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Introduction to Lagrange’s Linear Equation
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Today, we're going to explore Lagrange’s Linear Equation and how it serves as a powerful tool for solving partial differential equations. Can anyone remind me of the standard form of this equation?
Is it Pp + Qq = R?
Exactly! In simple terms, P, Q, and R are functions of x, y, and z. Now, why do we use Lagrange’s method? What are its advantages?
It simplifies PDEs into systems of ODEs?
Right! This transformation is crucial for finding solutions. By using characteristic equations, we can analyze the behavior of solutions more easily. Let’s remember: **L for Lagrange, L for Lines** — it helps us recall the linear nature of the method.
So, it’s like turning a complex problem into simpler parts?
Precisely! Now, let’s look at a specific example to see all this in action.
Example 1 - Solving PDE
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Let’s consider our Example 1: we need to solve ∂z/∂x + ∂z/∂y = z. Who can identify P, Q, and R for me?
P is 1, Q is 1, and R is z!
Great job! Now, what are our auxiliary equations?
They are dx = 1, dy = 1, dz = z.
Correct! Now let’s integrate to find our solutions. What can we derive from dx = dz?
We find out that x = ln(z) + c.
Exactly! And from this, we identify u and v. So, what is our final general solution?
It’s φ(x - y, ze^(-x)) = 0 or z = ef(x - y)!
Well done! This shows how we can systematically apply Lagrange's method to find solutions.
Example 2 - Another Application
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Now, let’s move on to Example 2, which deals with the equation ∂z/∂x - y∂z/∂y = 0. What can we identify for P and Q here?
P is y and Q is -x.
Correct! Here, there’s no R term, making it unique. Now, what do we get from our auxiliary equations?
We have dx = dy and dx = 0.
Right! This leads us to finding the general solution. What relationships do we derive?
We find x² + y² = c.
And z remains a constant, z = c.
Excellent! So our general solution would be φ(x² + y², z) = 0. This highlights how certain configurations in PDEs lead us to different forms of solutions.
Introduction & Overview
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Quick Overview
Standard
The section covers two detailed examples of using Lagrange's Linear Equation to solve partial differential equations. It walks through the formulation of the auxiliary equations and the derivation of general solutions, thereby enhancing the understanding of the characteristic method.
Detailed
Detailed Summary
This section focuses on solved examples that demonstrate the application of Lagrange’s Linear Equation, crucial for solving first-order partial differential equations (PDEs). The section presents two specific examples:
- Example 1 demonstrates how to solve the equation ∂z/∂x + ∂z/∂y = z. The teacher explains how to identify the coefficients P, Q, and R, subsequently formulating the auxiliary equations and extracting the general solution. Here, the characteristic curves u and v are derived, leading to the general solution of the PDE in terms of a function of u and v.
- Example 2 provides insights on solving the equation ∂z/∂x - y∂z/∂y = 0. Similar steps are followed, with a slight variation in the solutions due to different functions for P and Q, which results in a different characteristic curve relationship.
The examples solidify the reader's understanding of transforming PDEs into ODEs via the method of characteristics and provide practical solutions that exemplify the theoretical concepts introduced earlier in the chapter.
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Example 1
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Solve:
∂𝑧 ∂𝑧
𝑝 + 𝑞 = 𝑧 ⇒ + = 𝑧
∂𝑥 ∂𝑦
Solution: Here,
• 𝑃 = 1, 𝑄 = 1, 𝑅 = 𝑧
Auxiliary equations:
𝑑𝑥 𝑑𝑦 𝑑𝑧
= =
1 1 𝑧
𝑑𝑥 𝑑𝑦
From = ⇒ 𝑥 − 𝑦 = 𝑐_1
1 1
From 𝑑𝑥 = 𝑑𝑧 ⇒ ∫𝑑𝑥 = ∫ 𝑑𝑧 ⇒ 𝑥 = ln𝑧 +𝑐_2 ⇒ 𝑧 = 𝑒^{𝑥−𝑐_2}
1 𝑧 𝑧 2
Now,
• 𝑢 = 𝑥 − 𝑦 = 𝑐_1
• 𝑣 = 𝑧𝑒^{−𝑥} = 𝑐_2
General solution:
𝜙(𝑥−𝑦,𝑧𝑒^{−𝑥}) = 0 or 𝑧 = 𝑒^{𝑥}𝑓(𝑥− 𝑦)
Detailed Explanation
In this example, we start with a partial differential equation (PDE) where the relationship between changes in z with respect to x and y is defined. We set P, Q, and R from the equation, identifying them as P=1, Q=1, and R=z. Next, we form the auxiliary equations and solve them. From the equation dx/dy = 1, we find that x - y = c1. We also solve dx/dz, leading to an integration that gives us the relation between x and z. Finally, we identify two independent solutions and express the general solution in terms of u and v, yielding the general solution format.
Examples & Analogies
Imagine you are tracking temperature changes in a room based on time and external weather. If you know the temperature at two different times and can represent those changes with a formula, just as we did with z in the example, you can extrapolate what the temperature will be at any point in the future. This is similar to how we derive solutions in mathematics by looking at the relationships and changes defined in the problem.
Example 2
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Solve:
∂𝑧 ∂𝑧
𝑦𝑝 −𝑥𝑞 = 0 (i.e., 𝑦 −𝑥 = 0)
∂𝑥 ∂𝑦
Solution: Here,
• 𝑃 = 𝑦, 𝑄 = −𝑥, 𝑅 = 0
Auxiliary equations:
𝑑𝑥 𝑑𝑦 𝑑𝑧
= =
𝑦 −𝑥 0
From 𝑑𝑥 = 𝑑𝑦 ⇒ 𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0 ⇒ 𝑑(𝑥^2 + 𝑦^2) = 0 ⇒ 𝑥^2 + 𝑦^2 = 𝑐
𝑦 −𝑥 1
𝑑𝑥 𝑑𝑧
From = ⇒ 𝑑𝑧 = 0 ⇒ 𝑧 = 𝑐_2
𝑦 0 2
General solution:
𝜙(𝑥^2 + 𝑦^2, 𝑧) = 0 or 𝑧 = 𝑓(𝑥^2 + 𝑦^2)
Detailed Explanation
This example addresses a different kind of PDE. We again define P, Q, and R, with distinct values. The auxiliary equations lead us through a series of relationships, illustrating a system where we analyze the change in z by considering a circular symmetry in the x-y plane. We derive that changes in x and y are related to their squares, leading us to the conclusion that they form a circular equation. The final general solution represents a function of the radius squared, indicating that z remains constant along circles defined by the distance from the origin.
Examples & Analogies
Picture a forest as a circular area of trees surrounding a small pond. The further you stand from the center (the pond), the less likely you are to feel a breeze (represented by z in our example). As you walk in a circle around the pond, your position is determined by your distance from the center, similar to how we derived the circular symmetry in the second example. This analogy helps students visualize how variables related to position can affect outcomes in a mathematical sense.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Lagrange’s Linear Equation: A first-order PDE expressed in the format Pp + Qq = R.
-
Method of Characteristics: A technique to convert PDEs into ODEs using auxiliary equations.
-
General Solution: The family of solutions represented by arbitrary functions of the independent solutions derived.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
Example 1: Solve the PDE ∂z/∂x + ∂z/∂y = z, resulting in z = ef(x - y).
-
Example 2: Solve the PDE ∂z/∂x - y∂z/∂y = 0, leading to the solution z = f(x² + y²).
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
🎵 Rhymes Time
-
For Lagrange, we solve with ease, turn PDEs into ODEs, characteristic curves we do hold dear, for solutions, we can cheer!
📖 Fascinating Stories
-
Imagine a wise sage, Lagrange, who guides us through the maze of equations, showing us paths of characteristics that lead to treasures of solutions!
🧠 Other Memory Gems
-
Use the acronym 'CURE': C for Constants, U for Unraveling, R for Results, E for Equations — helps recall solving strategies.
🎯 Super Acronyms
The acronym 'SOLVE' for Lagrange
- S: for Standard form
- O: for Organizing variables
- L: for Lagrange's method
- V: for Variable substitution and E for Evaluate solutions.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Partial Differential Equation (PDE)
Definition:
An equation that involves partial derivatives of a function of several variables.
-
Term: Lagrange's Linear Equation
Definition:
A specific type of first-order PDE that can be solved using the method of characteristics.
-
Term: Auxiliary Equation
Definition:
The ordinary differential equations derived from a PDE to analyze characteristics.
-
Term: Characteristic Curves
Definition:
The curves along which the PDE reduces to an ordinary differential equation.
-
Term: General Solution
Definition:
The solution expressed in terms of arbitrary functions that encompass a family of solutions.