Problem 68 - University Calculus: Early Transcendentals

Question

Each of the following functions solves one of the initial value problems.Which function solves which problem? Give brief reasons for your answers. a. \(y=\int_{1}^{x} \frac{1}{t} d t-3\) b. \(y=\int_{0}^{x} \sec t d t+4\) c. \(y=\int_{-1}^{x} \sec t d t+4\) d. \(y=\int_{\pi}^{x} \frac{1}{t} d t-3\) $$y^{\prime}=\frac{1}{x}, \quad y(1)=-3$$

Step-by-Step Solution

Answer

Function (a) solves the initial value problem.

Step 1: Understand the Initial Value Problem

We need to identify which function corresponds to the initial value problem given by the differential equation \(y' = \frac{1}{x}\) and the initial condition \(y(1) = -3\). Remember, solving the problem involves ensuring that the derivative of the function corresponds to \(y' = \frac{1}{x}\) and the initial condition is satisfied.

Step 2: Verify Derivative Matches the Differential Equation

From the four given functions, we check which ensures \(y' = \frac{1}{x}\). This would typically involve checking the integral's derivative component. For both the functions \(a\) and \(d\), \(y = \int \frac{1}{t} dt\) so their derivatives with respect to \(x\) are \(\frac{1}{x}\), matching the differential equation.

Step 3: Confirm Initial Condition

Now we need to evaluate each candidate function at \(x = 1\) and verify if they meet the initial condition \(y(1) = -3\). - For function \(a\): Substitute \(x = 1\) in \(y = \int_{1}^{x} \frac{1}{t} dt - 3 = \int_{1}^{1} \frac{1}{t} dt - 3 = 0 - 3 = -3\), which satisfies the initial condition. - For function \(d\): Substitute \(x = 1\) in \(y = \int_{\pi}^{x} \frac{1}{t} dt - 3\). This integral cannot have an exact initial value \(y(1)\) as it requires knowing \(\int_{\pi}^{1} \frac{1}{t} dt\), which does not straightforwardly satisfy \(y(1) = -3\) without additional calculations.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that involves an unknown function and its derivatives. It describes the relationship between a function and its derivatives and is used to model real-world phenomena such as physics, engineering, and economics. In this exercise, the differential equation presented is \(y' = \frac{1}{x}\). This means that we are searching for a function \(y(x)\) whose derivative \(\frac{dy}{dx}\) is equal to \(\frac{1}{x}\).

Let's break down the steps needed to solve such problems:

• Identify the differential equation given in the problem. In this case, it's a first-order equation as it involves only the first derivative.
• Find functions whose derivatives satisfy this equation. Look for matching terms or known integral forms that simplify the problem.

Solving differential equations often involves integrating both sides, which we'll cover in the next section. It's essential to remember that the form of a differential equation dictates the methods used for finding solutions, including using initial value problems like the one in the exercise to pinpoint the correct particular solution.

Integration

Integration is a mathematical operation used to find the antiderivative or the area under a curve. In this context, it helps us solve differential equations by reversing the differentiation process. Given \(y' = \frac{1}{x}\), finding \(y(x)\) involves integrating \(\frac{1}{x}\) with respect to \(x\).

The integral of \(\frac{1}{x}\) is \(\ln|x| + C\), where \(C\) is the constant of integration. Why do we need this constant? Because when we take the derivative of a constant, it becomes zero, and any number could have been added initially before differentiation without changing the derivative.

In solving our exercise, you must process:

• Perform the integration for each function listed. Here, functions \(a\) and \(d\) involve integrating \(\frac{1}{t}\), matching our differential equation's requirement.
• Verify if their derived expressions include all terms, including any additional constant like \(-3\) in function \(a\), which shifts the entire function down.

Understanding integration is crucial in these contexts, as it transforms the derivative-based expressions back into the desired functions.

Initial Condition

An initial condition in a differential equation provides specific information on the value of a function at a particular point, which helps determine the constant \(C\) from integration. In this exercise, we're given \(y(1) = -3\).

This condition is crucial because it pinpoints one specific solution out of the many admissible solutions to a differential equation. To utilize the initial condition:

• Evaluate the integrated function at the given point, which in our exercise is \(x = 1\).
• Solve for any constants by setting the integrated function equal to the known function value. For instance, in function \(a\), \(y = \int_1^x \frac{1}{t} dt - 3\) naturally matches \(y(1) = -3\), confirming it is the right solution.

The initial condition ensures the solution not only functions correctly according to the differential equation but also matches any real-world constraints or specific requirements involved.