Problem 38 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { solve each differential equation. } $$ $$ \frac{d y}{d x}=\left(1-y^{2}\right) $$

Step-by-Step Solution

Answer

The solution is \( y = \tanh(x + C) \).

Step 1: Recognize the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = 1 - y^2 \). This is a first-order separable differential equation because it can be rewritten in a form where all the \( y \) terms are on one side and all the \( x \) terms are on the other side.

Step 2: Separate the Variables

Rearrange the equation to separate the variables. Rewrite it as: \[ \frac{dy}{1-y^2} = dx. \] This allows us to integrate both sides separately.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{dy}{1-y^2} = \int dx. \] The integral on the left side, \( \int \frac{dy}{1-y^2} \), can be recognized as the standard integral that results in the inverse hyperbolic tangent function, \( \tanh^{-1}(y) \), while the integral on the right side is simply \( x + C \), where \( C \) is the constant of integration.

Step 4: Apply the Antiderivative

After integrating, we have: \[ \tanh^{-1}(y) = x + C. \] This equation expresses \( y \) implicitly in terms of \( x \).

Step 5: Solve for \( y \)

To solve explicitly for \( y \), use the definition of the inverse hyperbolic tangent function: \[ y = \tanh(x + C). \] This is the explicit solution to the differential equation, subject to any initial conditions that might be given.

Key Concepts

Separable Equations

When tackling differential equations, one useful type is the separable equation. This type allows you to separate the variables, hence its name, to simplify the problem. In the context of a first-order separable differential equation, we aim to position all terms involving the dependent variable (usually denoted as \( y \)) on one side and those involving the independent variable (usually denoted as \( x \)) on the other.

Consider the differential equation we started with: \( \frac{dy}{dx} = 1 - y^2 \). To separate this, we rewrite it as \( \frac{dy}{1-y^2} = dx \). This allows us to isolate \( dy \) and \( dx \), effectively separating the variables involved. This separation is crucial because it sets the stage for integrating both sides of the equation independently.

• Move all \( y \)-related terms to one side
• Ensure \( x \)-related terms are on the opposite side
• Prepare to integrate each side separately

This simple rearrangement simplifies the integration process by allowing each variable to be handled individually, making separable equations a powerful tool in solving differential problems.

Integration Techniques

Once we have a separable differential equation, the next step involves integrating both sides. This is where understanding integration techniques becomes vital. Let's explore how these techniques apply to our expressions.

For the separated equation \( \frac{dy}{1-y^2} = dx \), we need to integrate both sides.

• Left Side: The integral \( \int \frac{dy}{1-y^2} \) can be tricky at first as it may not seem straightforward. However, recognizing the form \( \frac{1}{1-y^2} \) allows us to connect it with the \( \tanh^{-1}(y) \) function, a type of inverse hyperbolic function.
• Right Side: The integral of \( dx \) is relatively simple and results in \( x + C \).

This part of solving the equation relies heavily on being familiar with standard integrals and also recognizing less common functions like inverse hyperbolic functions. Mastery of these integration techniques not only helps in solving differential equations but also enriches your general calculus toolkit.

Inverse Hyperbolic Functions

A fascinating aspect of differential equations is their ability to connect with different mathematical functions, including inverse hyperbolic functions. In this exercise, we dealt with \( \int \frac{dy}{1-y^2} \), which equates to the inverse hyperbolic tangent function, expressed as \( \tanh^{-1}(y) \).

Inverse hyperbolic functions, like \( \tanh^{-1} \), are somewhat similar to the regular inverse trigonometric functions but are derived from hyperbolic functions. They can be handy when integrating particular types of rational functions.

• \( \tanh^{-1}(x) \): The inverse hyperbolic tangent function
• Derived from the hyperbolic tangent function: \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)
• Useful in solving different forms of integrals

Understanding these functions can be challenging, but they open up a wider range of solutions and techniques when dealing with complex integrations. Integrating with these functions provides deeper insights into the structure and solutions of differential equations.