Riccati equation

The Riccati equation is one of the most interesting first-order nonlinear differential equations. It is written in the form:

where a ( x ) , b ( x ) , c ( x ) are continuous functions depending on the variable x .

The Riccati equation is encountered in various fields of mathematics (for example, in algebraic geometry and in the theory of conformal mappings) and physics. It also frequently arises in applied mathematical problems.

The above equation is called the general Riccati equation. Its solution is based on the following theorem:

Theorem: If a particular solution y ₁ of the Riccati equation is known, then its general solution is determined by the formula

Indeed, substituting the solution y = y ₁ + u into the Riccati equation, we have:  

The underlined terms on the left and right sides can be canceled because y ₁ is a particular solution satisfying the equation. As a result, we obtain the differential equation for the function u ( x ) :

which is the Bernoulli equation. Substituting z = 1/ u transforms this Bernoulli equation into a linear differential equation that admits integration. In addition to the general Riccati equation, there are many special cases of the Riccati equation with coefficients a ( x ) , b ( x ) , c ​​( x ) of a certain form. Many of these special cases have integrable solutions. Returning to the general Riccati equation, we see that the general solution can be constructed if some special solution is known. Unfortunately, there is no rigorous algorithm for finding a special solution, which depends significantly on the form of the functions a ( x ) , b ( x ) and c ( x ) . Below we consider some well-known special cases of the Riccati equation. 

If the coefficients in a Riccati equation are constant, then such an equation can be reduced to an equation with separable variables. In this case, the general solution is described by the integral of a rational function with a quadratic trinomial in the denominator:

This integral is easily calculated for any values ​of a, b, and c (See more details on the page ” Integration of rational functions “).

Consider the Riccati equation of the form y’ = by ² + cx ⁿ, when the function a ( x ) in the linear term is zero, the coefficient b of y^2 is a constant, and c ( x ) is a power function:

This case of the Riccati equation has remarkable solutions!

First, note that if n = 0, then we again arrive at case 1, in which the variables are separated, and the equation can be integrated.

If n = −2, then the Riccati equation is transformed into a homogeneous equation by substituting y = 1/ z and can then also be integrated.

This differential equation can also be solved for

Here, the general solution is expressed in terms of cylindrical functions.

For all other values ​​of the exponent n,   the solution to the Riccati equation can be expressed in terms of integrals of elementary functions. This fact was established by the French mathematician Joseph Liouville (1809-1882) in 1841.

Many other special cases of the Riccati equation are presented on the EqWorld website.

Solve the differential equation y’ = y + y ² + 1.

This equation is the simplest Riccati equation with constant coefficients. The variables x and y are easily separated here, so the general solution of the equation is determined as follows:

      

Solve the Riccati equation .

We will look for a particular solution in the form:

      

Substituting this into the equation, we find:

      

We obtain a quadratic equation for c :

      

We can choose any value of c. For example, let c = 2. Now that the particular solution is known, let’s make the substitution:

      

Let’s substitute this back into the original Riccati equation:

      

As you can see, we have obtained the Bernoulli equation with parameter m = 2. Let’s make one more substitution:

      

Let us divide the Bernoulli equation by z ² (assuming that z ≠ 0 ) and write it in terms of the variable v :

      

The last equation is linear and can be easily solved using an integrating factor:

      

The function determines the general solution of a linear equation.

      

Now we will sequentially return to the previous variables. Since z = 1/ v, the general solution for z is written as follows:

      

Hence,

      

You can rename the constant: 3 C = C ₁ and write the answer as

      

where C ₁ is an arbitrary real number.

Find the general solution of the differential equation x ³ y’ + x ² y − y ² = 2 x ⁴.

Let’s bring the equation to standard form:

      

We’re dealing with the Riccati equation. Let’s try to find a particular solution in the form y ₁ = cx ². Substituting this into the differential equation, we can determine the coefficient c :

      

Solving the quadratic equation, we find the value of c :

      

So, we’ve obtained two particular solutions. Since it’s sufficient to know only one, we’ll choose, for example,   y ₁ = x ².

As a result, we can write the general solution to the Riccati equation in the form:

      

For the new function u (x), we obtain the following differential equation:

      

which is the Bernoulli equation. Substitution transforms it into a linear differential equation :

      

To solve this linear equation, we calculate the integrating factor:

      

We can take the function v ( x ) = x as an integrating factor. Indeed, we can verify that after multiplying by v ( x ) = x, the left-hand side of the equation becomes the derivative of the product z ( x ) v ( x ). The general solution of the linear differential equation is:

      

Since z = 1/ u , the function u ( x ) is defined by the formula

      

Therefore, the general solution of the original Riccati equation is expressed by the function.

      

where C is an arbitrary constant.

Solve the equation .

It is clear that this equation is a special case of the Riccati equation of the form y’ = by ² + cx ⁿ with degree n = −2.

By making the substitution y = 1/ z, we can transform this equation into a homogeneous one and then integrate it.
Let . Then

      

To solve the homogeneous equation, we make one more substitution:   z = tx ,   z ‘ = t ‘ x + t . Therefore,

      

The trinomial in the denominator of the fraction on the left side can be expanded as follows:

      

and then the rational fraction in the integrand can be decomposed into a sum of simple fractions using the method of undetermined coefficients:

      

As a result, we get:

      

Let us redesignate the constant: , so that the solution for the function t ( x ) will have the form:

      

Let us remember that . Therefore

      

Returning to the variable y, which is related to z by the relation , we find:

      

The last expression represents the general solution of the given Riccati equation in implicit form. Here, the constant C is any real number. Indeed, substituting C = 0, we see that this value also satisfies the differential equation:

      

Hence,