Problem 52 - Calculus and its applications

Question

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is directly proportional to the quotient of \(x\) divided by \(y\), and \(y=2\) when \(x=-1\).

Step-by-Step Solution

Answer

The differential equation is \( \frac{dy}{dx} = \frac{x}{y} \) and the particular solution is \( y = \sqrt{x^2 + 3} \).

Step 1: Formulating the Differential Equation

The problem states that the rate of change of \( y \) with respect to \( x \), \( \frac{dy}{dx} \), is directly proportional to the quotient of \( x \) divided by \( y \). This means we can write the equation as \( \frac{dy}{dx} = k \frac{x}{y} \). With \( k = 1 \), the differential equation becomes \( \frac{dy}{dx} = \frac{x}{y} \).

Step 2: Separating Variables

To solve the differential equation, first separate the variables. Multiply both sides by \( y \) to get \( y \frac{dy}{dx} = x \). Now, multiply both sides by \( dx \) which separates the variables: \( y \, dy = x \, dx \).

Step 3: Integrating Both Sides

Integrate both sides of the equation from Step 2. The integral of \( y \, dy \) is \( \frac{y^2}{2} + C_1 \) and the integral of \( x \, dx \) is \( \frac{x^2}{2} + C_2 \). We can combine the constants since they will cancel out in the process of finding \( C \): \[ \frac{y^2}{2} = \frac{x^2}{2} + C \].

Step 4: Solving for the General Solution

To find the general solution, solve the equation from Step 3 for \( y^2 \): \[ y^2 = x^2 + 2C \]. Taking the square root of both sides, we find that \[ y = \pm \sqrt{x^2 + 2C} \].

Step 5: Applying the Initial Condition

We know that \( y = 2 \) when \( x = -1 \). Substitute these values into the general solution to find \( C \): \[ 2 = \pm \sqrt{(-1)^2 + 2C} \]. This simplifies to \[ 2 = \pm \sqrt{1 + 2C} \]. Choosing the positive square root, we solve \( 2 = \sqrt{1 + 2C} \) for \( C \). Squaring both sides gives \( 4 = 1 + 2C \), so \( 2C = 3 \) and \( C = \frac{3}{2} \).

Step 6: Finding the Particular Solution

Substitute the value of \( C \) back into the general solution to find the particular solution: \[ y = \sqrt{x^2 + 3} \]. This satisfies the initial condition where \( y = 2 \) when \( x = -1 \).

Key Concepts

Initial Conditions

In the realm of differential equations, initial conditions are essential for determining a particular solution from a general solution. An initial condition provides a specific value for the dependent variable when the independent variable is specified. This specification allows us to zero in on a single solution, rather than a family of solutions that the differential equation might suggest without constraints.

In the context of the given problem, we have the initial condition that when \( x = -1 \), \( y = 2 \). This piece of information is crucial because without it, we're left with the general solution, \( y = \pm \sqrt{x^2 + 2C} \), which represents a range of possible curves.

By substituting the initial condition into the general solution, we can calculate the constant \( C \). The process involves plugging these specific values into the equation and solving for \( C \). In this way, the initial condition helps us pinpoint a unique curve from the infinite possibilities, making the solution applicable to the specific scenario presented in the problem.

Rate of Change

The concept of the rate of change is at the heart of differential equations. It describes how a quantity changes in relation to another quantity. Mathematically, the rate of change is expressed as a derivative, often denoted as \( \frac{dy}{dx} \), which is the rate at which \( y \) changes with respect to \( x \).

In this problem, the rate of change of \( y \) with respect to \( x \) is directly proportional to the quotient \( \frac{x}{y} \). This relationship is initially expressed as \( \frac{dy}{dx} = k \frac{x}{y} \), where \( k \) is the constant of proportionality. Here, the problem simplifies it by setting \( k = 1 \), leading to the differential equation \( \frac{dy}{dx} = \frac{x}{y} \).

This particular relationship means that as \( x \) increases or decreases, the change in \( y \) is influenced by the ratio of \( x \) to \( y \). Understanding the rate of change is crucial because it marks the starting point for deriving both general and particular solutions.

General Solution

The general solution of a differential equation is essentially a formula that encompasses a family of functions. These functions represent all potential solutions to the differential equation without any constraining initial conditions. In this scenario, to find the general solution, we separated variables and integrated both sides to solve for \( y \).

Initially, the differential equation was rearranged to allow integration: \( y \, dy = x \, dx \). After integrating, we reached \( \frac{y^2}{2} = \frac{x^2}{2} + C \), which simplifies to \( y^2 = x^2 + 2C \).

From this point, we took the square root to find \( y = \pm \sqrt{x^2 + 2C} \). This general solution represents numerous curves, all dictated by different values of the constant \( C \).

• The \( \pm \) indicates that \( y \) can be positive or negative, reflecting two branches of solutions.
• The constant \( C \) allows flexibility, tailoring the solution to meet specific conditions if provided.

Without an initial condition, the general solution remains flexible, offering a comprehensive picture of all potential solutions to the equation.