Problem 7 - Thomas Calculus

Question

In Exercises \(5-8,\) show that each function is a solution of the given initial value problem. (TABLE NOT COPY) $$ \begin{array}{l}{x y^{\prime}+y=-\sin x, \quad y\left(\frac{\pi}{2}\right)=0 \quad y=\frac{\cos x}{x}} \\ {x > 0}\end{array} $$

Step-by-Step Solution

Answer

The function \( y = \frac{\cos x}{x} \) satisfies both the differential equation and the initial condition.

Step 1: Verify the Differential Equation

The differential equation given is \( xy' + y = -\sin x \). We are given the function \( y = \frac{\cos x}{x} \), which we need to verify as a solution to the differential equation. First, we find its derivative: \( y' = \frac{d}{dx}\left(\frac{\cos x}{x}\right) \).

Step 2: Differentiate the Function

To find \( y' \), use the quotient rule: \[ y' = \frac{(x)(-\sin x) - (\cos x)(1)}{x^2} = \frac{-x \sin x - \cos x}{x^2} \].

Step 3: Substitute into the Differential Equation

Substitute \( y = \frac{\cos x}{x} \) and \( y' = \frac{-x \sin x - \cos x}{x^2} \) into the differential equation: \[ x\left(\frac{-x \sin x - \cos x}{x^2}\right) + \frac{\cos x}{x} \]. Simplify it: \[ \frac{-x^2 \sin x - x \cos x + \cos x}{x^2} = \frac{-x^2 \sin x}{x^2} = -\sin x \]. Since both sides are equal, it satisfies the equation.

Step 4: Verify the Initial Condition

Now, verify the initial condition \( y\left(\frac{\pi}{2}\right) = 0 \). Substitute \( x = \frac{\pi}{2} \) into \( y = \frac{\cos x}{x} \): \[ y\left(\frac{\pi}{2}\right) = \frac{\cos \left(\frac{\pi}{2}\right)}{{\frac{\pi}{2}}} = \frac{0}{\frac{\pi}{2}} = 0 \]. Thus, the initial condition is satisfied.

Key Concepts

Initial Value Problem

An initial value problem (IVP) in differential equations involves finding a function that not only satisfies a differential equation but also meets one or more specific conditions at a given point, called the initial condition. In this exercise, the initial value condition provided is \( y\left(\frac{\pi}{2}\right) = 0 \). This means that at \( x = \frac{\pi}{2} \), the function \( y \) should evaluate to zero.

To tackle an initial value problem, follow these steps:

• Solve the differential equation to find a general solution.
• Substitute the initial conditions into the general solution to find any constants or parameters involved.
• Use these results to pinpoint a specific solution that satisfies both the differential equation and the initial condition.

Initial value problems are fundamental in many scientific fields, as they help model real-world scenarios by determining specific behaviors of a system. In this case, showing that the function fits the initial condition ensures that the solution accurately starts from the specified point.

Quotient Rule

The Quotient Rule is a valuable tool in calculus, particularly when differentiating a function that is the quotient of two other functions. The general formula for the Quotient Rule is given by:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]where \( u \) and \( v \) are functions of \( x \), and \( u' \), \( v' \) are their respective derivatives.

In this exercise, the function \( y = \frac{\cos x}{x} \) requires differentiation. Here, \( u = \cos x \) and \( v = x \). Their derivatives are \( u' = -\sin x \) and \( v' = 1 \). Applying the Quotient Rule to find \( y' \) yields:\[y' = \frac{(-\sin x \cdot x) - (\cos x \cdot 1)}{x^2} = \frac{-x \sin x - \cos x}{x^2}.\]

By breaking down this process into clear steps, you can understand how to handle more complex functions and ensure accurate differentiation, crucial for solving and verifying differential equations.

Solution Verification

Solution verification involves checking whether a proposed solution actually satisfies the given differential equation and any initial conditions. This is a critical step in solving differential equations, ensuring that the derived function truly addresses the problem.

For this exercise, verify the solution in two stages:

• First, substitute the proposed solution function and its derivative back into the original differential equation: \( xy' + y = -\sin x \).
• Simplify and check if both sides of the equation equal each other. If they do, it confirms that the function is indeed a solution.

After confirming the function solves the differential equation, verify the function against the initial condition provided in the problem (e.g., \( y(\frac{\pi}{2})=0 \)). This verifies that the solution is not only correct but applicable to the specific scenario outlined in the initial value problem.

Solution verification is essential to avoid errors and guarantees that the function behaves as expected under given conditions.

Calculus

Calculus, the branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series, is crucial for solving differential equations. Understanding calculus concepts allows us to handle various operations needed to solve such equations efficiently.

Key topics in calculus related to differential equations include:

• Derivatives: Used to calculate rates of change or slopes of curves. Essential for solving differential equations, as they often involve derivatives of unknown functions.
• Integration: The inverse operation to differentiation, used to solve differential equations by finding functions with given rate behaviors.
• Limit concepts: Help understand the behavior of functions as they approach specific points or infinity, often utilized in verifying solutions.

In this exercise, calculus helps in performing the differentiation using the Quotient Rule and ensuring accurate solution verification. As students progress, a solid grasp of calculus principles becomes indispensable in advancing their understanding and solution strategies of more complex mathematical problems.