Problem 29 - Advanced Engineering Mathematics

Question

Find values of \(m\) so that the function \(y=e^{m x}\) is a solution of the given differential equation. $$ y^{\prime \prime}-5 y^{\prime}+6 y=0 $$

Step-by-Step Solution

Answer

The values of \( m \) are 3 and 2.

Step 1: Identify the form of the solution

We are given the function \( y = e^{mx} \). To find if it is a solution of the differential equation, we distinguish the function in terms of derivatives. The form of \( y(x) \) suggests using exponential functions.

Step 2: Compute the first derivative

First, compute \( y' \), the first derivative of \( y \):\( y' = \frac{d}{dx}(e^{mx}) = me^{mx} \).

Step 3: Compute the second derivative

Next, compute \( y'' \), the second derivative of \( y \):\( y'' = \frac{d}{dx}(me^{mx}) = m^2e^{mx} \).

Step 4: Substitute derivatives into the differential equation

Substitute \( y = e^{mx} \), \( y' = me^{mx} \), and \( y'' = m^2e^{mx} \) into the equation \( y'' - 5y' + 6y = 0 \):\[ m^2e^{mx} - 5me^{mx} + 6e^{mx} = 0. \]

Step 5: Factor out the common exponential term

Factor out \( e^{mx} \) as it is common to each term:\[ e^{mx}(m^2 - 5m + 6) = 0. \]Since \( e^{mx} eq 0 \) for any \( m \), we focus on:\[ m^2 - 5m + 6 = 0. \]

Step 6: Solve the quadratic equation

Solve the quadratic equation \( m^2 - 5m + 6 = 0 \) using the quadratic formula:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1 \), \( b = -5 \), \( c = 6 \):\[ m = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times 6}}{2 \times 1}. \]Calculate the discriminant: \( (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1 \).\[ m = \frac{5 \pm 1}{2}. \]The solutions are \( m = 3 \) and \( m = 2 \).

Key Concepts

Quadratic Equations

Quadratic equations are essential in various fields of mathematics and science. They appear frequently when solving problems involving areas, curves, and many other applications. A quadratic equation is typically of the form:\[ ax^2 + bx + c = 0 \]where \( a, b, \) and \( c \) are constants, and \( x \) is the variable. The solution to a quadratic equation can be found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula provides two solutions, often referred to as the roots of the equation. The roots can be either real or complex numbers, depending on the nature of the discriminant \( b^2 - 4ac \).

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root (also known as a repeated or double root).
• If the discriminant is negative, the roots are complex.

The quadratic formula is a powerful tool and a critical step in solving second-order differential equations.

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears in the exponent. They are essential in modeling a wide range of phenomena in science and engineering, such as population growth, radioactive decay, and many more. The general form of an exponential function is:\[ y = e^{mx} \]where \( e \) is the base of the natural logarithm, approximately equal to 2.71828, and \( m \) is a constant. One of the most significant properties of exponential functions is that their derivatives are proportional to the original function:

• The first derivative of \( e^{mx} \) is \( me^{mx} \).
• The second derivative of \( e^{mx} \) is \( m^2e^{mx} \).

This property makes exponential functions highly useful for finding solutions to differential equations, especially linear homogeneous equations with constant coefficients. Such functions can transform complex problems into simpler algebraic forms, which is invaluable for analysis.

Second-Order Differential Equations

Second-order differential equations play a fundamental role in modeling and solving real-world problems such as mechanical vibrations, electrical circuits, and many others. They generally take the form:\[ a y'' + b y' + c y = 0 \]These are called linear homogeneous differential equations with constant coefficients. In solving these equations, one common approach is to assume a solution of the form \( y = e^{mx} \), which leads to the characteristic equation:\[ am^2 + bm + c = 0 \]Finding \( m \) often involves solving a quadratic equation. The roots determine the general solution to the differential equation:

• If the roots are real and distinct, \( y = c_1 e^{m_1 x} + c_2 e^{m_2 x} \).
• If the roots are real and repeated, \( y = (c_1 + c_2x) e^{m x} \).
• If the roots are complex, the solution will involve sinusoids and exponentials.

This methodology simplifies the process of identifying solutions by converting the differential equation into an algebraic equation, thus making the problem more tractable.