Problem 15 - Advanced Engineering Mathematics

Question

Verify that the indicated function \(y=\phi(x)\) is an explicit solution of the given first-order differential equation. Proceed as in Example 5, by considering \(\phi\) simply as a function, give its domain. Then by considering \(\phi\) as a solution of the differential equation, give at least one interval \(I\) of definition. $$ (y-x) y^{\prime}=y-x+8 ; \quad y=x+4 \sqrt{x}+2 $$

Step-by-Step Solution

Answer

The function \(y = x + 4\sqrt{x} + 2\) is a solution on the interval \(x \geq 0\).

Step 1: Substitute the function into the differential equation

We have the differential equation \((y-x) y^{\prime} = y-x+8\) and the function \(y = x + 4\sqrt{x} + 2\). First, substitute the function \(y\) into the differential equation. The first step involves finding the derivative \(y'\).

Step 2: Calculate the derivative

Differentiate \(y = x + 4\sqrt{x} + 2\) with respect to \(x\). The derivative is \(y' = 1 + 2 x^{-1/2}\).

Step 3: Substitute the derivative and function into the equation

Substitute \(y\) and \(y'\) into the differential equation. Thus, \((y-x) y' = (x + 4\sqrt{x} + 2 - x)(1 + 2 x^{-1/2}) = y-x+8\). Simplify to verify if both sides match.

Step 4: Simplify the expressions

Simplify the left side: \((4\sqrt{x} + 2)(1 + \frac{2}{\sqrt{x}}) = 4\sqrt{x} + 2 + 8 = 4\sqrt{x} + 10\). The right side of the equation, \(y-x+8\), simplifies to \(4\sqrt{x} + 10\), showing both sides are equal.

Step 5: Identify the domain

The function \(y = x + 4\sqrt{x} + 2\) is defined for \(x \geq 0\) because \(\sqrt{x}\) is only defined for non-negative \(x\). Hence, the domain of \(\phi(x)\) is \(x \geq 0\).

Step 6: Determine interval of definition for the differential equation

Since the function is a solution on the domain \(x \geq 0\), and both the differential equation and its solution involve \(\sqrt{x}\), one interval of definition for the solution is any subinterval within \([0, \infty)\). Therefore, \(I = (0, \infty)\) is an interval of definition.

Key Concepts

Explicit Solution

An explicit solution to a differential equation allows us to express the dependent variable directly in terms of the independent variable. In other words, if we have a differential equation with dependent variable \(y\) and independent variable \(x\), an explicit solution provides \(y\) explicitly as a function of \(x\). For example, in the given exercise, the function \(y = x + 4\sqrt{x} + 2\) is an explicit solution to the differential equation \((y-x) y^{\prime} = y-x+8\). To verify an explicit solution, we differentiate it to find \( y' \), substitute \(y\) and \(y'\) back into the differential equation, and check if both sides are equal after simplification. Ensuring both sides match confirms that the function is indeed an explicit solution.

Domain of a Function

The domain of a function refers to all possible input values (usually \(x\)) for which the function is defined. In this exercise, the function \(y = x + 4\sqrt{x} + 2\) includes a square root, \(\sqrt{x}\), which is only valid for non-negative values of \(x\). - This means the smallest value \(x\) can be is 0.- Therefore, the domain of this function is \(x \geq 0\).It's very important to determine this, as it tells us over which intervals the function operates correctly without causing mathematical errors like taking the square root of a negative number. Identifying the domain ensures we are considering the function in its valid range, leading to accurate solutions and interpretations.

Interval of Definition

The interval of definition is the range of the independent variable \(x\) over which the solution to the differential equation is valid. For a solution to be valid, this interval needs to comply with the derived form of the function and any operations within it. In this exercise, both the differential equation and the solution function contain \(\sqrt{x}\), which implies - The variable \(x\) must be positive.- Therefore, an interval like \((0, \infty)\) is often chosen. Because we seek subintervals within the domain that do not lead to undefined expressions, ensure all functions involved remain valid throughout the interval. This guarantees the behavior of the solution function aligns with its mathematical construction and any practical applications.

Derivative Calculation

Calculating the derivative is crucial for verifying solutions to differential equations. In the exercise, to proceed with verifying the given function as a solution, we first need to find its derivative \(y'\). The function provided is \(y = x + 4\sqrt{x} + 2\).Here's how to calculate the derivative:- Differentiate each term with respect to \(x\).- The derivative of \(x\) is 1.- The derivative of \(4\sqrt{x}\) is \(4 \cdot 1/2 \cdot x^{-1/2}\) or \(2x^{-1/2}\).- The derivative of a constant \(2\) is 0.Thus, the derivative \(y' = 1 + 2x^{-1/2}\) is crucial for substituting back into the differential equation and verifying the explicit solution. Calculating derivatives accurately ensures the solution process aligns with the differential equation's requirements and fosters a comprehensive understanding of variable rates of change.