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Interactive Audio Lesson
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Introduction to Solved Examples
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Today, we'll be exploring solved examples that illustrate how to handle second-order homogeneous linear differential equations. These examples will help us understand the application of the characteristic equation.
What exactly do you mean by characteristic equation?
Great question! The characteristic equation is derived from the differential equation. It helps us find the roots that determine the general solution of the equation.
Are there different types of roots we should be aware of?
Yes! The types of roots can be distinct real roots, repeated real roots, or complex roots, each affecting the solution structure differently.
Let's move into our first example and apply these concepts.
Example 1: Real and Distinct Roots
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Let's solve the equation: d²y/dx² - 7dy/dx + 10y = 0 with initial conditions y(0) = 3 and y′(0) = 5.
How do we start?
First, we form the characteristic equation, which is r² - 7r + 10 = 0. Can anyone solve this?
The roots are 2 and 5, right?
Exactly! With distinct real roots, the general solution will be in the form y(x)=C₁e^(2x) + C₂e^(5x).
And we apply the initial conditions to find C₁ and C₂?
Correct! By substituting y(0) and y′(0), we can solve for the constants.
Now let's summarize this example. We derived a general solution based on distinct roots, applied initial conditions, and found specific constants.
Example 2: Repeated Roots
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In our second example, we solve the equation d²y/dx² - 4dy/dx + 4y = 0, with initial conditions y(0) = 2 and y′(0) = -1.
This looks different already; I see that the characteristic equation has repeated roots.
Yes! The roots are both 2, leading us to the general solution form y(x) = (C₁ + C₂x)e^(2x).
So how do you apply the initial conditions here?
We substitute y(0) to get C₁ = 2. Then we use y′(0) to solve for C₂.
In this case, we get a polynomial factor due to the repeated roots?
That's right! It’s crucial to note how different root types lead to distinct solution forms. Let's recap what we've learned in this example.
Introduction & Overview
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Quick Overview
Standard
This section delineates two significant examples: one involving real distinct roots and the other with repeated roots. The examples illustrate the process of utilizing initial conditions to derive the final solution of the differential equations.
Detailed
Solved Examples
This section covers the process of solving second-order homogeneous linear differential equations with constant coefficients through two key examples. The first example demonstrates the scenario where the roots of the characteristic equation are real and distinct, leading to an exponential solution that incorporates constants derived from initial conditions. The second example explores repeated roots, revealing a different structure for the solution that includes a polynomial factor. The primary steps involve setting up the characteristic equation, calculating roots, applying initial conditions, and ultimately expressing the general solution. This methodology is crucial in engineering applications where these equations commonly arise.
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Example 1: Real and Distinct Roots
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Example 1: Real and Distinct Roots
Solve:
d^2y/dx^2 - 7 dy/dx + 10y = 0, y(0) = 3, y'(0) = 5
Step 1: Characteristic equation:
r^2 - 7r + 10 = 0 ⇒ r = 2, 5
General solution:
y(x) = C₁ e^{2x} + C₂ e^{5x}
Apply initial conditions:
y(0) = C₁ + C₂ = 3 (i)
y'(x) = 2C₁ e^{2x} + 5C₂ e^{5x} ⇒ y'(0) = 2C₁ + 5C₂ = 5 (ii)
Solve equations (i) and (ii):
From (i): C₁ = 3 - C₂
Substitute into (ii):
2(3 - C₂) + 5C₂ = 5 ⇒ 6 - 2C₂ + 5C₂ = 5 ⇒ 3C₂ = -1 ⇒ C₂ = -1/3
Then C₁ = 3 + 1/3 = 10/3
Final solution:
y(x) = (10/3)e^{2x} - (1/3)e^{5x}
Detailed Explanation
In Example 1, we are solving a second-order differential equation with constant coefficients. The first step involves finding the characteristic equation derived from the original differential equation. This quadratic equation helps us to find the roots (in this case, r=2 and r=5). After determining the characteristic roots, we construct the general solution in the form of an exponential function. We then apply the initial conditions, y(0) and y'(0), which provide specific values that allow us to solve for the constants C₁ and C₂. By substituting the known values into two equations, we can isolate and calculate these constants, ultimately giving us the specific solution to the problem.
Examples & Analogies
Imagine a scenario where you're observing the population growth in a city. The population growth can be modeled by a differential equation, similar to the one we solved. Just like how we find specific constants from generic equations, some factors like birth rates and immigration can affect the overall population growth, which we can calculate using similar methods.
Example 2: Repeated Roots
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Example 2: Repeated Roots
Solve:
d^2y/dx^2 - 4dy/dx + 4y = 0, y(0) = 2, y'(0) = -1
Characteristic equation:
r^2 - 4r + 4 = 0 ⇒ r = 2 (repeated)
General solution:
y(x) = (C₁ + C₂ x)e^{2x}
Apply initial conditions:
y(0) = C₁ = 2 (i)
y'(x) = [C₂ + 2(C₁ + C₂ x)]e^{2x} ⇒ y'(0) = C₂ + 4 = -1 (ii)
From (i): C₁ = 2
Substitute into (ii): 2 + C₂ + 4 = -1 ⇒ C₂ = -7
Final solution:
y(x) = (2 - 7x)e^{2x}
Detailed Explanation
In Example 2, we deal with a second-order differential equation that leads to a repeated root in the characteristic equation. This means both roots are the same, which alters how we form the general solution. The presence of a repeated root introduces a polynomial component (in this case, C₂ x) multiplied by an exponential term. We again utilize the initial conditions to solve for the constants. By substituting the initial conditions into the equation formed from the general solution, we find the constants necessary to specify the solution uniquely.
Examples & Analogies
Consider a situation where a car's speed decreases as it approaches a stop sign. This deceleration can be compared to repeated roots in our differential equation. The car consistently applies brakes (like a constant damping effect) until it stops, like how a repeated root leads to a solution that decreases steadily over time. The terms in our equation mimic how different factors contribute to the overall behavior of the car's speed as it stops.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Characteristic Equation: A polynomial formed to determine roots of differential equations.
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Real Distinct Roots: Roots that are unique and lead to exponential solutions.
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Repeated Roots: Same roots that influence the form of the solution, introducing polynomial terms.
Examples & Real-Life Applications
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Examples
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Example of solving d²y/dx²−7dy/dx+10y=0 with initial conditions y(0)=3 and y′(0)=5.
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Example of solving d²y/dx²−4dy/dx+4y=0 with initial conditions y(0)=2 and y′(0)=−1.
Memory Aids
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🎵 Rhymes Time
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When roots are distinct, the solution is quick; e's and constants do the trick.
📖 Fascinating Stories
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In a land of equations, distinct roots danced in pairs, creating solutions that grew with grace, while repeated roots brought forth a polynomial heir.
🧠 Other Memory Gems
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DRR - Distinct Roots give Regular solutions while Repeated Roots yield Regularities.
🎯 Super Acronyms
ECR - Exponential for Complex roots, Regular for Repeated roots.
Flash Cards
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Glossary of Terms
Review the Definitions for terms.
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Term: Characteristic Equation
Definition:
A polynomial equation derived from a differential equation to determine its roots, critical for finding the general solution.
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Term: Homogeneous Differential Equation
Definition:
A differential equation in which the function and its derivatives appear in such a way that the equation equals zero.
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Term: Secondorder Differential Equation
Definition:
A differential equation involving the second derivative of a function.
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Term: Distinct Roots
Definition:
Roots of the characteristic equation that are different from each other.
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Term: Repeated Roots
Definition:
Roots that appear more than once in the characteristic equation.