Problem 109 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(109-114\) $$ \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}, \quad y(0)=0 $$

Step-by-Step Solution

Answer

The solution is \( y = \arcsin(x) \).

Step 1: Identify the Type of Differential Equation

The equation given is \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \). This is a first-order differential equation, which can be solved by integration since it is separable.

Step 2: Rewrite as an Integral

To solve the differential equation, rewrite it in an integral form: \( dy = \frac{1}{\sqrt{1-x^2}} dx \). This means we need to find \( y \) by integrating both sides with respect to their variables.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( y = \int \frac{1}{\sqrt{1-x^2}} \, dx \). The integral on the right-hand side is the standard form for the inverse sine function, \( \arcsin(x) \). Thus, \( y = \arcsin(x) + C \), where \( C \) is the integration constant.

Step 4: Apply the Initial Condition

Use the given initial condition \( y(0) = 0 \) to find \( C \). Substitute \( x = 0 \) and \( y = 0 \) into the equation \( 0 = \arcsin(0) + C \). Since \( \arcsin(0) = 0 \), we find that \( C = 0 \).

Step 5: Write the Final Solution

Substitute the value of \( C \) back into the equation for \( y \): \( y = \arcsin(x) \). This is the solution to the differential equation with the given initial condition.

Key Concepts

Understanding Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. These equations are used to describe various phenomena, ranging from physics and engineering to biology. They provide a way to model how things change. In our specific example, we dealt with the differential equation \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \). This is a first-order differential equation, meaning that it involves the first derivative of the function \( y(x) \). A differential equation can be seen as a tool to determine an unknown function based on its rate of change. The goal is to find a function \( y \) that satisfies this equation along with an initial condition which is provided to find the unique solution. Here, our initial condition was \( y(0) = 0 \), which helps in finding the particular solution.

Solving Using Separable Equations

Separable equations are a special class of differential equations where the variables can be separated on opposite sides of the equation. In simple terms, it means you can rearrange the equation so that one side only contains terms involving \( y \) and the other side only involves \( x \). For example, with \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \), we separated the variables by rewriting it as \( dy = \frac{1}{\sqrt{1-x^2}} dx \).

Once the equation is in this form, the next step is to integrate each side separately. This allows us to find the antiderivative or the original function that the derivative came from. The integrals often lead to a "constant of integration" \( C \), which is necessary because there are an infinite number of antiderivatives for any given function without additional information provided by initial conditions.

Mastering Integration in Differential Equations

Integration is a core process used to solve separable differential equations. Once the variables are separated, integration is used to determine the functions \( y(x) \) or \( x(y) \). In the integration step of our example, we computed \( y = \int \frac{1}{\sqrt{1-x^2}} \, dx \). This integral is a standard form that results in the inverse sine function, \( \arcsin(x) \).

When solving these types of problems, it's important to recognize common integrals and know how to apply integration techniques. Sometimes, tables of integrals or knowledge of inverse trigonometric functions can be crucial to finding solutions.

Once the integration is complete, the initial condition is applied to resolve the constant \( C \). In our initial value problem, using \( y(0) = 0 \) showed that \( C = 0 \), leading to the final solution, \( y = \arcsin(x) \). This integration process allows us to transition from the rate of change relationship described by the differential equation to the function \( y \) that satisfies it under given conditions.