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Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Finding the Homogeneous Solution
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Today, we will solve the differential equation \( y'' - y = e^x \). The first step is to find the general solution of the associated homogeneous equation, which is \( y'' - y = 0 \). Can anyone tell me how we can derive the characteristic equation?
Isn't it \( r^2 - 1 = 0 \)?
Correct! We factor that gives us roots \( r = 1 \) and \( r = -1 \). So the general solution to the homogeneous equation is...
Is it \( y_h(x) = C_1 e^x + C_2 e^{-x} \)?
Exactly! Great job. Now we have our first solution. Remember, we denote these solutions as \( y_1 \) and \( y_2 \). Now let’s move to the next step!
Computing the Wronskian
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Next, we need to find the Wronskian, which will help us evaluate the coefficients for the particular solution. Does anyone remember the formula for the Wronskian?
Yes, it’s \( W = y_1 y_2' - y_2 y_1' \).
That's right! Now can we define \( y_1 \) and \( y_2 \) to compute the Wronskian?
Sure, \( W = e^x(-e^{-x}) - e^{-x}(e^x) = -2 \).
Perfect! Now we have \( W = -2 \). It's essential for our next steps.
Finding the Particular Solution
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Now, onto calculating the derivatives for our particular solution! Who can recall the formulas for \( u_1' \) and \( u_2' \)?
I remember! It’s \( u_1' = -\frac{y_2 g(x)}{W} \) and \( u_2' = \frac{y_1 g(x)}{W} \).
Exactly! Great memory! Now substituting in our functions, what do we get for \( u_1' \)?
So we have \( u_1' = -\frac{e^{-x} imes e^x}{-2} = \frac{1}{2} \).
Excellent! Now let's find \( u_2' \) using the same steps.
Integrating for Coefficients
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Now, to find \( u_1 \) and \( u_2 \), we need to integrate the derivatives we calculated.
So, that means \( u_1 = \int \frac{1}{2} \, dx = \frac{x}{2} \) and \( u_2 = -\int \frac{e^{2x}}{2} \, dx = -\frac{e^{2x}}{4} \)?
Exactly, nice work! Now, how do we construct the particular solution?
We combine them: \( y_p = u_1 y_1 + u_2 y_2 = \frac{x}{2} e^x - \frac{e^{2x}}{4} e^{-x} \).
That's correct! Now we can finalize it and write down the general solution.
Finalizing the General Solution
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Finally, let's write the general solution for our differential equation. Does anyone want to summarize what we have?
We have \( y(x) = C_1 e^x + C_2 e^{-x} + \frac{x}{2} e^x - \frac{e^{x}}{4} \).
Excellent summary! This solution combines the homogeneous and particular solutions successfully.
I find it easier to follow the steps when we systematically break it down like this!
I'm glad to hear that! Remember, each step is important in the method of variation of parameters!
Introduction & Overview
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Quick Overview
Standard
In this section, a specific differential equation is solved using the variation of parameters method. By following systematic steps, including finding the homogeneous solution, computing the Wronskian, and constructing the particular solution, students can understand how to apply this powerful technique effectively.
Detailed
Example 1: Solving a Second-Order Differential Equation
In this section, the method of variation of parameters is applied to solve the equation \( y'' - y = e^x \). First, we identify the homogeneous equation associated with our differential equation, which is given by \( y'' - y = 0 \). The characteristic equation \( r^2 - 1 = 0 \) yields roots \( r = \pm 1 \), resulting in the general solution for the homogeneous equation:
\[y_h(x) = C_1 e^x + C_2 e^{-x}\]
The next step involves calculating the Wronskian \( W \) of the solutions:\n
\[ W = e^x e^{-x} - e^{-x} e^x = -2 \]\n
This Wronskian is crucial for determining the functions \( u_1 \) and \( u_2 \) in the particular solution. Using the formulas for the derivatives:
\[ u_1' = -\frac{y_2 g(x)}{W} \quad \text{and} \quad u_2' = \frac{y_1 g(x)}{W} \]
we find:
\[ u_1' = -\frac{e^{-x} e^x}{-2} = \frac{1}{2} \quad \text{and} \quad u_2' = \frac{e^x e^x}{-2} = -\frac{e^{2x}}{2} \]
After computing the integrals:
\[ u_1 = \int \frac{1}{2} \, dx = \frac{x}{2} \quad \text{and} \quad u_2 = -\int \frac{e^{2x}}{2} \, dx = -\frac{e^{2x}}{4} \]
the particular solution can be formed as follows:
\[ y_p = u_1 y_1 + u_2 y_2 = \frac{x}{2} e^x - \frac{e^{2x}}{4} e^{-x} = \frac{x}{2} e^x - \frac{e^x}{4} \]
Finally, combining the general solution yields:
\[ y(x) = C_1 e^x + C_2 e^{-x} + \frac{x}{2} e^x - \frac{e^x}{4} \]
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Audio Book
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Step 1: Homogeneous Equation
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Solve:
y′′−y =ex
Step 1: Homogeneous equation:
y′′−y =0⇒r2−1=0⇒r =±1
So:
y (x)=ex, y (x)=e−x
Detailed Explanation
In the first step, we start by rewriting the given differential equation, which is a second-order linear non-homogeneous equation. The first task is to solve the associated homogeneous equation, which is obtained by setting the right-hand side equal to zero. We solve for the roots of the characteristic equation resulting from the homogeneous form, r² - 1 = 0, which gives us the roots r = ±1. This means the two fundamental solutions to the homogeneous equation are e^x and e^(-x). Thus, the general solution to the homogeneous equation is a linear combination of these solutions.
Examples & Analogies
Think of the homogeneous equation as finding the natural behavior of a tuned musical instrument, where the two tuning frequencies correspond to the two solutions, e^x and e^(-x). Like different notes that resonate naturally, these solutions show how the system behaves without any external forces acting on it.
Step 2: Compute Wronskian
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Step 2: Compute Wronskian:
W =(cid:12) (cid:12)ex e−x (cid:12) (cid:12)=−2
(cid:12)ex −e−x(cid:12)
Detailed Explanation
In this step, we compute the Wronskian, which is a determinant used to check if our solutions are linearly independent. The Wronskian W is computed for the two solutions e^x and e^(-x). The formula for the Wronskian is W = y₁y₂' - y₂y₁'. Here, upon calculating, we find that W = -2, which is a non-zero value indicating that y₁ and y₂ are linearly independent and therefore valid for further calculations.
Examples & Analogies
Imagine the Wronskian as a measure of how different your notes sound when played together on a musical scale. If you strike the two keys (e^x and e^(-x)) and they sound distinct (the Wronskian is non-zero), it means they can continuously create beautiful harmonies, just like our solutions can combine to form the complete general solution.
Step 3: Compute Derivatives
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Step 3: Compute derivatives:
y g e−x·ex 1 x
u =− 2 =− = ⇒u =
1′ W −2 2 1 2
y g ex·ex e2x 1
u = 1 = =− ⇒u =− e2x
Detailed Explanation
Next, we compute the derivatives needed to find the particular solution using the variation of parameters method. For both u₁ and u₂, we use the formulas obtained for their derivatives with respect to the non-homogeneous term g = e^x. This leads us to the specific forms of u₁ and u₂ after substituting in the Wronskian calculated earlier. Here, we derive expressions for u₁ and u₂ that will later be integrated to find the particular solution.
Examples & Analogies
Think of computing these derivatives like preparing ingredients for a complex dish. Just as you measure out different spices according to your recipe, we are carefully calculating how much of each function will contribute to achieving a flavor profile (the correct particular solution) that will balance the overall taste (the final solution to our differential equation).
Step 4: Particular Solution
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Step 4: Particular solution:
y =u y +u y = ex− e2xe−x = ex− ex
Detailed Explanation
Finally, we construct the particular solution using the functions u₁ and u₂ that we derived previously. The particular solution yₚ is given by a combination of our independent solutions multiplied by the calculated u₁ and u₂. This results in yₚ = e^x - (1/4)e^(2x)e^(-x) which simplifies to the required form. This final step allows us to add this specific response to the overall behavior of our system, making it responsive to the external input defined by e^x.
Examples & Analogies
Picture this step as assembling the final dish after all individual components have been prepared. Just like in cooking, where you combine all the ingredients you’ve meticulously measured out to create a delicious meal, here we are combining our calculated contributions from u₁ and u₂ to form the complete particular solution that addresses the external influences defined in the original problem.
Step 5: General Solution
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So:
y(x)=C ex+C e−x+ − ex
Detailed Explanation
With our particular solution in hand, we can now write the general solution of our original equation. The general solution is a combination of the homogeneous solution and the particular solution. We express this as y(x) = C₁e^x + C₂e^(-x) + e^x/4, where C₁ and C₂ are constants determined by initial conditions. This encapsulates both the natural behavior of the system and its response to external forces.
Examples & Analogies
Think of this last assembly as having both your foundational recipe (how the dish should taste naturally without any additions) and the flavorful twists you added in the form of the particular solution. Just as the final dish incorporates all elements to represent the chef’s creative expression, the general solution captures the full response of our system to various influences.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Variation of Parameters: A technique for solving particular solutions of differential equations.
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Wronskian: Used to check linear independence of the solutions.
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Homogeneous Solution: The solution derived from the associated homogeneous equation.
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Particular Solution: The solution that satisfies the entire differential equation including the non-homogeneous part.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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In solving \( y'' + 4y = sin(x) \) using variation of parameters, you first find the homogeneous solution, then compute the Wronskian, and finally the particular solution utilizing the external forcing term \( g(x) = sin(x) \).
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When faced with \( y'' - y = e^x \), determine the characteristic roots, construct the homogeneous solution, calculate the Wronskian, and apply the derivatives to find the coefficients for the particular solution.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
🎵 Rhymes Time
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To find the Wronskian treat it right, e^x and e^{-x} will give you insight.
📖 Fascinating Stories
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Imagine a detective solving a case (the differential equation), first he looks for leads (the homogeneous solution), then checks the connections (the Wronskian), then puts together the final story (the particular solution).
🧠 Other Memory Gems
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HPC - Homogeneous, Parameter form, Combine solutions to finish.
🎯 Super Acronyms
HPC stands for Homogeneous, Particular, Combine, and it reminds students to follow these steps in order.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
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Term: Homogeneous Equation
Definition:
A differential equation where the right-hand side is zero.
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Term: Particular Solution
Definition:
A specific solution to a non-homogeneous differential equation that addresses the external input.
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Term: Wronskian
Definition:
A determinant used to determine the linear independence of solutions to differential equations.
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Term: Characteristic Equation
Definition:
An equation that is derived from a linear differential equation and used to find its solutions.
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Term: Variation of Parameters
Definition:
A method to find particular solutions to non-homogeneous differential equations.