Example - 1.5.3 | 1. Linear Differential Equations | Mathematics (Civil Engineering -1)
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Understanding the Auxiliary Equation

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0:00
Teacher
Teacher

Today, we're going to solve a second-order homogeneous linear differential equation. Can anyone tell me what the auxiliary equation is?

Student 1
Student 1

Is it the equation we get by replacing derivatives with powers of m?

Teacher
Teacher

Exactly! We convert the differential equation to a polynomial form. For our example, the differential equation is $$y'' - 5y' + 6y = 0$$. The auxiliary equation is $$m^{2} - 5m + 6 = 0$$. Does anyone know how to solve this quadratic?

Student 2
Student 2

We can factor it to find the roots.

Teacher
Teacher

Correct! The roots will help us form the general solution. Can anybody compute the roots?

Student 3
Student 3

The roots are $m_1 = 2$ and $m_2 = 3$.

Teacher
Teacher

Spot on! With these roots, how do we express the general solution?

Student 4
Student 4

It's $$y = C_1e^{2x} + C_2e^{3x}$$, right?

Teacher
Teacher

Exactly! And $C_1$ and $C_2$ are the constants determined by initial conditions. Great job, team!

Exploring the General Solution

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0:00
Teacher
Teacher

Now that we have our general solution, let's talk about the constants $C_1$ and $C_2$. What do they represent?

Student 1
Student 1

They are the constants that we find using initial conditions!

Teacher
Teacher

Exactly! So, if we have initial conditions like the value of $y$ at a specific point, we can determine these constants. Why is this important in engineering?

Student 2
Student 2

It helps us model real-life phenomena accurately!

Teacher
Teacher

Correct! Applying these concepts helps engineers predict behaviors in structures and materials.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section provides a practical example of solving a second-order homogeneous linear differential equation with constant coefficients.

Standard

The section illustrates the solution process of a second-order homogeneous linear differential equation through a detailed example, demonstrating the auxiliary equation method, which leads to finding the general solution based on the roots derived from the auxiliary equation.

Detailed

Example of Homogeneous Linear Differential Equation with Constant Coefficients

In this section, we take a closer look at a specific second-order homogeneous linear differential equation given in the format:
$$\frac{d^{2}y}{dx^{2}} - 5\frac{dy}{dx} + 6y = 0$$
To find the solution, we first need to construct the Auxiliary Equation (AE) related to the differential equation
$$m^{2} - 5m + 6 = 0$$
Solving this quadratic equation yields two distinct real roots, which are indicated as $m_1 = 2$ and $m_2 = 3$. With these roots, we can then infer the general solution of the differential equation as:
$$y = C_1e^{2x} + C_2e^{3x}$$
Where $C_1$ and $C_2$ are constants determined by initial or boundary conditions. This example highlights the process used in solving homogeneous linear differential equations with constant coefficients and the significance of identifying roots to form the general solution.

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Audio Book

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Problem Statement

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Solve:
d²y/dx² - 5 dy/dx + 6y = 0

Detailed Explanation

This chunk presents the differential equation that we need to solve. It is a second-order linear homogeneous differential equation, indicated by the presence of derivatives of y up to the second order (d²y/dx²) and no explicit external driving force (non-homogeneous term). To solve this type of equation, we need to find a solution that satisfies this equation for all values of x.

Examples & Analogies

Think of this equation as a representation of how a spring might behave when it's stretched or compressed. The terms in the equation describe the forces acting on the spring, with the goal of understanding how it will return to its original position after being disturbed.

Auxiliary Equation

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• AE: m² - 5m + 6 = 0
• Solve for roots m₁, m₂

Detailed Explanation

To solve the differential equation, we first form the Auxiliary Equation (AE) from the given differential equation. This transforms our problem of differential equations into an algebraic form. The AE helps us determine the roots m₁ and m₂ that will contribute to our general solution. We can solve this quadratic equation using the quadratic formula or by factoring.

Examples & Analogies

Imagine trying to find the points at which a parabolic shape (like a water fountain’s arc) crosses the ground (the x-axis). The solutions or roots to this quadratic equation tell us where those intersections (real values of x) occur.

Finding Roots

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• m = 2, 3

Detailed Explanation

By solving the Auxiliary Equation, we find that the roots are m = 2 and m = 3. These roots are the key to forming our general solution. Each root corresponds to a particular solution that we can combine for the complete solution to the original differential equation.

Examples & Analogies

Consider a race where two cars, one fast and one slower, leave the starting line at the same time. The points (roots) where they meet on the track can be seen as critical junctions that help us predict how the race (the solution) will progress.

General Solution

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• General solution: y = C₁e²ˣ + C₂e³ˣ

Detailed Explanation

With the found roots m₁ and m₂, we can now state the general solution to the differential equation. The general solution combines both exponential functions based on the roots we found, with C₁ and C₂ being arbitrary constants that represent the influence of initial conditions on the solution.

Examples & Analogies

Imagine creating a custom blend of tea by mixing two different types of tea leaves. C₁ and C₂ represent the proportions of each type in your blend. Depending on how much of each you add (initial conditions), the flavor (the shape of the solution) changes, but both types of leaves (the solution components) remain crucial.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Auxiliary Equation: The polynomial obtained from a differential equation that is used to find the roots.

  • General Solution: The complete solution to the differential equation given in terms of constants.

  • Homogeneous Equation: A type of differential equation where the output and derivatives are present and no additional functions exist.

  • Constant Coefficients: The fixed numerical values in front of the variables in a differential equation.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • For the equation $$\frac{d^{2}y}{dx^{2}} - 5\frac{dy}{dx} + 6y = 0$$, derive the auxiliary equation and find the general solution.

  • Given the equation $$\frac{d^{2}y}{dx^{2}} + 4y = 0$$, identify the roots and determine the general solution.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎵 Rhymes Time

  • Roots we find to solve the case, the auxiliary equation sets the pace.

📖 Fascinating Stories

  • In a math kingdom, the roots of the auxiliary equation become knights that help discover the general solution for each problem.

🧠 Other Memory Gems

  • Remember 'A G H' for 'Auxiliary General Homogeneous' to recall the four steps in solving differential equations.

🎯 Super Acronyms

Use 'R.E.A.D.' for 'Roots, Equation, Auxiliary, Derivatives' when tackling differential equations.

Flash Cards

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Glossary of Terms

Review the Definitions for terms.

  • Term: Auxiliary Equation

    Definition:

    A polynomial equation derived from a linear differential equation that helps in determining the general solution.

  • Term: General Solution

    Definition:

    The complete set of solutions to a differential equation, typically involving arbitrary constants representing initial conditions.

  • Term: Homogeneous Equation

    Definition:

    A differential equation in which all terms involve the dependent variable and its derivatives, with no free-standing functions.

  • Term: Real Distinct Roots

    Definition:

    Roots of a polynomial that are real numbers and are not equal to each other.

  • Term: Constant Coefficients

    Definition:

    Coefficients in a differential equation that do not change over time or space.