Problem 105 - Thomas Calculus

Question

Use a CAS to solve the initial problems in Exercises \(103-106 .\) Plot the solution curves. $$ y^{\prime}=\frac{1}{\sqrt{4-x^{2}}}, \quad y(0)=2 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \sin^{-1}\left(\frac{x}{2}\right) + 2 \).

Step 1: Identify the Differential Equation

The given differential equation is \( y' = \frac{1}{\sqrt{4-x^2}} \) with the initial condition \( y(0) = 2 \).

Step 2: Integrate the Differential Equation

To find \( y(x) \), we need to integrate the right-hand side of the equation. This means solving \[ y = \int \frac{1}{\sqrt{4-x^2}} \, dx. \]

Step 3: Find the Antiderivative

The integral \( \int \frac{1}{\sqrt{4-x^2}} \, dx \) is a standard form that can be solved using the arcsine function. The solution is \[ y = \sin^{-1}\left(\frac{x}{2}\right) + C, \] where \( C \) is the constant of integration.

Step 4: Apply the Initial Condition

Use the initial condition \( y(0) = 2 \) to find the constant \( C \). Substitute into the equation: \[ 2 = \sin^{-1}(0) + C. \] Since \( \sin^{-1}(0) = 0 \), it follows that \( C = 2 \).

Step 5: Write the Particular Solution

Substitute the value of \( C \) back into the equation of \( y(x) \). The particular solution of the differential equation is \[ y(x) = \sin^{-1}\left(\frac{x}{2}\right) + 2. \]

Step 6: Plot the Solution Curve

Using a CAS (Computer Algebra System), plot the solution curve given by \( y(x) = \sin^{-1}\left(\frac{x}{2}\right) + 2 \). This shows the behavior of the solution around the initial condition.

Key Concepts

Initial Conditions

Initial conditions are a crucial part of solving differential equations. They are values given to us that relate the independent variable, often time or space, to the dependent variable, typically a function or a curve.

In our exercise, the initial condition is given as \( y(0) = 2 \). This means that when \( x = 0 \), the value of \( y \) is 2.
This helps us find a specific solution to the differential equation, rather than a family of solutions.

Why are initial conditions important?

• They allow us to apply specific constraints on a general solution.
• They help us find the exact curve that fits the given scenario.
• They remove the ambiguity of the constant of integration.

Applying initial conditions can make a world of difference in problems involving physical phenomena or real-world data.

Antiderivative

Antiderivatives, also known as indefinite integrals, are a fundamental concept in calculus. They essentially reverse the process of differentiation.

When you're given a derivative and asked to find the original function, you're looking for the antiderivative.
In the context of our exercise, we needed to find the antiderivative of \( \frac{1}{\sqrt{4-x^2}} \).

How does it work?

• The antiderivative encompasses a range of functions, each differing by a constant \( C \).
• It helps us express the solution of a differential equation in terms of the original variables.
• The process typically involves recognizing standard integral forms or using substitution techniques.

For this exercise, recognizing that the antiderivative could involve an arcsine function was key.

Arcsine Function

The arcsine function, denoted \( \sin^{-1}(x) \), is the inverse of the sine function. It is used when we need to find an angle whose sine is a given number.

In integration, the arcsine function often appears when integrating expressions like \( \frac{1}{\sqrt{1-x^2}} \).

Here's a breakdown of its role in our exercise:

• Our integral was in the form \( \frac{1}{\sqrt{4-x^2}} \), which resembles \( \frac{1}{\sqrt{a^2 - x^2}} \).
• By factoring out constants, this form can integrate directly to an arcsine function.
• Using this, we concluded that \( y = \sin^{-1}\left(\frac{x}{2}\right) + C \).

Understanding the arcsine function helps us solve integrals that would otherwise be difficult to compute using basic techniques.

Particular Solution

A particular solution is a single, specific solution to a differential equation that satisfies the given initial conditions. This is different from the general solution, which includes arbitrary constants.

In the exercise, our goal was to find a particular solution for the equation using the initial condition \( y(0) = 2 \).

Steps to find a particular solution:

• First, solve the differential equation to find the general solution.
• Next, substitute the initial condition into the general solution to find the constant \( C \).
• Rewrite the solution with the new value of \( C \).

By doing these steps, we derived that the particular solution for our equation is \( y(x) = \sin^{-1}\left(\frac{x}{2}\right) + 2 \). The particular solution offers a unique curve that accurately describes the problem's constraints and behavior, especially around the initial condition.