Problem 7 In Exercises $5-8,$ show that ... [FREE SOLUTION]

Chapter 7: Problem 7

In Exercises $5-8,$ show that each function is a solution of the given initial value problem. $$\begin{array}{ll}{x y^{\prime}+y=-\sin x,} & {y\left(\frac{\pi}{2}\right)=0 \quad y=\frac{\cos x}{x}} \\ {x>0}\end{array}$$

Short Answer

Expert verified

The function $ y = \frac{\cos x}{x} $ solves the initial value problem.

Step by step solution

Understand the Initial Value Problem

The initial value problem is given by the differential equation $ x y' + y = -\sin x $ with the initial condition $ y\left(\frac{\pi}{2}\right) = 0 $. We need to show that $ y = \frac{\cos x}{x} $ satisfies this condition for $ x > 0 $.

Calculate the Derivative of y

Since $ y = \frac{\cos x}{x} $, we find $ y' $ using the quotient rule. Let $ u = \cos x $ and $ v = x $, then $ y = \frac{u}{v} $. The derivative is given by $ y' = \frac{v \cdot u' - u \cdot v'}{v^2} $. Calculate $ u' = -\sin x $ and $ v' = 1 $, then:\[ y' = \frac{x(-\sin x) - \cos x \cdot 1}{x^2} = \frac{-x \sin x - \cos x}{x^2}. \]

Verify the Differential Equation

Substitute $ y' = \frac{-x \sin x - \cos x}{x^2} $ and $ y = \frac{\cos x}{x} $ into the differential equation $ x y' + y = -\sin x $.Calculate the left side:\[ x \left( \frac{-x \sin x - \cos x}{x^2} \right) + \frac{\cos x}{x} = \frac{-x \sin x - \cos x}{x} + \frac{\cos x}{x}. \]Simplify:\[ \frac{-x \sin x - \cos x + \cos x}{x} = \frac{-x \sin x}{x} = -\sin x. \]The left side equals the right side $ -\sin x $, so the function satisfies the differential equation.

Check the Initial Condition

Evaluate the function at $ x = \frac{\pi}{2} $ to check the initial condition:\[ y \left( \frac{\pi}{2} \right) = \frac{\cos \frac{\pi}{2}}{\frac{\pi}{2}} = \frac{0}{\frac{\pi}{2}} = 0. \]The initial condition $ y \left( \frac{\pi}{2} \right) = 0 $ is satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation

A differential equation is a type of equation that involves an unknown function and its derivatives. In this context, it describes a relationship between a function and its rates of change. For our problem, the given differential equation is:
\[ x y' + y = -\sin x \]
This equation tells us how the function $ y $ and its derivative $ y' $ relate to the sine function over the interval $ x > 0 $. The objective is to verify that a specific function, $ y = \frac{\cos x}{x} $, satisfies this relationship, meaning it provides a consistent and true statement when substituted back into the equation.

Quotient Rule

The quotient rule is a method for finding the derivative of a function that is the quotient of two functions. If we have a function $ y = \frac{u}{v} $, its derivative $ y' $ can be calculated using the formula:

$ y' = \frac{v \cdot u' - u \cdot v'}{v^2} $

For our function $ y = \frac{\cos x}{x} $, we identify:

$ u = \cos x $, with derivative $ u' = -\sin x $
$ v = x $, with derivative $ v' = 1 $

Substituting these into the quotient rule gives:
\[ y' = \frac{x(-\sin x) - \cos x \cdot 1}{x^2} = \frac{-x \sin x - \cos x}{x^2} \]
We use this derivative in the process of verifying the original differential equation.

Initial Condition

An initial condition specifies the value of a function at a particular point and is essential in solving differential equations uniquely. In our problem, the initial condition is given by $ y(\frac{\pi}{2}) = 0 $. This means:

The function $ y $ must equal zero when $ x = \frac{\pi}{2} $.

This requirement ensures that our function not only satisfies the differential equation but does so in a way that matches the specific scenario described. Verifying this for our function involves substituting $ x = \frac{\pi}{2} $ into $ y = \frac{\cos x}{x} $:
\[ y\left(\frac{\pi}{2}\right) = \frac{\cos \frac{\pi}{2}}{\frac{\pi}{2}} = \frac{0}{\frac{\pi}{2}} = 0 \]
Hence, the function meets the initial condition.

Functional Verification

Functional verification is the process of confirming that a proposed function satisfies both the given differential equation and the initial condition. In this problem, we must ensure that our proposed solution $ y = \frac{\cos x}{x} $:

Solves the differential equation $ x y' + y = -\sin x $
Adheres to the initial condition $ y\left(\frac{\pi}{2}\right) = 0 $

We first verified the differential equation by substituting $ y $ and its derivative $ y' $ into it:
\[ x \left( \frac{-x \sin x - \cos x}{x^2} \right) + \frac{\cos x}{x} = -\sin x \]
Simplifying the expressions confirmed both sides were equal, proving the function's satisfaction of the equation. Finally, we confirmed the initial condition with a direct substitution, ensuring a correct solution throughout the specified interval.

91影视

In Exercises \(5-8,\) show that each function is a solution of the given initial value problem. $$\begin{array}{ll}{x y^{\prime}+y=-\sin x,} & {y\left(\frac{\pi}{2}\right)=0 \quad y=\frac{\cos x}{x}} \\ {x>0}\end{array}$$

Short Answer

Step by step solution

Understand the Initial Value Problem

Calculate the Derivative of y

Verify the Differential Equation

Check the Initial Condition

Key Concepts

Differential Equation

Quotient Rule

Initial Condition

Functional Verification

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