Problem 71 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=3 x^{-2 / 3}, \quad y(-1)=-5 $$

Step-by-Step Solution

Answer

The particular solution is \( y = 9x^{1/3} + 4 \).

Step 1: Integrate the differential equation

We start by integrating the given differential equation \( \frac{dy}{dx} = 3x^{-2/3} \). The integral of \( 3x^{-2/3} \) with respect to \( x \) is calculated as follows: \( \int 3x^{-2/3} \, dx = 3 \int x^{-2/3} \, dx = 3 \cdot \frac{x^{1/3}}{1/3} = 9x^{1/3} + C \), where \( C \) is the constant of integration.

Step 2: Apply the initial condition

We use the initial condition \( y(-1) = -5 \) to find the constant \( C \).Substitute \( x = -1 \) and \( y = -5 \) into the equation: \( -5 = 9(-1)^{1/3} + C \).We calculate \( (-1)^{1/3} = -1 \), thus:\( -5 = 9(-1) + C \).\( -5 = -9 + C \).Solving for \( C \) gives \( C = 4 \).

Step 3: State the particular solution

Now that we have found the constant \( C \), we can write the particular solution to the differential equation:\( y = 9x^{1/3} + 4 \). This is the solution that satisfies the given initial condition.

Key Concepts

Understanding Differential Equations

In mathematics, differential equations play a crucial role in describing a wide range of phenomena. They are equations that involve an unknown function and one or more of its derivatives. In the original problem, the differential equation is given as \( \frac{dy}{dx} = 3x^{-2/3} \), which involves the derivative of \( y \) with respect to \( x \).

These equations are used to model various situations where change is continuous, such as:

• Scientific phenomena (e.g., motion, growth)
• Engineering systems (e.g., electrical circuits)
• Economic models (e.g., interest rates)

To solve a differential equation implies finding the unknown function \( y \). This process requires understanding both the derivative involved and the integration process necessary to uncover \( y \). Often, additional information like initial conditions is used to find a unique solution.

The Role of Integration

Integration is a fundamental concept in calculus that helps us find a function when its rate of change (derivative) is known. In the context of differential equations, integration is used to solve for the unknown function.

In the given exercise, we integrate the equation \( \frac{dy}{dx} = 3x^{-2/3} \) to find \( y \). By integrating both sides, we effectively 'undo' the differentiation:

• We calculate \( \int 3x^{-2/3} \, dx \) to get \( 9x^{1/3} + C \).
• Here, \( C \) is introduced as the constant of integration.

This step is vital because it transforms the differential equation into a solvable function of \( y \). Integration allows us to shift from a description of change to a description of the total accumulation or value at any particular point, which is crucial for analyzing functions over an interval.

Importance of the Constant of Integration

When integrating, a constant of integration \( C \) always appears. This constant accounts for all the potential solutions, since multiple functions can have the same derivative. This reflects the fact that integration essentially reverses differentiation up to an unknown constant.

In the step-by-step solution, after integrating \( 3x^{-2/3} \), we include \( C \), resulting in \( 9x^{1/3} + C \).

This constant \( C \) is what shifts the general solution to become specific.

• If initial conditions are given, like \( y(-1) = -5 \), we can solve for \( C \).
• This process ensures that the solution fits perfectly with the initial conditions and the particular circumstances of the problem.

The constant of integration is a reminder of the infinite family of functions that can satisfy any antiderivative.

Applying Initial Conditions

Initial conditions are specific values given within a differential equation problem that help pin down one particular solution among the many. In our differential equation exercise, we have the initial condition \( y(-1) = -5 \). This particular piece of information allows us to solve for \( C \), the constant of integration.

Here's how we use them:

• Substitute \( x = -1 \) and \( y = -5 \) into the integrated function \( y = 9x^{1/3} + C \).
• Solving \( -5 = 9(-1)^{1/3} + C \) gives us a specific value for \( C \), which is \( 4 \).

Thus, the initial conditions transform the general solution into a particular solution that satisfies the given condition. Without these conditions, there's no way to determine the exact functional form necessary to describe the situation in the problem. Initial conditions make the solution applicable to real-world scenarios by anchoring it to a known point.