An equation of the form \(y^{\prime}=f(x) y^{2}+g(x) y+h(x)\) is called a Riccati equation. If we know one particular solution \(y_{p},\) then the substitution \(y=y_{p}+\frac{1}{2}\) gives a linear first-order equation for \(z\). We can solve this for \(z\) and substitute back to find a solution of the \(y\) equation containing one arbitrary constant (see Problem 26 ). Following this method, check the given \(y_{p},\) and then solve (a) \(\quad y^{\prime}=x y^{2}-\frac{2}{x} y-\frac{1}{x^{3}}, \quad y_{p}=\frac{1}{x^{2}}\) (b) \(\quad y^{\prime}=\frac{2}{x} y^{2}+\frac{1}{x} y-2 x, \quad y_{p}=x\) (c) \(\quad y^{\prime}=e^{-x} y^{2}+y-e^{x}, \quad y_{p}=e^{x}\)

Differential equations are equations that involve a function and its derivatives. They are essential in modeling real-world systems in fields like physics, engineering, and economics. For example, suppose we have a function y that depends on x. A differential equation can express how the rate of change of y (its derivative) is related to y itself and x. In this context, a Riccati equation is a specific type of first-order differential equation that looks like this: \[ y' = f(x)y^2 + g(x)y + h(x) \]. The apostrophe (') means the derivative with respect to x. Understanding and solving differential equations allows us to predict behaviors and understand the relationships between variables.

A particular solution is a specific solution to a differential equation that satisfies both the equation and an initial condition or specific requirement. In solving Riccati equations, knowing a particular solution, denoted as \( y_{p} \), is very useful. This solution helps transform the nonlinear Riccati equation into a linear first-order equation that is easier to solve. For example, given a particular solution \( y_{p} \) for the Riccati equation, we can use the substitution \( y = y_{p} + \frac{1}{z} \). This converts the problem into a simpler form, making it easier to find a general solution.

First-order linear equations are differential equations of the form \[ y' + P(x)y = Q(x) \]. These equations are easier to solve compared to nonlinear equations like the Riccati equation. By using the substitution method with a particular solution \( y_{p} \), we convert the complex Riccati equation into this simpler linear form. Once converted, we can solve it using standard techniques for linear equations. The resulting function \( z \) can then be used to find the general solution for the original equation by substituting back into \( y = y_{p} + \frac{1}{z} \). This process highlights the power and utility of first-order linear equations in tackling more complex differential equations.

The substitution method is a common technique used to simplify differential equations. For Riccati equations, we start with a known particular solution \( y_{p} \). We then use the substitution \( y = y_{p} + \frac{1}{z} \). This transforms the original nonlinear equation into a linear first-order equation in terms of \( z \).

Here’s a step-by-step outline of how it works:

• Identify the Riccati equation and particular solution.
• Substitute \( y = y_{p} + \frac{1}{z} \) into the Riccati equation.
• Simplify the resulting equation to express it in terms of \( z \) and \( z' \).
• Solve the new linear equation for \( z \).
• Finally, substitute back to find the general solution \( y = y_{p} + \frac{1}{z} \).

This method not only simplifies the problem but also helps in finding solutions that might not be easily evident from the original form.