Solving (D^2 - 4)y = E^(2x) + E^(-4x): Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of differential equations, specifically tackling the equation (D^2 - 4)y = e^(2x) + e^(-4x). If you're scratching your head wondering where to even begin, don't worry, we'll break it down step by step. This guide is designed to not only give you the solution but also to help you understand the underlying concepts. So, let’s get started!

Understanding the Problem

Before we jump into the solution, let's take a moment to understand what we're dealing with. The equation (D^2 - 4)y = e^(2x) + e^(-4x) is a second-order linear non-homogeneous differential equation. Sounds like a mouthful, right? Let's simplify that. "Second-order" means the highest derivative in the equation is the second derivative (D^2). "Linear" implies that the dependent variable (y) and its derivatives appear only to the first power and are not multiplied together. "Non-homogeneous" indicates that the right-hand side of the equation is not zero. Grasping these terms will make the entire solving process much smoother. Essentially, we are looking for a function y(x) that satisfies this equation. These types of equations pop up in various fields, including physics, engineering, and economics, often modeling systems that change over time. Think of the oscillation of a spring, the flow of current in a circuit, or even population growth – all can be described using differential equations similar to this one. Mastering these equations unlocks a powerful toolkit for analyzing and predicting the behavior of these systems.

Step 1: Finding the Homogeneous Solution

Our first mission is to find the homogeneous solution, often denoted as y_h. This involves setting the right-hand side of the equation to zero, giving us: (D^2 - 4)y = 0. To solve this, we'll use the auxiliary equation method. This is where we make the substitution D → m, transforming our differential equation into an algebraic equation. So, (D^2 - 4) becomes (m^2 - 4). Setting this equal to zero, we get the auxiliary equation: m^2 - 4 = 0. Now, we solve for m. This is a simple quadratic equation, and we can factor it as: (m - 2)(m + 2) = 0. This gives us two distinct real roots: m_1 = 2 and m_2 = -2. Because we have two distinct real roots, the homogeneous solution takes the form: y_h = C_1e^(2x) + C_2e^(-2x), where C_1 and C_2 are arbitrary constants. These constants will be determined later if we have initial conditions, but for now, they hang around as unknowns. The homogeneous solution represents the natural behavior of the system described by the differential equation, without considering any external forces or inputs (represented by the right-hand side of the original equation). It’s the foundation upon which we'll build the complete solution. Understanding how to find this homogeneous solution is crucial, as it forms a core component of solving many types of differential equations. We will find the particular solution in the next step which, when combined with the homogeneous solution, gives us the general solution to the original problem.

Step 2: Finding the Particular Solution

Now, let's tackle the particular solution, denoted as y_p. This is the trickier part, as we need to find a solution that satisfies the original non-homogeneous equation: (D^2 - 4)y = e^(2x) + e^(-4x). Since the right-hand side has two terms, we'll consider them separately and then combine the results. For the e^(2x) term, we might initially guess a solution of the form Ae^(2x). However, notice that e^(2x) is part of our homogeneous solution. This means that simply plugging Ae^(2x) into the equation will result in zero (because the homogeneous part is annihilated by the operator (D^2 - 4)). So, we need to modify our guess. A common technique is to multiply by x, so we'll try a particular solution of the form y_p1 = Axe^(2x). For the e^(-4x) term, we can guess a solution of the form Be^(-4x) since it’s not part of the homogeneous solution. So, y_p2 = Be^(-4x). Our total particular solution will be the sum of these: y_p = Axe^(2x) + Be^(-4x). Now, we need to find the values of A and B. This is where the heavy lifting comes in. We'll need to compute the first and second derivatives of y_p, substitute them into the original equation, and then solve for A and B. This process, while a bit lengthy, is a standard technique for finding particular solutions. It requires careful differentiation and algebraic manipulation, but the reward is a complete solution to the differential equation!

Step 3: Calculating Derivatives and Substituting

Okay, guys, time to roll up our sleeves and do some calculus! We need to find the first and second derivatives of our particular solution, y_p = Axe^(2x) + Be^(-4x). Let's start with the first derivative, y_p'. For the first term, Axe^(2x), we'll use the product rule: (uv)' = u'v + uv'. So, the derivative of Axe^(2x) is Ae^(2x) + 2Axe^(2x). For the second term, Be^(-4x), the derivative is simply -4Be^(-4x). Therefore, y_p' = Ae^(2x) + 2Axe^(2x) - 4Be^(-4x). Now, let's find the second derivative, y_p''. Differentiating y_p', we get: For Ae^(2x), the derivative is 2Ae^(2x). For 2Axe^(2x), we again use the product rule, giving us 2Ae^(2x) + 4Axe^(2x). For -4Be^(-4x), the derivative is 16Be^(-4x). So, y_p'' = 2Ae^(2x) + 2Ae^(2x) + 4Axe^(2x) + 16Be^(-4x) = 4Ae^(2x) + 4Axe^(2x) + 16Be^(-4x). Phew! That was a bit of a workout. Now comes the crucial step: substituting y_p and y_p'' into our original differential equation, (D^2 - 4)y = e^(2x) + e^(-4x). This means we'll replace D^2y with y_p'' and y with y_p. After substitution, we'll have an equation involving A, B, and x. The next step will be to equate the coefficients of like terms on both sides of the equation. This will give us a system of equations that we can solve for A and B. So, grab your pencils, and let's substitute these derivatives into the original equation and see what we get!

Step 4: Solving for the Constants A and B

Alright, we've substituted y_p and its derivatives into the original equation. Now we have: (4Ae^(2x) + 4Axe^(2x) + 16Be^(-4x)) - 4(Axe^(2x) + Be^(-4x)) = e^(2x) + e^(-4x). Let's simplify this equation by distributing the -4 and combining like terms: 4Ae^(2x) + 4Axe^(2x) + 16Be^(-4x) - 4Axe^(2x) - 4Be^(-4x) = e^(2x) + e^(-4x). Notice that the 4Axe^(2x) terms cancel out, leaving us with: 4Ae^(2x) + 12Be^(-4x) = e^(2x) + e^(-4x). Now, the magic happens! We equate the coefficients of the e^(2x) terms and the e^(-4x) terms on both sides of the equation. For the e^(2x) terms, we have: 4A = 1, which gives us A = 1/4. For the e^(-4x) terms, we have: 12B = 1, which gives us B = 1/12. Fantastic! We've found the values of A and B. This means we can now write our particular solution completely: y_p = (1/4)xe^(2x) + (1/12)e^(-4x). Remember, the particular solution is a specific solution that satisfies the non-homogeneous equation. We found it by making an educated guess, calculating derivatives, substituting, and then solving for the unknown coefficients. With y_p in hand, we're just one step away from the general solution!

Step 5: Forming the General Solution

We're in the home stretch, guys! We've found both the homogeneous solution, y_h = C_1e^(2x) + C_2e^(-2x), and the particular solution, y_p = (1/4)xe^(2x) + (1/12)e^(-4x). The general solution, y, is simply the sum of these two solutions: y = y_h + y_p. So, the general solution to our differential equation is: y = C_1e^(2x) + C_2e^(-2x) + (1/4)xe^(2x) + (1/12)e^(-4x). This is the complete solution! It represents a family of functions that satisfy the original differential equation. The constants C_1 and C_2 are arbitrary and can be determined if we have initial conditions (such as the value of y and its derivative at a specific point). The general solution combines the natural behavior of the system (represented by the homogeneous solution) with the response to the external input (represented by the particular solution). This solution is powerful because it gives us a comprehensive understanding of the behavior of the system described by the differential equation. Whether we're modeling the motion of a pendulum, the temperature distribution in a rod, or the concentration of a chemical in a reaction, this type of solution provides a complete picture.

Conclusion

And there you have it! We've successfully solved the differential equation (D^2 - 4)y = e^(2x) + e^(-4x). We navigated through finding the homogeneous solution, grappling with the particular solution, calculating derivatives, and finally, piecing it all together to form the general solution. I hope this step-by-step guide has made the process clearer and less daunting for you. Remember, differential equations might seem intimidating at first, but with practice and a systematic approach, you can conquer them. Keep practicing, and you'll become a differential equation whiz in no time! If you have any questions or want to dive deeper into other types of differential equations, feel free to ask. Happy solving!