Techniques for Solving Riccati Differential Equations with Worked Examples
Riccati differential equations have the property that, if a particular solution exists, the original nonlinear equation can be transformed into a linear differential equation via a substitution. In this guide on techniques for solving Riccati differential equations, you’ll learn how to recognize Riccati equations, understand the theory behind the solution method, and solve them step by step.
In this article, we will begin by defining Riccati differential equations. Next, we will derive the substitution used to transform a Riccati equation into a Bernoulli equation and ultimately a linear equation. We will then present a step-by-step procedure for solving these equations and work through several detailed examples that illustrate the entire process
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By the end of this guide, you will have a solid understanding of the most common techniques used to solve Riccati differential equations.
What Are Riccati Differential Equations?
Before learning how to solve a Riccati differential equation, we must first learn how to recognize a Riccati differential equation.
Definition
A Riccati differential equation is a differential equation that can be written in the form
$$\frac{dy}{dx} = a(x)y^2 + b(x)y + c(x),$$
where \( a(x) \), \( b(x) \), and \( c(x) \) are functions of \( x \).
Derivation
Before proceeding with the derivation, you might find it helpful to review the article How to Differentiate a Function Step by Step: A Beginner’s Guide.
Consider the Riccati differential equation
$$\frac{dy}{dx} = a(x)y^2 + b(x)y + c(x),$$
where \( a(x) \), \( b(x) \), and \( c(x) \) are functions of \( x \).
Suppose that we already know a particular solution \( y_p(x) \) that satisfies the Riccati equation. Then
$$\frac{dy_p}{dx} = a(x)y_p^2 + b(x)y_p + c(x).$$
Let
$$y = y_p + v.$$
Differentiating both sides with respect to \( x \) gives
$$\frac{dy}{dx} = \frac{dy_p}{dx} + \frac{dv}{dx}.$$
Substituting into the Riccati equation yields
$$\frac{dy_p}{dx} + \frac{dv}{dx} = a(x)(y_p + v)^2 + b(x)(y_p + v) + c(x).$$
Expanding the squared term gives
$$\frac{dy_p}{dx} + \frac{dv}{dx} = a(x)(y_p^2 + 2y_pv + v^2) + b(x)(y_p + v) + c(x).$$
Expanding further gives
$$\frac{dy_p}{dx} + \frac{dv}{dx} = a(x)y_p^2 + 2a(x)y_pv + a(x)v^2 + b(x)y_p + b(x)v + c(x).$$
Recall that \( y_p \) satisfies
$$\frac{dy_p}{dx} = a(x)y_p^2 + b(x)y_p + c(x).$$
Substituting this relationship into the previous equation gives
$$\frac{dy_p}{dx} + \frac{dv}{dx} = \frac{dy_p}{dx} + 2a(x)y_pv + a(x)v^2 + b(x)v.$$
This Implies
$$\frac{dv}{dx} – 2a(x)y_pv – b(x)v = a(x)v^2.$$
Factoring ( -v ) yields
$$\frac{dv}{dx} – (2a(x)y_p – b(x))v = a(x)v^2.$$
This is a Bernoulli equation that can be reduced to a linear differential equation using substitution, as discussed in the article Bernoulli Differential Equations for Beginners: Everything You Need to Know. Once the equation is reduced to a linear differential equation, it can be solved using integrating factors as discussed in the article A Step-by-Step Tutorial on Linear Differential Equations with Examples and Solutions.
This derivation provides a procedure for solving Riccati differential equations when a particular solution is known. In the next section, we will summarize this procedure into a simple, step-by-step method applicable to a wide variety of problems.
Steps for Solving Riccati Differential Equations
Now that we have derived the method for solving Riccati equations, let’s summarize the process into a step-by-step procedure.
Step 1: Write the Differential Equation in Standard Form
Begin by rewriting the differential equation in the form
$$\frac{dy}{dx} = a(x)y^2 + b(x)y + c(x).$$
Step 2: Find a Particular Solution
The Riccati solution method requires a known particular solution \( y_p \). Some good guesses are:
$$C.$$
$$at^n.$$
$$ae^{bt}.$$
Step 3: Apply the Substitution
Once a particular solution has been identified, introduce the substitution
$$y = y_p + v.$$
Step 4: Differentiate the Substitution
Differentiate both sides of the substitution with respect to \( x \) to get
$$\frac{dy}{dx} = \frac{dy_p}{dx} + \frac{dv}{dx}.$$
Step 5: Substitute into the Original Equation
Replace every occurrence of \( y \) and \( \frac{dy}{dx} \) with the expressions obtained from the substitution.
Step 6: Solve the Resulting Bernoulli Differential Equation
After simplifying, the Riccati equation becomes a Bernoulli equation that can be solved using another substitution.
Step 7: Convert Back to the Original Variable
Back substitute.
This solves the original Riccati differential equation.
Step 8: Apply the Initial Condition (If Given)
If an initial condition is provided, substitute it into the general solution to determine the constant C
Worked Out Example
The best way to learn how to solve Riccati differential equations is by working through several examples. In this example, we will follow the procedure developed in the previous section. This example also requires the techniques covered in the articles Basic Integration Problems for Beginners and The Ultimate Step-by-Step Guide to Solving Integrals Using Substitution.
Example 1: Solve \( \frac{dy}{dx} = y^2 – 3y + 2 \).
Solution: The equation is already in standard form. Suppose there exists a particular solution of the form
$$y_p = C.$$
Differentiating gives
$$\frac{y_p}{dx} = 0.$$
Substituting into the differential equation, we obtain
$$0 = C^2- 3C + 2.$$
Factoring this is
$$0 = (C – 2)(C – 1).$$
Hence,
$$C – 2 = 0, C – 1 = 0.$$
Which Implies
$$C = 2, C = 1.$$
Thus \( y_p = 2 \), and \( y_p = 1 \) are particular solutions. Let
$$y = v + 1.$$
Differentiating we get
$$\frac{dy}{dx} = \frac{dv}{dx}.$$
Substituting into the differential equation, we find
$$\frac{dv}{dx} = (v + 1)^2 – 3(v + 1) + 2.$$
Expanding \( (v + 1)^2 \) this is
$$\frac{dv}{dx} = v^2 + 2v + 1 – 3(v + 1) + 2.$$
Distributing gives
$$\frac{dv}{dx} = v^2 + 2v + 1 – 3v – 3 + 2.$$
Simplifying we obtain
$$\frac{dv}{dx} = v^2 – v.$$
Which implies
$$\frac{dv}{dx} + v = v^2.$$
Multiplying each term of the equation by \( v^{-2} \), we obtain
$$v^{-2}\frac{dv}{dx} + v^{-1} = 1.$$
Next we let
$$w = v^{-1}.$$
Using the chain rule to differentiate this, we get
$$\frac{dw}{dx} = -v^{-2}\frac{dv}{dx}.$$
Multiplying both sides by -1 then gives
$$-\frac{dw}{dx} = v^{-2}\frac{dv}{dx}.$$
Substituting into the differential equation, we find
$$-\frac{dw}{dx} + w = 1.$$
Multiplying each term of the equation by \( -1 \), we obtain
$$\frac{dw}{dx} – w = 1.$$
The integrating factor is
$$\mu(x) = e^{-\int dx}.$$
Integrating we get
$$\mu(x) = e^{-x}.$$
Multiplying each term of the equation by the integrating factor, we obtain
$$e^{-x}\frac{dw}{dx} – e^{-x}w = e^{-x}.$$
Rewriting the left-hand side as a product rule, this becomes
$$\frac{d}{dx}(e^{-x}w) = e^{-x}.$$
Integrating both sides, we find
$$\int \frac{d}{dx}(e^{-x}w) dx = \int e^{-x} dx.$$
To evaluate the integral on the right-hand side, let \( u = -x \), then \( du = -dx \) which implies \( -du = dx \). Substituting into the integral then gives
$$\int \frac{d}{dx}(e^{-x}w) dx = –\int e^u du.$$
Integrating both sides, we obtain
$$e^{-x}w = –e^u + C.$$
Therefore,
$$e^{-x}w = –e^{-x} + C.$$
Multiplying each term of the equation by \( e^x \), we obtain
$$e^xe^{-x}w = –e^xe^{-x} + Ce^x.$$
Using exponential properties, we find
$$e^{x – x}w = -e^{x – x} + Ce^x.$$
Simplifying this is equivalent to
$$e^0w = -e^0 + Ce^x.$$
Hence,
$$1w = -1 + Ce^x.$$
Which implies
$$w = –1 + Ce^x.$$
Back substituting gives
$$v^{-1} = –1 + Ce^x.$$
Using exponential properties, this is
$$\frac{1}{v} = –1 + Ce^x.$$
Reciprocating both sides, we find
$$v = \frac{1}{–1 + Ce^x}.$$
Our final solution is then
$$y = \frac{1}{–1 + Ce^x} + 1.$$
This example demonstrates that once a particular solution is known, the Riccati equation can be transformed into a Bernoulli differential equation and ultimately a linear differential equation.
Conclusion
In this guide on Techniques for solving Riccati differential equations, we began by defining Riccati differential equations and discussing their key characteristics. We then derived the solution method, developed a step-by-step procedure for solving these equations, and worked through an example that illustrated the entire process from start to finish.
Further Reading
Homogeneous Differential Equations Explained Step by Step for Beginners – Like Riccati differential equations, homogeneous differential equations are also solved using substitution.
