Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d y}{d x}=x^{-3}+2 ; y=3\) at \(x=1\)

Step-by-Step Solution

Answer

The particular solution is \(y = -\frac{1}{2} x^{-2} + 2x + \frac{3}{2}\).

Step 1: Identify the Differential Equation

The differential equation given is \(\frac{d y}{d x}=x^{-3}+2\). Our task is to find the general solution first, which involves integrating each term.

Step 2: Integrate the Differential Equation

Integrate both sides of the equation. For the left-hand side, it becomes \(\int d y = \int (x^{-3} + 2) \, d x\). This results in \(y = \int x^{-3} \, d x + \int 2 \, d x\).

Step 3: Integrate Each Term Separately

First, integrate \(x^{-3}\), which yields \(-\frac{1}{2} x^{-2}\) and integrate \(2\) to obtain \(2x\). Therefore, the integrated left-hand side becomes \(y = -\frac{1}{2} x^{-2} + 2x + C\), where \(C\) is the constant of integration.

Step 4: Write the General Solution

The complete general solution, incorporating the constant \(C\), is \(y = -\frac{1}{2} x^{-2} + 2x + C\).

Step 5: Apply Initial Condition to Find Particular Solution

Use the condition that \(y=3\) when \(x=1\). Substitute \(x=1\) and \(y=3\) into the general solution: \(3 = -\frac{1}{2} (1)^{-2} + 2(1) + C\).

Step 6: Solve for Constant C

Calculate: \(3 = -\frac{1}{2} \times 1 + 2 + C\). This simplifies to \(3 = -\frac{1}{2} + 2 + C\). Thus, \(3 = \frac{3}{2} + C\). Solve for \(C\) to obtain \(C = 3 - \frac{3}{2} = \frac{3}{2}\).

Step 7: Write the Particular Solution

Substitute \(C = \frac{3}{2}\) back into the general solution: \(y = -\frac{1}{2} x^{-2} + 2x + \frac{3}{2}\). This represents the particular solution satisfying the initial condition.

Key Concepts

Integration

In solving differential equations, integration plays a pivotal role. When faced with a differential equation like \( \frac{d y}{d x}=x^{-3}+2 \), integration helps us find the antiderivative or the original function. The goal is to reverse the process of differentiation, converting the derivative back into its function.

• **Integrating Individual Terms:** Start by integrating each term separately. Terms like \( x^{-3} \) integrate to \( -\frac{1}{2} x^{-2} \), whereas constants such as \( 2 \) integrate to \( 2x \).
• **Combining Results:** Once each term is integrated, we combine them to form the solution. Also remember to introduce the constant of integration \( C \), which represents any constant that could be added to the antiderivative, since differentiation of a constant results in zero.

Understanding integration is key to unlocking a differential equation's general form, serving as the foundation for further applications, like finding particular solutions.

Initial Conditions

After obtaining the general solution of a differential equation, initial conditions are used to find a specific, particular solution that fits a given situation. In our exercise, we were given that \( y = 3 \) when \( x = 1 \).

• **Purpose of Initial Conditions:** These conditions help identify a unique solution from the infinite possibilities provided by the general solution. They specify where the solution should pass through, allowing us to calculate the constant \( C \).
• **Using Initial Conditions:** Substitute the provided values into the general solution. This substitution validates the conditions and is used to solve for \( C \).

By effectively applying the initial conditions, we ensure that the solution not only solves the equation but also fits the required scenario.

General Solution

The general solution of a differential equation encompasses all possible solutions. After integrating, we get this form that typically includes an arbitrary constant \( C \).

• **Form of General Solutions:** For our given equation, the general solution looks like: \( y = -\frac{1}{2} x^{-2} + 2x + C \). Here, \( C \) signifies any constant value that could be present, due to indefinite integration.
• **Wide Applicability:** Because it includes \( C \), the general solution is flexible. It represents a family of curves, each corresponding to a different value of \( C \).

The general solution is a broad expression, offering insight into the behavior of the differential equation under a wide range of conditions.

Particular Solution

A particular solution is derived from a general solution by applying specific initial conditions, yielding a precise solution that fits a given scenario.

• **Deriving Particular Solutions:** From the general solution \( y = -\frac{1}{2} x^{-2} + 2x + C \), and using the initial condition \( y=3 \) when \( x=1 \), we determine the exact value of \( C \). In this case, \( C = \frac{3}{2} \).
• **Form of the Particular Solution:** With \( C \) known, we substitute back into the general solution to get the particular solution: \( y = -\frac{1}{2} x^{-2} + 2x + \frac{3}{2} \).

This specific solution precisely addresses the problem's conditions, showing how the function behaves within the provided constraints. It pinpoints exactly one curve from the family's possibilities in the general solution.