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Chapter 1: Problem 63

The equation \(d y / d x=A(x) y^{2}+B(x) y+C(x)\) is called a Riccati equation. Suppose that one particular solution \(y_{1}(x)\) of this equation is known. Show that the substitution $$ y=y_{1}+\frac{1}{v} $$ transforms the Riccati equation into the linear equation $$ \frac{d v}{d x}+\left(B+2 A y_{1}\right) v=-A $$

Short Answer

Expert verified
By substituting \( y = y_1 + \frac{1}{v} \), the Riccati equation simplifies to \( \frac{d v}{d x} + (B + 2Ay_1) v = -A \).

Step by step solution

01

Write the original equation

Start with the given Riccati equation: \( \frac{d y}{d x} = A(x) y^2 + B(x) y + C(x) \).
02

Substitute y with y_1 + 1/v

Replace \( y \) with \( y_1 + \frac{1}{v} \) in the differential equation. This gives: \( \frac{d}{d x} \left( y_1 + \frac{1}{v} \right) = A(x) \left(y_1 + \frac{1}{v}\right)^2 + B(x) \left(y_1 + \frac{1}{v}\right) + C(x) \).
03

Differentiate the substitution

Calculate \( \frac{d}{d x} \left( y_1 + \frac{1}{v} \right) \) as \( \frac{d y_1}{d x} - \frac{1}{v^2} \frac{d v}{d x} \).
04

Substitute y_1 into the Riccati equation

Since \( y_1(x) \) is a solution of the Riccati equation, substitute \( y_1 \) to express \( \frac{d y_1}{d x} = A(x) y_1^2 + B(x) y_1 + C(x) \).
05

Expand the squared term

Expand \( \left(y_1 + \frac{1}{v}\right)^2 \) as \( y_1^2 + \frac{2y_1}{v} + \frac{1}{v^2} \).
06

Substitute expanded terms into the equation

Plug in the expanded terms: \( A(x)\left(y_1^2 + \frac{2y_1}{v} + \frac{1}{v^2}\right) + B(x)\left(y_1 + \frac{1}{v}\right) + C(x) \).
07

Simplify the equation

Combine terms to simplify the result: \( A(x) y_1^2 + B(x) y_1 + C(x) + \left(2 A(x) y_1 + B(x) \right)\frac{1}{v} + \frac{A(x)}{v^2}\).
08

Cancel out terms using equality

Since \( y_1(x) \) satisfies the original equation, the terms \( A(x) y_1^2 + B(x) y_1 + C(x) \) will cancel with \( \frac{d y_1}{d x} \), leaving \( \frac{dv}{dx} - \frac{1}{v^2} \frac{dv}{dx} = -A(x) - \left(2A(x) y_1 + B(x) \right)\frac{1}{v} \).
09

Simplify to the linear equation

Rearrange the terms to form: \( \frac{dv}{dx} + \left(B(x) + 2A(x)y_1\right)v = -A(x) \). This shows the transformation to the linear equation is successful.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are equations involving derivatives, which represent rates of change. They are important in mathematics because they help model and solve problems involving dynamic systems.
For instance, a differential equation can describe how a quantity evolves over time based on a given rate of change. In the context of a Riccati equation, the equation involves a derivative of a variable, often capturing complex behaviors in physics, engineering, and economics.
  • In the exercise, the Riccati equation is a particular type of nonlinear differential equation.
  • It generally has the form \( \frac{dy}{dx} = A(x) y^2 + B(x) y + C(x) \).
  • Here, \( y \) is the dependent variable, while \( x \) is the independent variable.
Understanding these equations is crucial for problem-solving in various fields, making them fundamental in the study of dynamics and changes.
Particular Solution
A particular solution to a differential equation is a specific solution that satisfies the equation. When working with differential equations, sometimes we find general solutions involving constants.
A particular solution provides a concrete set of values that fits the differential equation based on given initial conditions.
  • In the context of the Riccati equation, \( y_1(x) \) is given as a particular solution.
  • This means that if you substitute \( y_1(x) \) into the Riccati equation, it will satisfy the equation.
Recognizing a particular solution allows us to simplify equations and perhaps find additional solutions through subsequent transformations.
Substitution Method
The substitution method in differential equations refers to replacing variables with new expressions to simplify the equation. This technique can often transform complicated equations into simpler forms, making them easier to solve.
The given Riccati equation employs substitution by setting \( y = y_1 + \frac{1}{v} \).
  • This substitution transforms the nonlinear Riccati equation into a linear one.
  • By calculating the derivative and plugging it back into the original equation, the expression simplifies.
This approach shows the versatility of substitutions in solving complex problems, revealing hidden structures within the equations.
Linear Equations
Linear equations are equations of the first degree, meaning they involve no products or powers of the unknown variables. In differential equations, linearity simplifies both solving the equation and interpreting the results.
After substitution, the Riccati equation transforms into a linear differential equation: \( \frac{dv}{dx} + \left(B(x) + 2A(x)y_1\right)v = -A(x) \).
  • This linear form is more straightforward because it is easier to analyze and solve using standard techniques.
  • The coefficients of \( v \) are functions of \( x \), which fits the general linear differential equation form.
Linear equations are easier to solve systematically, making them crucial for mathematical analysis and practical applications.

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