Problem 42 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The solution to the differential equation is \( y(x) = \frac{1}{2}\tan^{-1}(\frac{x}{2}) + \frac{7π}{8} \)

Step 1: Integrate the given differential equation

Integrate \(\frac{d y}{d x}=\frac{1}{4+x^{2}}\) with respect to \(x\) on both sides, which implies that the equation will become: \(y(x) = \int \frac{1}{4+x^{2}} dx + C\). Here, \(C\) is the constant of integration.

Step 2: Evaluate the integral

The integral \(\int \frac{1}{4+x^{2}} dx\) can be evaluated using the formula \(\int \frac{1}{a^{2}+x^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C\), where \(a=2\) (since \(a^{2}=4\)). So, \(y(x) = \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C\).

Step 3: Use the given initial condition y(2)=π to determine C

Substitute \(x=2\) and \(y(2)=\pi\) in the equation from Step 2: \(\pi = \frac{1}{2} \tan^{-1}(1) + C\). Now, \(\tan^{-1}(1) = \frac{π}{4}\), therefore, \(C = \pi - \frac{π}{8} = \frac{7π}{8}\).

Step 4: Write down the final solution

Replace constant \(C\) in the equation from Step 2 with the value we got, so the solution of the differential equation is: \(y(x) = \frac{1}{2}\tan^{-1}(\frac{x}{2}) + \frac{7π}{8}\)

Key Concepts

Integration

Integration is an essential process in solving differential equations. It involves finding the antiderivative, which essentially reverses the process of differentiation. By integrating both sides of the equation \(\frac{dy}{dx} = \frac{1}{4 + x^2}\), we are determining the function \(y(x)\). This step helps in finding a general solution of the differential equation without any constraints.

When integrating, we include a constant of integration \(C\). This constant appears because the derivative of any constant is zero, meaning many functions could serve as our antiderivative, each differing by a constant. Thus, the integration is expressed as:

• \( y(x) = \int \frac{1}{4 + x^2} \, dx + C \)

Recognizing when and how to integrate is key to solving many types of differential equations.

Initial Conditions

Initial conditions play an important role in finding particular solutions to differential equations. These conditions are given values, usually specific points, of the solution that fit the physical or theoretical problem we aim to solve.

In this exercise, the initial condition is \(y(2) = \pi\). This means that when \(x = 2\), the value of \(y\) is \(\pi\). Such conditions allow us to solve for the constant of integration \(C\), resulting in a specific solution rather than a family of solutions. By substituting into our integral solution before solving for \(C\), we obtain:

• \(\pi = \frac{1}{2} \tan^{-1}(1) + C\)

. The initial condition is crucial in modeling because it customizes the general solution to meet the specific requirements of the scenario.

Inverse Trigonometric Functions

Inverse trigonometric functions like \(\tan^{-1}\) (also called arctan) are essential in evaluating certain integrals. These functions help transform certain algebraic expressions into more workable forms, often seen during integration.

In this problem, the integral \(\int \frac{1}{4+x^{2}} \, dx\) uses the inverse tangent function because its standard form \(\int \frac{1}{a^2 + x^2} \, dx\) translates into an inverse tangent. Specifically:

• \( \int \frac{1}{a^2+x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C\)

For our problem, where \(a = 2\) (thus \(a^2 = 4\)), the use of \(\tan^{-1}(\frac{x}{2})\) simplifies our integral. Understanding these functions is vital in dealing with integrals that involve quadratic expressions in the denominator, transforming them into expressions involving inverse trigonometric functions.