Problem 7 - Calculus: Early Transcendentals

Question

(a) What can you say about a solution of the equation \( y{'} = - y^{2} \) just by looking at the differential equation? (b) Verify that all members of the family \( y = 1/(x + C) \) are solutions of the equation in part (a). (c) Can you think of a solution of the differential equation \( y^{'} = - y^{2} \) that is not a member of the family in part (b)? (d) Find a solution of the initial-value problem. \( y^{'} = - y^2 \) \( y (0) = 0.5 \)

Step-by-Step Solution

Answer

The solution decreases, and \( y = \frac{1}{x+2} \) solves the initial-value problem.

Step 1: Analyzing the Differential Equation

The given differential equation is \( y' = -y^2 \). Observing the equation, we can say that the rate of change of \( y \) is negative, implying that \( y \) decreases as it grows larger, because \( y^2 \) is always non-negative. Therefore, every solution \( y(x) \) is decreasing except possibly at critical points where \( y = 0 \).

Step 2: Verifying the Family of Solutions

We are given the family of functions \( y = \frac{1}{x + C} \). We need to verify if each function in this family satisfies the differential equation. First, differentiate the function: \( y' = -\frac{1}{(x + C)^2} \). Substitute \( y = \frac{1}{x + C} \) and \( y' = -\frac{1}{(x + C)^2} \) into the equation \( y' = -y^2 \): \(-\frac{1}{(x + C)^2} = - \left(\frac{1}{x + C}\right)^2\). Both sides match, hence the family \( y = \frac{1}{x + C} \) satisfies the differential equation.

Step 3: Finding Non-family Solutions

Consider a constant solution \( y = 0 \). Then \( y' = 0 \) and \( -y^2 = 0 \), which satisfies the differential equation. However, \( y = 0 \) cannot be expressed in the form \( y = \frac{1}{x + C} \) for any real constant \( C \), indicating that \( y = 0 \) is a solution not included in the given family.

Step 4: Solving the Initial-value Problem

To solve \( y' = -y^2 \) with initial condition \( y(0) = 0.5 \), start with the known general solution \( y = \frac{1}{x + C} \). Using the initial condition \( y(0) = 0.5 \), substitute into the solution: \( 0.5 = \frac{1}{0 + C} \), so \( C = 2 \). Thus, the specific solution satisfying the initial condition is \( y = \frac{1}{x + 2} \).

Key Concepts

Initial-Value Problem

When dealing with differential equations, an initial-value problem is a type of problem where, not only is the differential equation provided, but also the value of the variable at a specific point. This specific value is known as the initial condition. For example, in our exercise, the problem states that we need to solve the differential equation \( y' = -y^2 \) with the initial condition \( y(0) = 0.5 \).

• The initial condition \( y(0) = 0.5 \) tells us the specific state of the function \( y \) when \( x = 0 \).
• This information helps in finding a particular solution from a general family of solutions, effectively "anchoring" the equation to a point and directing its trajectory.

By incorporating the initial condition, we can determine a unique solution that connects seamlessly to the condition specified. This method is crucial in practical applications, where the context usually provides initial values.

Solutions of Differential Equations

Solutions to differential equations provide a function or set of functions that satisfy the given equation. For the equation \( y' = -y^2 \), the step-by-step solution explores different possible solutions, including both general and particular forms.

• The given family of solutions for our problem is \( y = \frac{1}{x+C} \), where \( C \) is a constant.
• This family encompasses a wide range of solutions by varying the constant \( C \), providing flexibility to match different conditions.

The verification of this family involves substituting both \( y \) and its derivative \( y' \) back into the original differential equation to confirm they match. Such verification ensures the mathematical integrity of the solution and confirms that \( y = \frac{1}{x + C} \) appropriately satisfies the differential equation under various scenarios.

Rate of Change

In the context of differential equations, the rate of change refers to the derivative of a function, which indicates how the function's value changes as the input variable changes. The differential equation \( y' = -y^2 \) directly describes the rate of change of \( y \) with respect to \( x \).

• Here, \( y' = -y^2 \) tells us the change in \( y \) decreases as \( y \) becomes larger.
• The negative sign implies that \( y \) is decreasing overall, meaning as \( y \) grows, it will tend to reduce or slow its growth due to the squaring.

Recognizing the rate of change helps in predicting the behavior of solutions over time. In many real-world scenarios, understanding this change is key to forecasting future states based on current data.

Family of Solutions

A family of solutions in differential equations refers to a set of solutions that satisfy the equation under varied conditions, specifically by changing a parameter such as \( C \) in this case. For the equation \( y' = -y^2 \), the family is given as \( y = \frac{1}{x + C} \).

• The constant \( C \) can be varied to obtain different curves or solutions, adjusting for different initial conditions or requirements.
• To verify the family, you substitute \( y \) and the derivative \( y' \) back into the differential equation and check its validity over the range of possible \( C \) values.

This concept shows that one differential equation can often produce an entire set of solutions, each fitting different specific conditions or parameters. It's like having a blueprint that adapts to form various structures depending on the context.