 9-15-98

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