Problem 112 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

We are given the differential equation \( \frac{d y}{d x} = \frac{1}{1+x^{2}} - \frac{2}{\sqrt{1-x^{2}}} \) and an initial condition \( y(0) = 2 \). Our task is to find the function \( y(x) \) that satisfies both the differential equation and the initial condition.

Step 2: Integrate the First Term

The first term of the right-hand side is \( \frac{1}{1+x^2} \). This is the derivative of the arctangent function. Therefore, \( \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C_1 \), where \( C_1 \) is a constant of integration.

Step 3: Integrate the Second Term

The second term is \( -\frac{2}{\sqrt{1-x^2}} \). This resembles the derivative of an arcsine function. Thus, \( \int -\frac{2}{\sqrt{1-x^2}} \, dx = -2\sin^{-1}(x) + C_2 \), where \( C_2 \) is another constant of integration.

Step 4: Combine Results to Get General Solution

Combine the results from Step 2 and Step 3 to form the general solution: \( y(x) = \tan^{-1}(x) - 2\sin^{-1}(x) + C \), where \( C = C_1 + C_2 \) is a constant that you'll determine using the initial condition.

Step 5: Apply Initial Condition

Use the initial condition \( y(0) = 2 \) to find \( C \). Substitute \( x = 0 \) and \( y = 2 \) into the general solution: \( 2 = \tan^{-1}(0) - 2\sin^{-1}(0) + C \). This simplifies to \( 2 = 0 + 0 + C \), so \( C = 2 \).

Step 6: Write the Particular Solution

Now substitute \( C = 2 \) into the general solution to get the particular solution: \( y(x) = \tan^{-1}(x) - 2\sin^{-1}(x) + 2 \).

Key Concepts

Initial Value Problem

An initial value problem involves a differential equation with an accompanying specific value, called an initial condition. These problems seek a particular solution that satisfies both the differential equation and this initial condition. In our exercise, the differential equation we tackled was \( \frac{dy}{dx} = \frac{1}{1+x^2} - \frac{2}{\sqrt{1-x^2}} \). The initial condition provided was \( y(0) = 2 \).

When solving such problems, the initial condition helps to determine the constants that result from the integration of the differential equation. This process ensures the solution meets the unique constraints set by the initial condition. Here's a simple way to think about it:

• You find the general solution by integrating the differential equation.
• Apply the initial condition to solve for any constants in your solution.
• This leads to the unique, particular solution for the problem.

Understanding the concept of initial value problems is crucial, as they allow for precise modeling of real-world situations where we know the state of a system at a particular point in time.

Integration Techniques

When dealing with differential equations, integration techniques are indispensable tools that transform derivatives into functions. In the given exercise, we used integration to resolve the differential equation into a form that captures the relationship between \( x \) and \( y \).

Different terms in the differential equation often require different integration techniques. For our problem:

• The term \( \frac{1}{1+x^2} \) required the recognition that it equals the derivative of the arctangent function, allowing us to directly integrate it to \( \tan^{-1}(x) \).
• The term \( -\frac{2}{\sqrt{1-x^2}} \) needed familiarity with the derivative of the arcsine function, which guided us to integrate it to \(-2\sin^{-1}(x) \).

Mastering integration techniques means recognizing common forms and knowing the antiderivatives. These skills are a core part of solving differential equations and are frequently applied in calculus problems.

Arctangent Function

The arctangent function, often written as \( \tan^{-1}(x) \) or \( \text{atan}(x) \), is the inverse operation of the tangent function. Its derivative is particularly useful in calculus and solving differential equations.

In our exercise, we encountered \( \frac{1}{1+x^2} \), the derivative of \( \tan^{-1}(x) \). By recognizing this form, we integrated it directly to obtain the arctangent function as a part of the general solution.

The arctangent function has some interesting properties:

• It provides the angle for a given tangent ratio, restricted usually between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• It is a smooth, continuous curve, which makes it ideal for modeling and solving problems involving angles or periodic functions.
• Understanding how to integrate the derivative of arctangent helps in identifying solutions for differential equations that involve this term.

The proficiency in manipulating the arctangent is valuable in various fields of mathematics and applied sciences.

Arcsine Function

The arcsine function, symbolized as \( \sin^{-1}(x) \) or \( \text{asin}(x) \), is the inverse of the sine function. Recognizing and integrating the derivative of the arcsine function is important in tackling differential equations.

In our problem, we had to integrate the expression \( -\frac{2}{\sqrt{1-x^2}} \). The derivative of the arcsine function \( \frac{1}{\sqrt{1-x^2}} \) was modified by a constant factor, simplifying to \(-2\sin^{-1}(x)\) upon integration.

Some key properties of the arcsine function include:

• It returns the angle corresponding to a given sine value, typically within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
• The graph of arcsine is also smooth and continuous, characteristics which aid in modeling spherical and trigonometric scenarios.
• A firm grasp of arcsine integration allows refined approaches to solving integrals and helps in deriving solutions within trigonometric contexts.

Understanding how to work with the arcsine function in mathematics opens the door to resolving many equations involving trigonometric identities.