Problem 22 - Differential Equations and Linear Algebra

Question

Determine all values of the constant \(r\) such that the given function solves the given differential equation. $$y(x)=e^{r x}, \quad y^{\prime \prime}-y^{\prime}-6 y=0$$.

Step-by-Step Solution

Answer

The constant values of \(r\) that satisfy the given differential equation \(y'' - y' -6y = 0\) for the function \(y(x) = e^{rx}\) are \(r = 3\) and \(r = -2\).

Step 1: Find the Derivatives

Find the first and second derivatives of y(x) = e^(rx) with respect to x: First derivative, y'(x): \(y'(x) = \frac{d}{dx}(e^{rx}) = re^{rx}\) Second derivative, y''(x): \(y''(x) = \frac{d^2}{dx^2}(e^{rx}) = r^2e^{rx}\) Now that we have the first and second derivatives, let's substitute them into the given differential equation.

Step 2: Substitute the derivatives and the original function into the differential equation

Substitute y(x), y'(x), and y''(x) into the given equation, y'' - y' - 6y = 0 and simplify: \(r^2 e^{rx} - r e^{rx} - 6 e^{rx} = 0\)

Step 3: Factor the simplified equation

Factor by removing the common factor e^(rx) from all the terms: \(e^{rx} (r^2 - r - 6) = 0\) Now, since e^(rx) cannot be zero, we can focus on solving the quadratic equation: \(r^2 - r - 6 = 0\)

Step 4: Solve the quadratic equation

To find the values of r, we can either use factoring, the quadratic formula, or other algebraic methods to solve the equation: Factoring the equation \(r^2 - r - 6\), we get: \((r - 3)(r + 2) = 0\) Now, we can find two values for r by setting each factor equal to 0: \(r - 3 = 0 \Rightarrow r = 3\) \(r + 2 = 0 \Rightarrow r = -2\) Thus, the values of the constant r for which the given function y(x) = e^(rx) solves the given differential equation are: \(r = 3\) and \(r = -2\)

Key Concepts

Exponential Functions

Exponential functions are equations where a constant base, often denoted as "e" in calculus, is raised to a variable exponent. In our exercise, the function is given as \( y(x) = e^{rx} \). Here, "e" is the base and "rx" is the exponent. Exponential functions are essential in modeling growth and decay in natural processes.

One of their unique properties is that the rate of change (derivative) of the function is proportional to the value of the function itself. Thus, differentiating an exponential function like \( e^{rx} \) results in another exponential function \( re^{rx} \), where "r" is a constant. This makes exponential functions frequently used to solve differential equations.

They are always positive and asymptotically approach zero but never actually reach it, unless modified by the constant "r". As "r" changes, the steepness of the exponential curve will either increase (if positive) or decrease (if negative). They are critical in many scientific fields, including physics, engineering, and finance, due to their predictable patterns and behavior.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). In our step-by-step solution, we encounter a specific quadratic equation: \( r^2 - r - 6 = 0 \).

This equation arises after substituting the derivatives into the differential equation and factoring out the exponential term. It's called a quadratic equation because the highest power of the variable (in this case, "r") is 2. Quadratic equations are significant because they help describe a wide range of phenomena in physics and engineering, including motion and equilibrium.

Solving quadratic equations can be done through several methods, including:

• Factoring, if the quadratic can be expressed as a product of its factors (e.g., \((r - 3)(r + 2) = 0\)).
• Using the quadratic formula, \( r = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), which gives the roots of any quadratic equation.
• Completing the square, a method that involves rewriting the quadratic in a perfect square form.

These roots, \( r = 3 \) and \( r = -2 \), show where the quadratic curve intersects the x-axis and represent the values solved by our equation.

Factorization Techniques

Factorization is the process of breaking down an expression into a product of its simpler factors. It's a central technique in algebra that simplifies complex equations, making them easier to solve. In our solution, we use factorization to break down the quadratic equation \( r^2 - r - 6 = 0 \) into simpler expressions.

One factorization method involves identifying two numbers that multiply to the constant term ("-6" in this case) and add up to the middle coefficient ("-1" here). For our equation, those numbers are 3 and -2. Thus, the equation can be written as \( (r - 3)(r + 2) = 0 \).

By factorization, we also reduce these expressions to their simplest forms, allowing us to find the solutions more readily. After factorizing, each factor can be set to zero to solve for "r". In our context, these roots (or solutions) are \( r = 3 \) and \( r = -2 \).

Factorization is crucial not only in solving quadratics but also in decomposition strategies for more complex polynomials or ensuring rational expressions are in simplest form, making it a versatile tool in algebra and calculus.