Problem 41 - Essential Calculus

Question

Use the differential equation and the specified initial condition to find \(y .\) $$ \begin{array}{l} \frac{d y}{d x}=\frac{1}{\sqrt{4-x^{2}}} \\ y(0)=\pi \end{array} $$

Step-by-Step Solution

Answer

The solution to the differential equation is \(y(x) = \sin^{-1}\frac{x}{2} + \pi\).

Step 1: Identify the Function f(x)

The function \(f(x)\) is given by the equation \(\frac{1}{\sqrt{4-x^{2}}}\). This function is joined to \(\frac{dy}{dx}\) in the differential equation \(\frac{dy}{dx}=\frac{1}{\sqrt{4-x^{2}}}\).

Step 2: Perform Antiderivation

The antiderivative of \(f(x)\) is necessary to find the function \(y(x)\). For our function \(\frac{1}{\sqrt{4-x^{2}}}\), the antiderivative is \(\sin^{-1}\frac{x}{2} + C\). The '+ C' denotes the constant of integration.

Step 3: Apply Initial Condition

We're given the initial condition \(y(0) = \pi\). Plugging in these values into \(y = \sin^{-1}\frac{x}{2} + C\) gives: \(\pi = \sin^{-1}\frac{0}{2} + C\). Solving this equation for \(C\) yields \(C = \pi\).

Step 4: Formulate Solution

Putting everything together, we find that the solution to the differential equation is \(y(x) = \sin^{-1}\frac{x}{2} + \pi\).

Key Concepts

Antiderivatives

An antiderivative is a function whose derivative is the original function we started with. You can think of it as working backwards from a derivative to find the original function. When dealing with differential equations, finding the antiderivative helps us determine the solution to the equation.

For instance, in our problem, we have the function \( f(x) = \frac{1}{\sqrt{4-x^{2}}} \) which represents \( \frac{dy}{dx} \). To solve this, we need to find a function \( y(x) \) such that when differentiated, it gives back \( f(x) \). This is where the concept of antiderivatives comes in handy.

• Antiderivatives help solve differential equations.
• The process involves finding a function that differentiates to the given derivative.

In our example, the antiderivative of \( \frac{1}{\sqrt{4-x^{2}}} \) is \( \sin^{-1}\left(\frac{x}{2}\right) + C \). "\( C \)" is known as the constant of integration and represents an infinite number of possible solutions, each differing by a constant.

Initial Conditions

Initial conditions are specific values assigned to an equation, helping us find a unique solution among many possibilities. Without initial conditions, solving a differential equation only provides a general solution due to the constant \( C \) picked up during antiderivation.

In our exercise, the initial condition given is \( y(0) = \pi \). This means that at \( x = 0 \), \( y \) should equal \( \pi \). Using this condition, we substitute into our general solution \( y = \sin^{-1}\left(\frac{x}{2}\right) + C \).

• This helps us find \( C \).
• It changes the general solution into a specific one.

Therefore, substituting into the initial condition equation: \[ y = \sin^{-1}\left(\frac{0}{2}\right) + C = \pi \] gives us \( C = \pi \).
Finally, we have determined a particular solution: \( y(x) = \sin^{-1}\left(\frac{x}{2}\right) + \pi \).

Integration

Integration is a fundamental concept in calculus used to find antiderivatives. While closely related to differentiation, it essentially reverses the process of finding derivatives. Solving a differential equation involves integration, which is why it's a crucial step in our exercise.

In the equation \( \frac{dy}{dx} = \frac{1}{\sqrt{4-x^{2}}} \), integration tells us to find the function \( y(x) \) that, when we differentiate it, will result in \( \frac{1}{\sqrt{4-x^{2}}} \).

• Integration provides the antiderivative.
• It reverts a derivative to its original form.

The antiderivation step we executed is technically an integration process, where we focused on the specific interval or conditions represented by the problem. This leads us to find \( y(x) = \sin^{-1}\left(\frac{x}{2}\right) + C \) as the antiderived function.
Moreover, integration can yield many possible antiderivatives, clarified to one exact function with initial conditions, turning indefinite integration into a precise solution.