Problem 17 - Thomas Calculus in SI Units

Question

Solve the differential equations. $$\frac{d y}{d x}=2 x \sqrt{1-y^{2}}, \quad-1

Step-by-Step Solution

Answer

The solution is \( y = \sin(x^2 + C) \), \(-1 < y < 1\).

Step 1: Rewrite the differential equation

The given differential equation is \( \frac{dy}{dx} = 2x \sqrt{1 - y^2} \). We will first rewrite it in a form that separates variables: \( \frac{dy}{\sqrt{1 - y^2}} = 2x \, dx \). This step allows us to separate the variables for integration.

Step 2: Integrate both sides

Integrate both sides of the equation. Start by integrating the left side: \( \int \frac{dy}{\sqrt{1-y^2}} = \arcsin(y) + C_1 \). Now integrate the right side: \( \int 2x \, dx = x^2 + C_2 \). Thus, we have the equation \( \arcsin(y) = x^2 + C \) where \( C = C_2 - C_1 \).

Step 3: Solve for \( y \)

To find \( y \), take the sine of both sides of the equation \( \arcsin(y) = x^2 + C \). Thus, \( y = \sin(x^2 + C) \).

Step 4: Consider the range of y

Ensure that the solution satisfies the condition \(-1 < y < 1\), which matches the range of the sine function \( y = \sin(x^2 + C) \) since \( -1 \leq \sin(x^2 + C) \leq 1 \). This confirms that our solution fits the given range constraint of \( y \).

Step 5: General solution

The general solution to the differential equation is \( y = \sin(x^2 + C) \), where \( C \) is an arbitrary constant, and \(-1 < y < 1\) holds true.

Key Concepts

Separation of Variables

On your journey to solving differential equations, one of the foundational techniques you'll encounter is "Separation of Variables." This method is particularly useful for solving first-order ordinary differential equations, where the variables can be separated so that all terms involving one variable (like \( y \)) appear on one side of the equation, and all terms involving another variable (like \( x \)) appear on the opposite side. In the given problem, we have:

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

We will separate the variables by rearranging the equation to:

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

This allows us to integrate each side of the equation independently, transforming the problem into one that can be tackled with basic integration techniques. By separating variables, we make the differential equation more manageable, converting it into integrals we can solve separately. Remember, the ultimate goal of separation of variables is to make the equation easier to solve by isolating different parts of the expression.

Integration Techniques

Once variables are separated in a differential equation, as seen in the separation of \( \frac{dy}{\sqrt{1 - y^2}} = 2x \, dx \), we apply integration techniques to find a solution. The integration process involves handling each side independently:

• The left side integrates to \( \int \frac{dy}{\sqrt{1-y^2}} = \arcsin(y) + C_1 \). This integral leads us to the arcsine function, a known antiderivative of \( \frac{1}{\sqrt{1-y^2}} \).
• The right side simplifies to \( \int 2x \, dx = x^2 + C_2 \), a straightforward application of the power rule for integration.

Combining these results, we form the equation \( \arcsin(y) = x^2 + C \) where \( C = C_2 - C_1 \). Integration is a powerful tool, transforming differential expressions into solvable equations. Mastery of integration techniques is crucial in solving differential equations efficiently, especially when dealing with separable equations.

Arcsine Function

The arcsine function appears when integrating expressions like \( \frac{1}{\sqrt{1-y^2}} \). This function, denoted as \( \arcsin(y) \), is the inverse of the sine function and returns an angle whose sine is \( y \). Understanding its range and properties is key, especially when solving differential equations involving trigonometric identities.

• The range of the arcsine function typically spans from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), meaning it only outputs values within this interval, which corresponds to \(-1 \leq y \leq 1\).
• Thus, when solving for \( y \) and checking conditions like \(-1 < y < 1\), arcsine will naturally adhere to these boundaries.

In practice, as seen in the exercise, once you solve for \( \arcsin(y) = x^2 + C \), to find \( y \), you take the sine of both sides, yielding \( y = \sin(x^2 + C) \). This respects the restriction \(-1 < y < 1\), ensuring that the solution fits the problem's constraints. The arcsine function, therefore, plays a pivotal role in linking the solution of integrals back to valid function outputs.