Solving Y'' - 2y' + Y = 0 Verifying Solutions And Correct Statements Guide

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Hey guys! Today, we're diving deep into the world of differential equations, specifically focusing on solving the homogeneous second-order linear differential equation y'' - 2y' + y = 0. This type of equation pops up everywhere in physics and engineering, so mastering it is super important. We'll not only walk through the solution process step-by-step but also emphasize how to verify your solutions and understand the nuances of what makes a statement about these solutions correct. Let's get started!

Understanding the Problem: A Second-Order Homogeneous Linear Differential Equation

Before we jump into solving, let's break down what we're dealing with. The equation y'' - 2y' + y = 0 is classified as a second-order homogeneous linear differential equation with constant coefficients. Whew, that’s a mouthful! Let's unpack it:

  • Second-order: This means the highest derivative of the unknown function y that appears in the equation is the second derivative, denoted as y''.
  • Homogeneous: This term indicates that the equation is set equal to zero. If it were equal to some non-zero function of x, it would be non-homogeneous.
  • Linear: This means that y and its derivatives appear only to the first power, and there are no products of y and its derivatives (e.g., no y y', no (y')²).
  • Constant coefficients: The coefficients of y'', y', and y are constants (in our case, 1, -2, and 1, respectively).

Recognizing these characteristics is crucial because it tells us which solution methods are applicable. For this type of equation, we can use a clever trick involving an auxiliary equation.

The Auxiliary Equation: Our Key to Unlocking the Solution

The core idea behind solving homogeneous linear differential equations with constant coefficients is to assume a solution of the form y = e^(rx), where r is a constant. This might seem like magic, but it works because the exponential function has the unique property that its derivatives are multiples of itself. When we substitute this assumed solution into the differential equation, we get:

  • y = e^(rx)
  • y' = re^(rx)
  • y'' = r²e^(rx)

Plugging these into our equation y'' - 2y' + y = 0, we obtain:

r²e^(rx) - 2re^(rx) + e^(rx) = 0

Now, we can factor out the common term e^(rx):

e^(rx)(r² - 2r + 1) = 0

Since e^(rx) is never zero for any real x, we can divide both sides by it, leaving us with the auxiliary equation:

r² - 2r + 1 = 0

This is a quadratic equation, which we can solve for r. The roots of this equation will dictate the form of our solutions.

Solving the Auxiliary Equation and the Case of Repeated Roots

The auxiliary equation r² - 2r + 1 = 0 is a classic quadratic. We can solve it by factoring, using the quadratic formula, or recognizing it as a perfect square. In this case, it factors nicely:

(r - 1)² = 0

This gives us a repeated root: r = 1. This is a crucial point because repeated roots require a slightly different approach when constructing the general solution.

When we have two distinct real roots (say, r₁ and r₂), the general solution is of the form:

y(x) = c₁e^(r₁x) + c₂e^(r₂x)

where c₁ and c₂ are arbitrary constants. However, with a repeated root, this doesn't quite work. If we just wrote y(x) = c₁e^(x) + c₂e^(x), we'd essentially have only one independent solution (since c₁e^(x) + c₂e^(x) = (c₁ + c₂)e^(x) = Ce^(x), where C is another constant).

To get a second linearly independent solution, we multiply the first solution by x. So, when we have a repeated root r, the general solution takes the form:

y(x) = c₁e^(rx) + c₂xe^(rx)

In our case, with r = 1, the general solution is:

y(x) = c₁e^(x) + c₂xe^(x)

This is the general solution to our differential equation. It represents a family of solutions, each corresponding to different values of the constants c₁ and c₂. These constants are determined by initial conditions, if given.

Verifying the Solution: Ensuring Accuracy

Okay, we've found a solution, but how do we know it's correct? This is where verification comes in. It's a crucial step to ensure we haven't made any mistakes along the way. The best way to verify is to plug our solution back into the original differential equation and see if it satisfies the equation.

Our general solution is y(x) = c₁e^(x) + c₂xe^(x). Let's find its first and second derivatives:

y'(x) = c₁e^(x) + c₂(e^(x) + xe^(x)) = c₁e^(x) + c₂e^(x) + c₂xe^(x)

y''(x) = c₁e^(x) + c₂e^(x) + c₂(e^(x) + xe^(x)) = c₁e^(x) + 2c₂e^(x) + c₂xe^(x)

Now, let's substitute y(x), y'(x), and y''(x) into our original equation y'' - 2y' + y = 0:

(c₁e^(x) + 2c₂e^(x) + c₂xe^(x)) - 2(c₁e^(x) + c₂e^(x) + c₂xe^(x)) + (c₁e^(x) + c₂xe^(x)) = 0

Let's simplify this expression. Grouping like terms, we get:

(c₁ - 2c₁ + c₁)e^(x) + (2c₂ - 2c₂)e^(x) + (c₂ - 2c₂ + c₂)xe^(x) = 0

This simplifies to:

0e^(x) + 0e^(x) + 0xe^(x) = 0

0 = 0

Since the equation holds true, our solution is verified! This gives us confidence that we've found the correct general solution. This verification step is super important, guys, because it can catch any errors we might have made in the differentiation or substitution steps.

Correct Statements and Misconceptions: Deepening Our Understanding

Beyond just finding the solution, it's essential to understand the correct statements we can make about solutions to differential equations. Here are a few key concepts and potential misconceptions to clarify:

  • Linear Independence: Two solutions, y₁(x) and y₂(x), are linearly independent if neither is a constant multiple of the other. In our case, e^(x) and xe^(x) are linearly independent. This means that neither can be written as a constant times the other. Linear independence is crucial because it ensures that our general solution (c₁y₁(x) + c₂y₂(x)) can represent all possible solutions to the differential equation.
  • General Solution vs. Particular Solution: The general solution, which we found as y(x) = c₁e^(x) + c₂xe^(x), represents a family of solutions. Each specific choice of the constants c₁ and c₂ gives us a particular solution. To find a particular solution, we need additional information, typically in the form of initial conditions (e.g., y(0) = 1, y'(0) = 0).
  • Uniqueness: If we have a second-order linear homogeneous differential equation and we specify two initial conditions (e.g., y(x₀) = y₀, y'(x₀) = y₁), then there exists a unique solution that satisfies both the differential equation and the initial conditions. This is a fundamental theorem in the theory of differential equations.
  • Superposition Principle: For linear homogeneous differential equations, the superposition principle states that if y₁(x) and y₂(x) are solutions, then any linear combination of them (i.e., c₁y₁(x) + c₂y₂(x), where c₁ and c₂ are constants) is also a solution. This is why the general solution is a linear combination of linearly independent solutions.

Common Misconceptions

Let's address some common misunderstandings that students often have:

  • Assuming all solutions are exponentials: While we assume a solution of the form e^(rx) to find the auxiliary equation, this doesn't mean all solutions are simply exponentials. The repeated root case demonstrates that we sometimes need to modify our assumed form (by multiplying by x) to find a complete set of linearly independent solutions.
  • Confusing general and particular solutions: It's crucial to understand the difference. The general solution is a formula containing arbitrary constants, while a particular solution is a specific solution obtained by plugging in values for those constants (usually determined by initial conditions).
  • Forgetting to verify: Always verify your solution! It's a simple step that can save you a lot of headaches. Plugging the solution back into the original equation is the best way to catch errors.

Conclusion: Mastering Differential Equations

So, guys, we've journeyed through solving the differential equation y'' - 2y' + y = 0. We've seen how to use the auxiliary equation, handle repeated roots, verify our solution, and understand the correct statements about solutions. This is a fundamental type of differential equation, and the techniques we've discussed are applicable to a wide range of similar problems.

The key takeaways are:

  • Understand the characteristics of the differential equation (order, homogeneity, linearity, constant coefficients).
  • Use the auxiliary equation to find the roots, and pay attention to repeated roots.
  • Construct the general solution based on the roots of the auxiliary equation.
  • Always verify your solution by plugging it back into the original equation.
  • Grasp the concepts of linear independence, general vs. particular solutions, uniqueness, and the superposition principle.

Keep practicing, and you'll become a differential equation master in no time! Remember, these equations are powerful tools for modeling real-world phenomena, so the effort you put in now will pay off big time. Happy solving!