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Interactive Audio Lesson
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Understanding Problem Types
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Today we're going to explore some practice problems that apply the method of undetermined coefficients. But before we dive in, can anyone remind me what types of functions this method can be applied to?
I know it works for polynomials, exponentials, and trigonometric functions.
Exactly! Great job, Student_1. Now, what about products of these functions?
Yes! If the right-hand side is a combination like x^2e^x, we can still use this method.
Correct! Remember, we cannot use this method for logarithmic or other non-standard functions.
Exploring Problem 1
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Let’s begin with our first practice problem. Solve y'' + 5y' + 6y = 3e^{-x}. What is our first step?
We need to solve the homogeneous equation first, right?
Right! So we set up the auxiliary equation. What do we get from that?
The auxiliary equation is r² + 5r + 6 = 0, which factors to (r + 2)(r + 3) = 0. So r = -2, -3.
Exactly! Now, what’s our complementary function?
y_c = C₁e^{-2x} + C₂e^{-3x}.
Perfect! Now, how about our particular integral?
Working Through Problem 2
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Now, let's tackle another problem: y'' - 4y = x². Can anyone summarize the steps we discussed before?
First, find the complementary function by solving the auxiliary equation.
Correct! And what’s our auxiliary equation for this case?
It's r² - 4 = 0, which gives us r = ±2.
Well done! That leads us to our complementary solution. What is it?
y_c = C₁e^{2x} + C₂e^{-2x}.
Now for the particular solution, how do we guess the form?
Since the right-hand side is x², we can try y_p = Ax² + Bx + C.
Correct! Don’t forget to substitute back into the equation after computing the derivatives.
Introduction & Overview
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Quick Overview
Standard
The practice problems challenge students to apply the method of undetermined coefficients in various contexts, reinforcing their understanding of how to determine particular integrals for different types of forcing functions.
Detailed
Practice Problems
This section includes a series of practice problems designed to test and reinforce the application of the method of undetermined coefficients to solve non-homogeneous linear differential equations. The problems presented here are structured to encourage students to utilize critical steps such as deriving the complementary function, proposing trial solutions based on the form of the forcing function, and substituting to find undetermined coefficients. Each problem has been chosen to reflect common scenarios students may encounter when using this method, thereby preparing them for real-world applications and further studies in differential equations.
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Audio Book
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Problem 1
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Solve y′′+5y′+6y =3e−x
Detailed Explanation
This problem requires solving a second-order linear non-homogeneous differential equation. The equation consists of a homogeneous part (y′′ + 5y′ + 6y = 0) and a non-homogeneous part with a forcing function (3e−x). Step 1 will involve finding the complementary function by solving the characteristic equation associated with the homogeneous part. Step 2 will involve guessing a particular integral that fits the form of the non-homogeneous term (3e−x). After substituting this guess into the original equation and solving for unknown coefficients, we will combine the results to obtain the general solution.
Examples & Analogies
Consider a scenario where you're trying to model how an object behaves when an external force (like a spring in damping motion) is applied. The homogeneous part represents the natural behavior of the system without external force, while the forcing function represents how the system responds to that external push.
Problem 2
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Solve y′′−4y = x²
Detailed Explanation
For this problem, we also have a second-order linear non-homogeneous differential equation. The first step is to find the complementary function by solving the homogeneous equation (y′′ - 4y = 0). The characteristic polynomial will be r² - 4 = 0, which yields roots r = 2 and r = -2. This leads to the complementary function y_c = C₁e^(2x) + C₂e^(-2x). Next, we guess a particular solution for the non-homogeneous term (x²) by assuming a trial solution that fits the polynomial form. Once substituted into the original equation and simplified, we solve for coefficients to find the particular integral. Combining both solutions gives the general solution.
Examples & Analogies
Think of this as modeling the displacement of a beam under a specific load – here, the load can be represented as the polynomial term x². The first part of the problem shows how the beam moves without any load, allowing us to see its natural deflection before understanding how additional loads like x² modify its behavior.
Problem 3
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Solve y′′+y = xsin(x)
Detailed Explanation
This differential equation features a forced term involving a product of functions (xsin(x)). The complementary function is first found by solving the associated homogeneous equation (y′′ + y = 0). We find y_c = C₁cos(x) + C₂sin(x). Next, for the particular solution, we guess a trial function reflecting the non-homogeneous term. Since both x and sin(x) are present, we could assume a form such as x(Acos(x) + Bsin(x)). After determining the derivatives and substituting into the original equation, we solve for coefficients A and B by matching against the non-homogeneous term.
Examples & Analogies
Imagine swinging on a swing while someone is pushing you periodically. Your natural swinging motion represents the homogeneous solution, while the multiple pushes during your swing can be modeled by the xsin(x) term. This analogy helps visualize how we can analyze both the natural behavior and responses due to external influences.
Problem 4
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Solve y′′+2y′+y = e−x
Hint: The complementary function here will include e−x, so use modification.
Detailed Explanation
In this problem, the method starts similarly by deriving the complementary function from the homogeneous part (y′′ + 2y′ + y = 0), leading us to a characteristic equation with roots that provide a solution of the form y_c = C₁e^(-x) + C₂xe^(-x). Notably, we have to modify the form of our trial solution for the particular integral because e^(-x) appears in y_c. Therefore, our trial guess must be x²e^(-x) (to eliminate duplication). This approach avoids overlap with the complementary function when we substitute this into the original equation to solve for coefficients and find the complete general solution.
Examples & Analogies
This can relate to how a damped oscillator behaves under a steady external force. You have the natural decay of the oscillator's movement represented by the e^(-x) form, while the external push must be modeled correctly to ensure we capture the whole behavior of the system, ensuring we handle overlaps effectively.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Method of Undetermined Coefficients: A technique to find particular integrals of non-homogeneous linear differential equations.
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Complementary Function: The solution to the associated homogeneous equation.
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Particular Integral: The solution that takes into account the non-homogeneous term.
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Auxiliary Equation: The equation obtained from the homogeneous part of a differential equation.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example: Solve y'' - 4y = x², applying the method of undetermined coefficients.
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Example: Solve y'' + 5y' + 6y = 3e^{-x}, using trial solutions.
Flash Cards
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Glossary of Terms
Review the Definitions for terms.
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Term: Complementary Function (y_c)
Definition:
The solution to the homogeneous part of a differential equation.
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Term: Particular Integral (y_p)
Definition:
The solution to the non-homogeneous part of a differential equation.
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Term: Auxiliary Equation
Definition:
A polynomial equation derived from the differential equation used to find the complementary function.
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Term: NonHomogeneous Term
Definition:
The term in a differential equation that is not a function of the dependent variable and its derivatives.