Problem 11 - Advanced Engineering Mathematics

Question

Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ 2 y^{\prime}+y=0 ; \quad y=e^{-x / 2} $$

Step-by-Step Solution

Answer

The function \(y = e^{-x/2}\) is an explicit solution of the differential equation \(2y' + y = 0\).

Step 1: Identify the Given Function and Differential Equation

In this exercise, the given function is \(y = e^{-x / 2}\) and the differential equation is \(2y' + y = 0\). We need to verify that this function is a solution to the differential equation.

Step 2: Compute the Derivative of the Given Function

First, we find the derivative of \(y\) with respect to \(x\). The exponential rule gives us:\[ y' = \frac{d}{dx}(e^{-x/2}) = -\frac{1}{2} e^{-x/2}\]

Step 3: Substitute the Function and Its Derivative into the Differential Equation

Now, substitute \(y = e^{-x/2}\) and \(y' = -\frac{1}{2} e^{-x/2}\) into the differential equation:\[2\left(-\frac{1}{2} e^{-x/2}\right) + e^{-x/2} = 0\]

Step 4: Simplify the Expression

Simplify the expression obtained from the substitution:\[-e^{-x/2} + e^{-x/2} = 0\] which equals 0.

Step 5: Conclude the Verification

Since the left-hand side of the equation simplifies to zero, which matches the right-hand side, the function \(y = e^{-x/2}\) is an explicit solution to the differential equation \(2y' + y = 0\).

Key Concepts

Explicit Solution

In the realm of differential equations, the term "explicit solution" refers to an expression where the dependent variable (typically denoted as \( y \)) is presented in terms of the independent variable (often \( x \)) and constants. In simpler terms, an explicit solution makes it straightforward to see how \( y \) changes with respect to \( x \).

For example, for the differential equation \( 2y' + y = 0 \), the solution \( y = e^{-x/2} \) is an explicit solution because \( y \) is directly expressed in terms of \( x \). The simplicity of explicit solutions makes them particularly user-friendly; they provide a clear perspective on the relationship between variables without needing further manipulation.

Thus, verifying that a function like \( y = e^{-x/2} \) is an explicit solution involves showing that it satisfies the original differential equation. Once the derivatives are computed and substituted back into the differential equation, simplification should lead to both sides of the equation matching, confirming the solution's validity.

Exponential Functions

Exponential functions are a cornerstone of calculus, especially within the study of differential equations. These functions often have the form \( y = ae^{bx} \), where \( a \) and \( b \) are constants. They play a crucial role because of their properties, such as their specific rates of change.

In our differential equation \( 2y' + y = 0 \), the given solution is \( y = e^{-x/2} \). This is an exponential function where the base is the natural number \( e \) (approximately 2.718), and it's raised to a power that involves the independent variable \( x \).

Exponential functions like these exhibit unique behaviors:

• Their derivative forms are notably related to the functions themselves, as we see here where the derivative of \( y = e^{-x/2} \) is \( y' = -\frac{1}{2} e^{-x/2} \).
• They can model significant real-world phenomena, such as population growth, radioactive decay, and cooling processes.

Recognizing the exponential nature of a solution can pave the way to understanding the underlying dynamics of a differential equation.

Interval of Definition

When discussing differential equations, the interval of definition is a vital concept. It describes the range of the independent variable \( x \) over which the solution is applicable and valid. This interval can vary greatly, depending on the nature of the differential equation and the solution involved.

For the solution \( y = e^{-x/2} \) to the differential equation \( 2y' + y = 0 \), we assume that the interval of definition \( I \) is reasonably unrestricted given the nature of the exponential function. Since \( e^{-x/2} \) is defined for all real numbers, the interval can extend over the entire set of real numbers \( (-\infty, \infty) \).

However, practical considerations, initial conditions, or specific problem constraints may limit \( I \) to a more finite range. Always consider the context of the problem when determining the appropriate interval of definition. Understanding this aspect ensures that our solutions are not only mathematically correct but also practically applicable.