Problem 17 - Thomas Calculus

Question

Solve the differential equations in Exercises \(9-22\) $$\frac{d y}{d x}=2 x \sqrt{1-y^{2}}, \quad-1 < y < 1$$

Step-by-Step Solution

Answer

The solution is \( y = \sin(x^2 + C) \).

Step 1: Recognize the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = 2x \sqrt{1-y^2} \). This is a first-order ordinary differential equation.

Step 2: Attempt Separation of Variables

To solve the equation, we'll start by trying to separate variables. Rearrange the terms to get the equation in the form \( \frac{dy}{\sqrt{1-y^2}} = 2x \, dx \).

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{dy}{\sqrt{1-y^2}} = \int 2x \, dx \). The integral of the left side is \( \arcsin(y) + C_1 \), and the right side is \( x^2 + C_2 \).

Step 4: Equate the Integrals

Since both integrals should equal each other, we have \( \arcsin(y) = x^2 + C \), where \( C = C_2 - C_1 \) is the constant of integration.

Step 5: Solve for y

To solve for \( y \), take the sine of both sides: \( y = \sin(x^2 + C) \). This is the general solution of the differential equation.

Step 6: Verification

To verify, substitute \( y = \sin(x^2 + C) \) back into the original differential equation to check if both sides are equal.

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique in solving first-order ordinary differential equations (ODEs). It involves rearranging an equation so that each variable appears on a different side of the equals sign. The aim is to isolate the dependent variable and its differential on one side, and the independent variable with its differential on the other.

In the given problem, we start with the differential equation \( \frac{dy}{dx} = 2x \sqrt{1-y^2} \). To separate the variables, we rearrange it to get:

• \( \frac{dy}{\sqrt{1-y^2}} = 2x \, dx \)

This process allows us to integrate each side independently. The left side only involves the variable \( y \), and the right side only involves the variable \( x \). Successfully separating variables sets the stage for applying integration techniques to solve the differential equation.

Integration Techniques

Once variables are separated, integration techniques come into play to find the antiderivatives of each side. For the separated equation \( \frac{dy}{\sqrt{1-y^2}} = 2x \, dx \), we need to integrate both sides.

• The left side, \( \int \frac{dy}{\sqrt{1-y^2}} \), resembles the standard integral that results in the arcsine function, which is \( \arcsin y + C_1 \).
• The right side, \( \int 2x \, dx \), is a standard polynomial integral resulting in \( x^2 + C_2 \).

After integration, we equate the two results to form \( \arcsin(y) = x^2 + C \), where \( C = C_2 - C_1 \). This step allows us to relate the two variables through integration constants and builds toward solving for \( y \).

Sine Function

The sine function, denoted as \( \sin \), is a trigonometric function that arises naturally from the integration performed in our solution. Let's break down its role in the context of the differential equation.

After integrating and equating, we obtained \( \arcsin(y) = x^2 + C \). To solve for \( y \), we take the sine of both sides, exploiting the inverse relationship between sine and arcsine:

• \( y = \sin(x^2 + C) \)

This expression represents the general solution to the differential equation. It's important to remember that the sine function maps any given angle to its corresponding ratio on the unit circle, enabling a cyclical range of values which fits well within our given conditions \(-1 < y < 1\). This cyclic nature of the sine function is crucial in ensuring the solution stays within the specified interval.

Arcsine Function

The arcsine function, expressed as \( \arcsin \), is the inverse of the sine function. It takes a value between \(-1\) and \(1\) and returns an angle, usually measured in radians, whose sine is the original value.

When addressing the equation \( \frac{dy}{dx} = 2x \sqrt{1-y^2} \), integrating the separated left side gives us:\

• \( \arcsin(y) + C_1 \)

Here, \( \arcsin(y) \) solves the integral of \( \frac{1}{\sqrt{1-y^2}} \), which is characteristic of inverse trigonometric identities. This result plays a crucial role in shaping the relationship between \( y \) and \( x \), ultimately aiding in deriving \( y = \sin(x^2 + C) \).

Understanding the arcsine function is important because it allows us to transition elegantly from the integral form to the solution, capturing the essential inverse trigonometric nature needed to solve our differential equation.