Problem 30 - Advanced Engineering Mathematics

Question

Find values of \(m\) so that the function \(y=e^{m x}\) is a solution of the given differential equation. $$ 2 y^{\prime \prime}+9 y^{\prime}-5 y=0 $$

Step-by-Step Solution

Answer

The values of \(m\) are \(\frac{1}{2}\) and \(-5\).

Step 1: Differentiate the Function

Find the first derivative of the function \(y = e^{mx}\). The derivative is given by \(y' = m e^{mx}\).

Step 2: Differentiate the Function Again

Find the second derivative of the function \(y = e^{mx}\). The second derivative is \(y'' = m^2 e^{mx}\).

Step 3: Substitute Derivatives into the Differential Equation

Substitute \(y = e^{mx}\), \(y' = m e^{mx}\), and \(y'' = m^2 e^{mx}\) into the differential equation. The equation becomes: \[2(m^2 e^{mx}) + 9(m e^{mx}) - 5(e^{mx}) = 0\].

Step 4: Factor Out Exponential Term

Factor out \(e^{mx}\) from each term. The equation becomes: \[e^{mx}(2m^2 + 9m - 5) = 0\]. Since \(e^{mx} eq 0\) for any \(x\), we only need to solve the quadratic equation \(2m^2 + 9m - 5 = 0\).

Step 5: Solve the Quadratic Equation

Use the quadratic formula to solve \(2m^2 + 9m - 5 = 0\). The quadratic formula is \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = 9\), and \(c = -5\). Substitute these into the formula to find \(m\).

Step 6: Compute the Discriminant

Calculate the discriminant \(b^2 - 4ac\) using \(b = 9\), \(a = 2\), and \(c = -5\). This gives \(9^2 - 4(2)(-5) = 81 + 40 = 121\).

Step 7: Solve for m Values

Finish solving for \(m\) using the quadratic formula: \[m = \frac{-9 \pm \sqrt{121}}{4}\]. This results in \(m = \frac{-9 + 11}{4} = \frac{1}{2}\) and \(m = \frac{-9 - 11}{4} = -5\).

Key Concepts

Second Derivative

The second derivative of a function measures how the rate of change of the function is changing. It's a powerful tool in calculus, particularly in solving differential equations like the one provided. To find the second derivative of the exponential function \(y = e^{mx}\), we first differentiate once to get \(y' = m e^{mx}\), reflecting the rate of change of \(y\). Then, by differentiating \(y'\) again, we calculate the second derivative \(y'' = m^2 e^{mx}\).
This second derivative represents how the acceleration or curvature of the function \(e^{mx}\) behaves. When solving the differential equation \(2y'' + 9y' - 5y = 0\), the second derivative \(y''\) forms a crucial part of the substitution process to find values of \(m\) for which the function satisfies the equation. Understanding the significance of second derivatives helps in evaluating the concavity and stability of solutions in differential equations.

Exponential Function

An exponential function is one where a constant base is raised to the power of a variable, like \(y = e^{mx}\). This type of function features prominently in areas involving growth and decay processes. With a base \(e\) (approximately 2.718), the exponential function has unique properties, such as each derivative being a scaled version of the function itself.

• First derivative: \(y' = m e^{mx}\)
• Second derivative: \(y'' = m^2 e^{mx}\)

These properties make exponential functions particularly suitable for differential equations. Solutions of the form \(y = e^{mx}\) often arise in linear differential equations with constant coefficients. Their symmetry simplifies the calculation, as seen when we substituted the function and its derivatives into our given differential equation.

Quadratic Equation

A quadratic equation is any equation that can be rewritten in the form \(ax^2 + bx + c = 0\). In the context of this problem, the form \(2m^2 + 9m - 5 = 0\) was derived after substituting the function \(y = e^{mx}\) and its derivatives into the differential equation. Factoring out the common exponential base \(e^{mx}\), which never equals zero, leads us directly to this quadratic equation.
Solving quadratic equations is a fundamental skill in algebra and calculus, particularly because they frequently arise when finding characteristic roots of linear differential equations. The solutions of these quadratics, like \(2m^2 + 9m - 5 = 0\), determine key properties of the differential equation solutions.

Quadratic Formula

The quadratic formula is a universal method to find the roots of a quadratic equation, provided as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\). In our problem, substituting \(a = 2\), \(b = 9\), and \(c = -5\) into this formula allows us to determine the values of \(m\) for which \(y = e^{mx}\) satisfies the differential equation.
The step-by-step process:

• Calculate the discriminant: \(b^2 - 4ac = 121\)
• Substitute into the formula: \(m = \frac{-9 \pm \sqrt{121}}{4}\)
• Solve for \(m\): \(m = \frac{1}{2}\) and \(m = -5\)

These roots provide the \(m\) values that ensure the original function serves as a solution to the differential equation, demonstrating the quadratic formula's power and utility.