# Characteristic equation with real distinct roots

##### Intros

##### Examples

###### Lessons

**Using the Characteristic Equation with Real Distinct Roots**

Find the particular solution to the following differential equation:

$y''-9y=0$

With initial values $y(0)=2, y' (0)=1$- Find the particular solution to the following differential equation:

$6y''+8y'-8y=0$

With initial values $y(2)=1, y' (2)=4$ - Find the particular solution to the following differential equation:

$y''-5y'+3y=0$

With initial values $y(0)=0, y' (0)=1$

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###### Topic Notes

## Introduction to Characteristic Equation with Real Distinct Roots

The characteristic equation with real distinct roots is a fundamental concept in solving linear, homogenous, constant coefficient second-order differential equations. This topic builds upon previous knowledge and is crucial for understanding more complex mathematical problems. Our introduction video provides a comprehensive overview of this concept, highlighting its significance in various applications of characteristic equations. The characteristic equation is a powerful tool that allows us to determine the general solution of differential equations by finding their roots. When these roots are real and distinct, it leads to a specific form of solution that is particularly useful in modeling real-world phenomena. By mastering differential equations concepts, students gain a solid foundation for tackling more advanced topics in differential equations. The video explains the step-by-step process of deriving and solving the characteristic equation, emphasizing the importance of recognizing real distinct roots and their implications for the overall solution.

## Understanding the Characteristic Equation

The characteristic equation is a fundamental concept in the study of second-order differential equations, playing a crucial role in determining the nature of solutions. This equation is derived from the general form of a second-order differential equations with constant coefficients: ay'' + by' + cy = 0, where a, b, and c are constants, and y is a function of an independent variable (usually time or space).

To understand the characteristic equation, let's break down the components of the general form:

- a: The coefficient of the second derivative (y'')
- b: The coefficient of the first derivative (y')
- c: The coefficient of y
- y: The dependent variable

The characteristic equation is formed by substituting y = e^(rx) into the general equation, where r is an unknown constant. This substitution leads to the characteristic equation: ar^2 + br + c = 0. Solving this quadratic equation for r gives us the characteristic roots, which are crucial in determining the general solution of the original differential equation.

The discriminant, given by the formula b^2 - 4ac, plays a pivotal role in analyzing the nature of the roots of the characteristic equation. There are three possible scenarios based on the value of the discriminant:

- If b^2 - 4ac > 0, the equation has two distinct real roots.
- If b^2 - 4ac = 0, the equation has one repeated real root.
- If b^2 - 4ac < 0, the equation has two complex conjugate roots.

The nature of these roots directly influences the form of the general solution to the original differential equation. For instance, distinct real roots lead to a solution involving exponential functions, while complex roots result in solutions with sine and cosine functions.

The characteristic equation is particularly significant in various fields of science and engineering. In physics, it helps describe oscillatory systems, such as spring-mass systems or electrical circuits. In control theory, it's used to analyze the stability of systems. In structural engineering, it aids in understanding the vibration modes of structures.

Solving second-order differential equations using the characteristic equation method involves several steps:

- Form the characteristic equation from the given differential equation.
- Solve the characteristic equation to find the roots.
- Use the roots to construct the general solution of the differential equation.
- Apply initial or boundary conditions to determine specific solutions.

The power of the characteristic equation lies in its ability to transform a complex differential equation into a more manageable algebraic problem. This transformation simplifies the process of finding solutions and provides insights into the behavior of the system described by the differential equation.

In conclusion, the characteristic equation is an indispensable tool in the analysis of second-order differential equations. Its relationship with the discriminant provides a clear pathway to understanding the nature of solutions, making it a cornerstone concept in differential equations. Mastering this concept opens doors to solving a wide range of problems in mathematics, physics, and engineering, underlining its importance in both theoretical and applied sciences.

## Real Distinct Roots: Conditions and Significance

In the realm of quadratic equations solutions and second-order linear differential equations, the concept of real distinct roots plays a crucial role. The condition for real distinct roots is determined by the discriminant, which is expressed as b^2 > 4ac. This inequality is fundamental in understanding the nature of solutions to quadratic equations solutions and the behavior of certain real-world applications differential equations.

The discriminant, given by b^2 - 4ac, is a powerful tool in determining the nature of roots for a quadratic equation ax^2 + bx + c = 0. When b^2 > 4ac, the equation has two real and distinct roots. This means that the parabola representing the quadratic function intersects the x-axis at two separate points. The significance of this condition extends beyond basic algebra and into the realm of second-order linear differential equations.

In the context of second-order linear differential equations, particularly second-order linear differential equations, the nature of roots of the characteristic equation roots determines the behavior of solutions. When the discriminant is positive (b^2 > 4ac), the characteristic equation yields two distinct real roots. This scenario corresponds to solutions that are combinations of two exponential functions, each with a different growth or decay rate.

For example, consider the differential equation y'' + 4y' + 3y = 0. The characteristic equation is r^2 + 4r + 3 = 0. Here, a = 1, b = 4, and c = 3. Calculating the discriminant: b^2 - 4ac = 4^2 - 4(1)(3) = 16 - 12 = 4. Since 4 > 0, we have real distinct roots. Solving the equation gives us r = -1 and r = -3. Consequently, the general solution to this differential equation is y = C1e^(-t) + C2e^(-3t), where C1 and C2 are arbitrary constants.

The presence of real distinct roots in differential equations often indicates a system that exhibits non-oscillatory behavior. Each root represents a different rate of exponential growth decay solutions in the solution. This characteristic is particularly important in fields such as physics and engineering, where understanding the long-term behavior of systems is crucial.

It's worth noting that the discriminant not only tells us about the existence of real distinct roots but also provides information about their separation. A larger positive discriminant indicates roots that are further apart, while a smaller positive discriminant suggests roots that are closer together. This insight can be valuable in analyzing the sensitivity of systems to parameter changes.

In conclusion, the condition b^2 > 4ac for real distinct roots is a fundamental concept in algebra and differential equations. It provides crucial information about the nature of solutions, helps in predicting system behavior, and forms the basis for more advanced analysis in various scientific and engineering applications. Understanding this condition and its implications is essential for anyone delving into the study of quadratic equations, real-world applications differential equations, and their practical applications in real-world scenarios.

## Solving the Characteristic Equation

Solving the characteristic equation is a crucial step in finding the general solution to a second-order linear differential equation. This guide will walk you through the process of solving the characteristic equation to find real distinct roots, using the quadratic formula and demonstrating with examples how to find r1 and r2.

### Step 1: Identify the Characteristic Equation

The characteristic equation is derived from the homogeneous part of a second-order linear differential equation. For an equation in the form ay'' + by' + cy = 0, the characteristic equation is ar² + br + c = 0.

### Step 2: Apply the Quadratic Formula

The quadratic formula is essential for solving the characteristic equation. It is given by:

r = [-b ± (b² - 4ac)] / (2a)

Where a, b, and c are the coefficients from the characteristic equation.

### Step 3: Calculate the Discriminant

The discriminant, given by b² - 4ac, determines the nature of the roots:

- If b² - 4ac > 0, there are two real distinct roots
- If b² - 4ac = 0, there is one repeated real root
- If b² - 4ac < 0, there are two complex conjugate roots

### Step 4: Find r1 and r2

For real distinct roots, calculate r1 and r2 using the quadratic formula:

r1 = [-b + (b² - 4ac)] / (2a)

r2 = [-b - (b² - 4ac)] / (2a)

### Example:

Let's solve the characteristic equation 2r² + 5r - 3 = 0

- Identify a = 2, b = 5, c = -3
- Calculate the discriminant: b² - 4ac = 5² - 4(2)(-3) = 25 + 24 = 49
- Since the discriminant is positive, we have real distinct roots
- Apply the quadratic formula:

r1 = [-5 + 49] / (2(2)) = (-5 + 7) / 4 = 1/2

r2 = [-5 - 49] / (2(2)) = (-5 - 7) / 4 = -3

### Importance of r1 and r2 in the General Solution

The roots r1 and r2 are crucial in forming the general solution of the differential equation. For real distinct roots, the general solution takes the form:

y = ce^(rx) + ce^(rx)

Where c and c are arbitrary constants determined by initial conditions.

### Conclusion

Solving the characteristic equation to find real distinct roots is a fundamental skill in differential equations. By mastering the quadratic formula and understanding how to find r1 and r2, you can confidently approach a wide range of second-order linear differential equations. Remember that these roots are the building blocks of the general solution, allowing you to describe the behavior of various physical and mathematical systems. Practice with different examples to reinforce your understanding of this critical process.

## Forming the General Solution

The general solution for a second-order linear homogeneous differential equation with real distinct roots is a crucial concept in differential equations. This solution takes the form y = c1e^(r1x) + c2e^(r2x), where r1 and r2 are the distinct roots of the characteristic equation. Understanding how to form and interpret this solution is essential for solving a wide range of differential equations.

To begin, let's break down the components of the general solution. The exponential functions e^(r1x) and e^(r2x) represent the fundamental solutions of the differential equation. These are derived from the characteristic equation, which is obtained by substituting y = e^(rx) into the original differential equation. The constants c1 and c2 are arbitrary constants that play a crucial role in tailoring the solution to specific initial conditions.

The significance of c1 and c2 cannot be overstated. These constants allow the general solution for a second-order linear homogeneous differential equation to be adapted to particular scenarios. By adjusting their values, we can satisfy various initial conditions and boundary values. Essentially, c1 and c2 represent the "weights" or "contributions" of each fundamental solution of the differential equation to the overall solution.

To illustrate how this general solution satisfies the original differential equation, let's consider an example. Suppose we have the differential equation y'' - 5y' + 6y = 0. The characteristic equation would be r^2 - 5r + 6 = 0, which factors to (r - 2)(r - 3) = 0. This gives us the roots r1 = 2 and r2 = 3.

Our general solution is thus y = c1e^(2x) + c2e^(3x). To verify that this satisfies the original equation, we can substitute it back in:

y' = 2c1e^(2x) + 3c2e^(3x)

y'' = 4c1e^(2x) + 9c2e^(3x)

Substituting these into the original equation:

(4c1e^(2x) + 9c2e^(3x)) - 5(2c1e^(2x) + 3c2e^(3x)) + 6(c1e^(2x) + c2e^(3x)) = 0

Simplifying:

(4c1e^(2x) - 10c1e^(2x) + 6c1e^(2x)) + (9c2e^(3x) - 15c2e^(3x) + 6c2e^(3x)) = 0

0 + 0 = 0

This demonstrates that our general solution indeed satisfies the original differential equation for any values of c1 and c2.

The power of this general solution lies in its flexibility. By choosing appropriate values for c1 and c2, we can meet specific initial conditions. For instance, if we know the value of y and y' at x = 0, we can set up a system of equations to solve for c1 and c2, thus obtaining a particular solution that satisfies both the differential equation and the given initial conditions.

In conclusion, the general solution y = c1e^(r1x) + c2e^(r2x) provides a powerful framework for solving second-order linear homogeneous differential equations with real distinct roots. By understanding the roles of the exponential functions e^(r1x) and e^(r2x) and the constants c1 and c2, we can adapt this solution to a wide range of problems in physics, engineering, and other fields where differential equations model real-world phenomena.

## Initial Value Problems and Particular Solutions

Initial value problems (IVPs) play a crucial role in the study of second-order differential equations. These problems involve finding a specific solution to a differential equation that satisfies given initial conditions. In the context of second-order differential equations, two initial conditions are necessary to uniquely determine a particular solution.

Second-order differential equations are characterized by the presence of the second derivative of the unknown function. The general form of a second-order linear differential equation is:

a(x)y'' + b(x)y' + c(x)y = f(x)

Where y is the unknown function, x is the independent variable, and a(x), b(x), c(x), and f(x) are given functions.

Two initial conditions are required for second-order differential equations because the general solution contains two arbitrary constants. These constants need to be determined to find a particular solution. Typically, the initial conditions specify the value of the function (y) and its first derivative (y') at a given point.

For example, initial conditions might be given as:

y(x0) = y0 and y'(x0) = y1

Where x0 is the initial point, y0 is the initial value of the function, and y1 is the initial value of the first derivative.

To solve an initial value problem for a second-order differential equation, follow these steps:

- Find the general solution of the differential equation.
- Apply the first initial condition to the general solution.
- Apply the second initial condition to the derivative of the general solution.
- Solve the resulting system of equations to determine the values of the arbitrary constants.
- Substitute the constants back into the general solution to obtain the particular solution.

Let's consider an example to illustrate this process:

Solve the initial value problem: y'' - 4y = 0, y(0) = 2, y'(0) = -3

Step 1: Find the general solution

The characteristic equation is r^2 - 4 = 0, which gives r = ±2

The general solution is y = C1e^(2x) + C2e^(-2x)

Step 2: Apply the first initial condition

y(0) = 2, so C1 + C2 = 2

Step 3: Apply the second initial condition

y' = 2C1e^(2x) - 2C2e^(-2x)

y'(0) = -3, so 2C1 - 2C2 = -3

Step 4: Solve for C1 and C2

From the equations C1 + C2 = 2 and 2C1 - 2C2 = -3, we get:

C1 = 1/4 and C2 = 7/4

Step 5: Substitute back into the general solution

The particular solution is y = (1/4)e^(2x) + (7/4)e^(-2x)

This example demonstrates how initial conditions are used to find a particular solution to a second-order differential equation. By providing two specific pieces of information about the function at a given point, we can determine the unique solution that satisfies these conditions.

Initial value problems have numerous applications in physics, engineering, and other sciences. They are used to model real-world phenomena where the initial state of a system is known, and we want to

## Applications and Examples

Differential equations with real distinct roots have numerous real-world applications across various fields, including physics, engineering, and biology. These equations are crucial in modeling and solving complex problems that involve rates of change. Let's explore some practical examples and solve a complete problem to illustrate their importance.

In physics, differential equations are used to describe motion, heat transfer, and wave propagation. For instance, the motion of a damped harmonic oscillator, such as a spring-mass system with friction, is modeled using a second-order differential equation. The solution to this equation helps predict the position of the mass over time, accounting for factors like initial displacement, velocity, and damping coefficient.

In electrical engineering, RC (resistor-capacitor) circuits are analyzed using first-order differential equations. These equations describe how the voltage across a capacitor changes over time when connected to a resistor and a voltage source. The solution provides insights into the charging and discharging behavior of the circuit, which is crucial for designing timing circuits and filters.

Environmental scientists use population growth differential equations to model population growth and decay. The exponential growth model or decay of a population can be described by a first-order differential equation, where the rate of change is proportional to the current population size. This model helps in predicting future population sizes and understanding the factors affecting growth rates.

Now, let's solve a complete problem to demonstrate the application of differential equations with real distinct roots:

Problem: A 2-kg mass is attached to a spring with a spring constant of 8 N/m. The system is placed in a medium that exerts a damping force proportional to the velocity, with a damping coefficient of 4 N·s/m. If the mass is displaced 0.5 m from its equilibrium position and released from rest, find the equation of motion for damped systems.

Step 1: Set up the differential equation

The equation of motion for damped systems is:

m(d²x/dt²) + c(dx/dt) + kx = 0

Where m is mass, c is the damping coefficient, k is the spring constant, and x is displacement.

Step 2: Substitute the given values

2(d²x/dt²) + 4(dx/dt) + 8x = 0

Step 3: Rearrange the equation

d²x/dt² + 2(dx/dt) + 4x = 0

Step 4: Find the characteristic equation

r² + 2r + 4 = 0

Step 5: Solve for the roots

Using the quadratic formula: r = [-2 ± (2² - 4(1)(4))] / 2(1)

r = -1 ± -3

r = -1 + i3 and r = -1 - i3

Step 6: Write the general solution

x(t) = e(Ccos(3t) + Csin(3t))

Step 7: Apply initial conditions

At t = 0, x = 0.5 and dx/dt = 0

0.5 = C

0 = -C + 3C

Solving these equations: C = 0.5 and C = 0.5/3

Step 8: Write the particular solution

x(t) = e(0.5cos(3t) + (0.5/3)sin(3t))

This solution describes the position of the mass as a function of time, accounting for factors like initial displacement, velocity, and damping coefficient. Such population growth differential equations are essential in various scientific and engineering applications.

## Conclusion

In summary, characteristic equations with real distinct roots play a crucial role in solving differential equations. The introduction video provides a solid foundation for understanding this topic, highlighting key concepts and solution methods. To reinforce your knowledge, it's essential to practice solving problems involving characteristic equations. This hands-on approach will help you develop intuition and problem-solving skills. As you progress, consider exploring related topics in differential equations, such as complex roots and repeated roots, to broaden your understanding. Remember that mastering characteristic equations is fundamental to tackling more advanced concepts in differential equations. By combining the insights from the video with regular practice and further study, you'll be well-equipped to handle increasingly complex problems in this field. Don't hesitate to revisit the video and seek additional resources to solidify your grasp of this important mathematical concept.

### FAQs

Here are some frequently asked questions about characteristic equations with real distinct roots:

#### 1. What are distinct roots in mathematics?

Distinct roots refer to different solutions of an equation. In the context of characteristic equations, distinct roots mean that the equation has two different real number solutions. These roots are unique and do not repeat.

#### 2. How do you find real and distinct roots?

To find real and distinct roots, you need to solve the characteristic equation. This typically involves using the quadratic formula: x = [-b ± (b² - 4ac)] / (2a). If the discriminant (b² - 4ac) is positive, the equation has two real and distinct roots.

#### 3. What is an example of two distinct real roots?

Consider the equation x² - 5x + 6 = 0. Using the quadratic formula, we find that the roots are x = 3 and x = 2. These are two distinct real roots because they are different real numbers.

#### 4. When are roots real and distinct?

Roots are real and distinct when the discriminant (b² - 4ac) of the quadratic equation ax² + bx + c = 0 is positive. This means that b² - 4ac > 0. In this case, the quadratic formula yields two different real number solutions.

#### 5. How do you find the distinct real roots of a polynomial?

For a quadratic polynomial, use the quadratic formula. For higher-degree polynomials, you may need to use factoring, synthetic division, or numerical methods like the Newton-Raphson method. The key is to find all real solutions and ensure they are different from each other.

### Prerequisite Topics

Understanding the characteristic equation with real distinct roots is a crucial concept in advanced mathematics, particularly in differential equations. To fully grasp this topic, it's essential to have a solid foundation in several prerequisite areas. Let's explore how these fundamental concepts contribute to your understanding of characteristic equations.

First and foremost, a strong grasp of applications of linear equations is vital. This knowledge forms the basis for understanding how characteristic equations are derived and applied in real-world scenarios. Similarly, distance and time related questions in linear equations provide practical context for the application of characteristic equations in physics and engineering problems.

As we delve deeper into the topic, modeling with differential equations becomes increasingly important. This skill allows you to translate real-world phenomena into mathematical models, which often involve characteristic equations. For instance, population growth models frequently utilize differential equations with characteristic equations that have real distinct roots.

Understanding parallel line equations may seem unrelated at first, but it actually helps in visualizing the behavior of solutions to differential equations. This geometric perspective is invaluable when interpreting the results of characteristic equations.

A crucial skill in this journey is determining the number of solutions to linear equations. This concept directly translates to understanding the nature of roots in characteristic equations, which is fundamental to our main topic.

The study of continuous growth and decay provides real-world applications for characteristic equations with real distinct roots. Many natural and economic phenomena follow exponential patterns, which are often modeled using differential equations with such characteristic equations.

Perhaps one of the most directly relevant prerequisites is using the quadratic formula to solve quadratic equations. The characteristic equation is often a quadratic equation, and the ability to solve it efficiently is crucial for finding the roots that determine the behavior of the differential equation's solutions.

Lastly, graphing from slope-intercept form provides a visual understanding of linear functions, which is helpful when interpreting the solutions derived from characteristic equations with real distinct roots.

By mastering these prerequisite topics, you'll build a strong foundation for understanding characteristic equations with real distinct roots. This knowledge will enable you to solve complex differential equations, model real-world phenomena, and gain deeper insights into the behavior of dynamic systems in various fields of science and engineering.

$Ay''+By'+Cy=0$

By using the characteristic equation:By using the characteristic equation:

$Ar^2+Br+C=0$

And solving this quadratic will yield two roots, $r_1,r_2$. Let's suppose that both $r_1$ and $r_2$ are distinct and real.

So the solution will be:

$y_1 (x)=e^{r_1 x}$ $y_2 (x)=e^{r_2 x}$

Or in full generality:

$y(x)=c_1 e^{r_1 x}+c_2 e^{r_2 x}$

This is the general solution. We can find a particular solution with initial parameters

$y(x_0 )=y_0, y' (x_1 )=y_1$.

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