Question

Solve the initial-value problems. The equation $$ \frac{d y}{d x}=A(x) y^{2}+B(x) y+C(x) $$ is called Riccati's equation. (a) Show that if \(A(x)=0\) for all \(x\), then Equation (A) is a linear equation, whereas if \(C(x)=0\) for all \(x\), then Equation (A) is a Bernoulli equation. (b) Show that if \(f\) is any solution of Equation (A), then the transformation $$ y=f+\frac{1}{v} $$ reduces (A) to a linear equation in \(v\). In each of Exercises \(39-41\), use the result of Exercise 38 (b) and the given solution to find a one-parameter family of solutions of the given Riccati

Short answer

In summary, we showed that if \(A(x) = 0\), Riccati's equation becomes a linear differential equation, and if \(C(x) = 0\), it becomes a Bernoulli equation. Moreover, we demonstrated that the transformation \(y = f + \frac{1}{v}\) reduces Riccati's equation to a linear equation in v.

Step-by-step solution

Part a: Show A(x) = 0 makes Equation(A) linear equation

If we have A(x) = 0 for all x, then the given Riccati equation becomes: \[ \frac{dy}{dx} = B(x)y + C(x) \] This equation is a first-order linear differential equation since there's no term with a power of y greater than 1.

Part a: Show C(x) = 0 makes Equation(A) a Bernoulli equation

If we have C(x) = 0 for all x, then the given Riccati equation becomes: \[ \frac{dy}{dx} = A(x)y^2 + B(x)y \] This equation is a Bernoulli equation which takes the general form: \[ \frac{dy}{dx} + P(x)y = Q(x)y^n, \quad n \neq 0,1 \] In this case, we have \(P(x) = B(x)\), \(Q(x) = A(x)\), and \(n = 2\).

Part b: Show the given transformation reduces Equation (A) to a linear equation

Now let's consider the transformation: \[ y = f + \frac{1}{v} \] Taking the derivative of y with respect to x: \[ \frac{dy}{dx} = \frac{dv}{dx} \left(-\frac{1}{v^2}\right) \] Substituting y and dy/dx into the Riccati equation, we get: \[ \frac{dv}{dx} \left(-\frac{1}{v^2}\right) = A(x)\left(f + \frac{1}{v}\right)^2 + B(x)\left(f + \frac{1}{v}\right) + C(x) \] Now, since f is a solution of the Riccati equation, it satisfies the following condition: \[ \frac{df}{dx} = A(x)f^2 + B(x)f + C(x) \] Subtracting the equation for f from the equation with the substitution, we obtain: \[ \frac{dv}{dx}\left(-\frac{1}{v^2}\right) - \frac{df}{dx} = A(x)\left(\frac{1}{v}\right)^2 \] This is a linear equation in v, as it takes the general form: \[ \frac{dv}{dx} + P(x)v = Q(x) \] So, the transformation turns Riccati's equation into a linear equation in v.

Key concepts

Initial-Value Problems

An initial-value problem in the context of differential equations is a scenario where you are not only tasked with finding the general solution to the equation but also with identifying a specific solution that meets initial conditions provided. These conditions typically specify the value of the unknown function and, possibly, its derivatives at a certain point. For instance, a problem might stipulate that y(0) = 1, which means you're looking for a solution curve that passes through the point (0,1).

When solving initial-value problems involving Riccati's equation, one must integrate the transformed linear equation and then apply the initial conditions to find the particular solution. This often requires a bit of skill with integration techniques and sometimes familiarity with specialized functions, depending on the coefficients A(x), B(x), and C(x).

It's important to correctly apply the initial conditions only after the general solution is found, as premature application can lead to incorrect or incomplete solutions.

Linear Differential Equation

A linear differential equation is an equation that involves an unknown function and its derivatives, which enters the equation linearly; this means that the unknown function and its derivatives are each raised to the first power and are not multiplied by each other. The general form of a first-order linear differential equation is:

\[\begin{equation}\frac{dy}{dx} + P(x)y = Q(x)\end{equation}\]

In the context of Riccati's equation, when A(x) = 0, we are left with a first-order linear differential equation where P(x) is represented by the coefficient B(x) and Q(x) by C(x). Such equations are generally easier to solve than nonlinear equations because they often yield to simple integration techniques or straightforward application of integrating factors.

Bernoulli Equation

The Bernoulli equation is a specific type of nonlinear differential equation that has the standard form:

\[\begin{equation}\frac{dy}{dx} + P(x)y = Q(x)y^n\end{equation}\]

where n is a real number that is not equal to 0 or 1. This particular form allows for a certain transformation involving a power of y, which turns the Bernoulli equation into a linear differential equation, enabling us to use linear techniques to solve an originally nonlinear problem. In the case where C(x) = 0 in Riccati’s equation, we recognize it as a Bernoulli equation with n = 2. Solving a Bernoulli equation usually involves dividing through by y^n, which in this case would mean dividing by y^2. Then, one introduces a new variable, often v, that is a suitable function of y, aiming to make the equation linear in terms of v.

This technique ultimately enables the equation to be tackled by the same methods as linear differential equations, which is a significant simplification.