1.5.3 - Example
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Understanding the Auxiliary Equation
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Today, we're going to solve a second-order homogeneous linear differential equation. Can anyone tell me what the auxiliary equation is?
Is it the equation we get by replacing derivatives with powers of m?
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?
We can factor it to find the roots.
Correct! The roots will help us form the general solution. Can anybody compute the roots?
The roots are $m_1 = 2$ and $m_2 = 3$.
Spot on! With these roots, how do we express the general solution?
It's $$y = C_1e^{2x} + C_2e^{3x}$$, right?
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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Now that we have our general solution, let's talk about the constants $C_1$ and $C_2$. What do they represent?
They are the constants that we find using initial conditions!
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?
It helps us model real-life phenomena accurately!
Correct! Applying these concepts helps engineers predict behaviors in structures and materials.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
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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Problem Statement
Chapter 1 of 4
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Chapter Content
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
Chapter 2 of 4
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Chapter Content
• 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
Chapter 3 of 4
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Chapter Content
• 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
Chapter 4 of 4
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Chapter Content
• 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.
Key Concepts
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Auxiliary Equation: The polynomial obtained from a differential equation that is used to find the roots.
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General Solution: The complete solution to the differential equation given in terms of constants.
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Homogeneous Equation: A type of differential equation where the output and derivatives are present and no additional functions exist.
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Constant Coefficients: The fixed numerical values in front of the variables in a differential equation.
Examples & Applications
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
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Rhymes
Roots we find to solve the case, the auxiliary equation sets the pace.
Stories
In a math kingdom, the roots of the auxiliary equation become knights that help discover the general solution for each problem.
Memory Tools
Remember 'A G H' for 'Auxiliary General Homogeneous' to recall the four steps in solving differential equations.
Acronyms
Use 'R.E.A.D.' for 'Roots, Equation, Auxiliary, Derivatives' when tackling differential equations.
Flash Cards
Glossary
- Auxiliary Equation
A polynomial equation derived from a linear differential equation that helps in determining the general solution.
- General Solution
The complete set of solutions to a differential equation, typically involving arbitrary constants representing initial conditions.
- Homogeneous Equation
A differential equation in which all terms involve the dependent variable and its derivatives, with no free-standing functions.
- Real Distinct Roots
Roots of a polynomial that are real numbers and are not equal to each other.
- Constant Coefficients
Coefficients in a differential equation that do not change over time or space.
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