Problem 27 - Advanced Engineering Mathematics

Question

Find values of \(m\) so that the function \(y=e^{m x}\) is a solution of the given differential equation. $$ y^{\prime}+2 y=0 $$

Step-by-Step Solution

Answer

The value of m is -2.

Step 1: Understand the Given Differential Equation

The problem involves the differential equation \(y' + 2y = 0\). Our goal is to find the values of \(m\) such that \(y = e^{mx}\) is a solution to this equation.

Step 2: Differentiate the Proposed Solution

To proceed, we need the derivative of the proposed solution \(y = e^{mx}\). Differentiate this with respect to \(x\), yielding: \(y' = me^{mx}\).

Step 3: Substitute into the Differential Equation

Substitute \(y = e^{mx}\) and \(y' = me^{mx}\) into the differential equation. This results in the equation: \(me^{mx} + 2e^{mx} = 0\).

Step 4: Factor and Simplify the Equation

Factor out \(e^{mx}\) from the left side: \((m + 2)e^{mx} = 0\). Since \(e^{mx} eq 0\) for any real \(x\), the equation simplifies to \(m + 2 = 0\).

Step 5: Solve for \(m\)

Solve the equation \(m + 2 = 0\) to find \(m = -2\).

Key Concepts

Understanding Exponential Solutions

Exponential solutions often appear in differential equations, particularly when dealing with growth or decay processes. In the context of the equation given, we consider the function \( y = e^{mx} \), which is an exponential function where \( e \) is the base of the natural logarithm, and \( m \) serves as a parameter determining the rate of exponential growth or decay. Exponential functions are very useful in solving differential equations because:

• They simplify differentiation and integration.
• Their derivatives are proportional to themselves, making them ideal candidates for finding solutions to linear differential equations.
• They have straightforward mathematical properties that facilitate algebraic manipulations, such as differentiation and substitution.

In this case, by expressing the function in terms of \( e^{mx} \), we can directly employ its properties to deduce a suitable value for \( m \) that resolves the differential equation problem at hand. The elegance of exponential solutions aids in breaking complex equations into more manageable forms.

Decoding First-Order Linear Differential Equations

First-order linear differential equations have the general form:\[ y' + P(x)y = Q(x) \]Here, \( y' \) represents the first derivative of \( y \), and \( P(x) \) and \( Q(x) \) are functions of \( x \). For the problem in question, the given equation is:\[ y' + 2y = 0 \]This is a homogeneous equation (notice the zero on the right side), meaning it has no external forcing term, so \( Q(x) = 0 \). Deciphering this type of equation involves finding functions that change at each point in a way that meets certain set constraints.

• In this equation, the derivative of \( y \) is essentially related to \( y \) itself, multiplied by a constant.
• The characteristic of linearity implies superposition; in other words, linearity allows us to consider combinations of solutions to understand the equation’s behavior.
• Solving this involves using methods like the integrating factor method, or inspection, especially when exponential forms are directly examined as potential solutions.

First-order linear differential equations like this are foundational in understanding more complex systems, making them key to exploring applications in physics, engineering, and other sciences.

Finding Parameters in Solutions

When dealing with differential equations, particularly those involving functions like \( e^{mx} \), finding the correct parameter \( m \) is crucial for determining when the function satisfies the equation. Here is how you would typically approach it:

• Differentiate the function: Start by calculating the derivative of your proposed solution, which involves using basic differentiation rules. For \( y = e^{mx} \), the derivative \( y' \) turns out to be \( me^{mx} \).
• Substitute back into the original equation: This step involves plugging \( y \) and \( y' \) back into the differential equation. For our case, substitute to get \( me^{mx} + 2e^{mx} = 0 \).
• Simplify and solve for \( m \): Factor out the common exponential term \( e^{mx} \), and since this cannot be zero, solve the remaining algebraic equation for \( m \), leading to \( m + 2 = 0 \). The solution yields \( m = -2 \).

Solving for parameters like \( m \) is a vital part of formulating the exact solution that fits the dynamics described by the differential equation. It ensures that the proposed solution is indeed a valid candidate, offering insight into the underlying system's behavior.