Problem 109 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the differential equation and initial condition

We have the differential equation \( y' = \frac{1}{\sqrt{4-x^2}} \) with the initial condition \( y(0) = 2 \). This tells us we need to find a function \( y(x) \) such that its derivative matches the given function, and it passes through the point (0, 2).

Step 2: Integrate the differential equation

To solve for \( y(x) \), we integrate both sides of the equation. The integral becomes \( \int dy = \int \frac{1}{\sqrt{4-x^2}} \, dx \). The right-hand side is a known integral that results in \( \sin^{-1}(\frac{x}{2}) \). Hence, \( y = \sin^{-1}(\frac{x}{2}) + C \), where \( C \) is the constant of integration.

Step 3: Use the initial condition to solve for the constant

Apply the initial condition \( y(0) = 2 \) to find \( C \). Substitute \( x = 0 \) and \( y = 2 \) into the equation: \( 2 = \sin^{-1}( \frac{0}{2} ) + C \). Since \( \sin^{-1}(0) = 0 \), we find that \( C = 2 \).

Step 4: Write the particular solution

Incorporate the value of \( C \) back into the equation to obtain the particular solution \( y = \sin^{-1}\left(\frac{x}{2}\right) + 2 \). This is the function that satisfies both the differential equation and the initial condition.

Step 5: Plot the solution curve

Plot the function \( y = \sin^{-1}\left(\frac{x}{2}\right) + 2 \) over the relevant domain of \( x \). Ensure the plot clearly shows the initial point \( (0, 2) \) and displays the behavior of the function in relation to the differential equation. Use a computer algebra system (CAS) or graphing software to visualize the curve.

Key Concepts

Differential Equations

Differential equations are equations that involve unknown functions and their derivatives. These equations are fundamental in expressing a range of physical laws and phenomena. In our problem, we have a differential equation given by \( y' = \frac{1}{\sqrt{4-x^2}} \). This notation \( y' \) represents the derivative of \( y \), indicating how \( y \) changes with respect to \( x \). The right-hand side of the equation \( \frac{1}{\sqrt{4-x^2}} \) specifies the rate of change as a function of \( x \).
This type of equation is called a first-order differential equation because it involves the first derivative. To solve it, we seek a function \( y(x) \) whose derivative matches this given function. These equations are vital because they model real-world systems where quantities change continually over time or space, such as motion, growth, decay, and many more.~Understanding how to solve them allows us to predict and describe these dynamic processes.

Integration Techniques

Integration techniques are essential tools for solving differential equations. In our exercise, integration is used to find the function \( y(x) \) that satisfies the specified differential equation. To do this, we integrate both sides of the equation:
\[ \int dy = \int \frac{1}{\sqrt{4-x^2}} \, dx \]
Integrating the left side with respect to \( y \) is direct, resulting in \( y \). The challenge lies on the right side, where we recognize the integral as a well-known form. This particular integral evaluates to \( \sin^{-1}(\frac{x}{2}) \).
Upon integrating, the general solution becomes \( y = \sin^{-1}(\frac{x}{2}) + C \), where \( C \) is a constant of integration. This constant represents all the solutions to the differential equation before considering initial conditions. Mastery of different integration techniques is crucial for efficiently solving a variety of differential equations.

Solution Curves

Solution curves are graphical representations of the solutions to differential equations. They help visualize how the solutions behave over a given domain. In this exercise, the particular solution found was \( y = \sin^{-1}(\frac{x}{2}) + 2 \). To plot this solution curve, you would graph \( y \) as a function of \( x \).
The curve shows how \( y \) changes as \( x \) varies, and it's crucial in understanding the behavior of the function across different values. With the initial condition \( y(0) = 2 \), the graph will pass through the point \( (0, 2) \).

• The curve's shape is dictated by the expression \( \sin^{-1}(\frac{x}{2}) \), which maps each \( x \) within its domain to a corresponding \( y \).
• Visualizing the solution helps in interpreting the differential equation's implications, providing a concrete view of abstract results obtained mathematically.

Using tools such as CAS (Computer Algebra Systems), these curves can be plotted quickly, aiding both in understanding and in practical problem-solving.

Initial Conditions

Initial conditions are values that specify the state of a system at the start of an observation. They are essential for determining specific solutions to differential equations known as particular solutions. In this exercise, we were given the initial condition \( y(0) = 2 \). This condition tells us that when \( x = 0 \), \( y \) must equal 2.
By applying the initial condition to the general solution \( y = \sin^{-1}(\frac{x}{2}) + C \), we set \( x = 0 \) and \( y = 2 \) yielding:

• \( 2 = \sin^{-1}(\frac{0}{2}) + C \)
• \( 2 = 0 + C \)
• Thus, \( C = 2 \)

This calculation pinpoints the exact \( C \) that satisfies both the differential equation and the initial conditions. Without initial conditions, we would have only a general solution encompassing a family of curves. With them, we specify a unique curve that exactly fits the initial scenario described.