operator was a total *game-changer*. It allowed us to transform intimidating calculus expressions into more manageable algebraic forms, making the manipulation and `elimination method` much smoother and less prone to errors. It's truly a _superhero_ tool for these kinds of problems!\n* ***Systematic Elimination:*** Our `game plan` involved strategically eliminating one variable to derive a single, higher-order differential equation. This process is the heart of solving `coupled systems` and requires careful, step-by-step application of `operator algebra`.\n* ***Homogeneous and Particular Solutions:*** Once we had a single ODE for `x(t)`, we leveraged our knowledge of finding both the `homogeneous solution` (from the characteristic equation) and the `particular solution` (using the method of undetermined coefficients for `e^t`). This is a foundational skill in solving _any_ linear non-homogeneous differential equation.\n* ***Back-Substitution for the Complete Solution:*** Finally, we used one of our derived operator equations to find `y(t)` by substituting our hard-earned `x(t)` and its derivatives. This step completed our **solution to the system**, providing a pair of functions that satisfy both original equations simultaneously.\n\nThe **solution to the system** we found, involving various exponential terms and arbitrary constants, elegantly describes the long-term behavior of `x` and `y` under the influence of the `e^t` forcing function. These constants (`C1`, `C2`, `C3`) would be determined by initial conditions, which weren't provided in this problem but are crucial in real-world applications to pinpoint a unique solution.\n\nMastering **coupled differential equations** like these opens up a whole new level of understanding in various scientific and engineering fields. Whether you're modeling circuits, population dynamics, or mechanical vibrations, the ability to solve such systems is incredibly valuable. So, give yourselves a pat on the back! You've tackled a challenging problem with a robust and elegant approach. Keep practicing, keep exploring, and remember that with the right tools and a systematic approach, even the most complex mathematical puzzles can be solved. You've got this! Keep rocking those equations!" alt="Mastering Coupled Differential Equations: A Step-by-Step Guide\n\nHey there, awesome math enthusiasts! 👋 Ever looked at a system of differential equations and thought, "_Whoa, where do I even begin?_" Well, you're in luck, because today we're going to demystify one such beast! We're diving deep into the world of **coupled differential equations**, specifically tackling a system that might seem a bit intimidating at first glance. But trust me, by the end of this guide, you'll have a clear roadmap and the confidence to tackle similar problems head-on. Our goal is to transform what looks like a complex mathematical puzzle into a straightforward, step-by-step process. We're talking about finding the *suitable solution* to a system that includes both first-order and second-order derivatives, and even an exponential term on the right-hand side. This isn't just about crunching numbers; it's about understanding the logic, the tools, and the strategies that seasoned mathematicians use. So, buckle up, grab your favorite beverage, and let's unlock the secrets of these fascinating equations together! We’ll be breaking down each part, making sure you grasp *why* we’re taking specific steps and *how* each piece contributes to the final solution. This isn't just some theoretical exercise either; *coupled differential equations* are everywhere in the real world, modeling everything from oscillating circuits and predator-prey dynamics to the spread of diseases and complex mechanical systems. Understanding how to solve them gives you a powerful toolset for analyzing and predicting the behavior of interconnected systems, making this journey not just academically fulfilling but also incredibly practical. We'll be using a cool trick called the **operator method**, which really simplifies the manipulation of these equations, turning intimidating calculus into a more algebraic-like process. It's a game-changer, especially when you're dealing with multiple variables and derivatives. Get ready to level up your differential equation skills!\n\n## Understanding Our System: The Equations We're Tackling\n\nAlright, guys, before we jump into solving anything, let's properly introduce our main characters—the **system of differential equations** we're here to conquer. We've got two equations that are *coupled*, meaning they're interconnected and depend on each other. You can't solve one without considering the other, which is precisely what makes them so interesting (and sometimes a bit tricky!). Our system looks like this:\n\n1. `dx/dt + dy/dt = e^t`\n2. `-d^2x/dt^2 + dx/dt = -2x - y`\n\nLet's break down what these *differential equations* are telling us. The first equation involves the *first derivatives* of two functions, `x(t)` and `y(t)`, with respect to time `t`. It states that their sum is equal to `e^t`, a common exponential function. This `e^t` term is super important because it acts as an "input" or "forcing function" to our system, and it will play a key role when we look for particular solutions. The second equation is even more interesting because it involves a *second derivative* of `x(t)`, specifically `-d^2x/dt^2`, along with `dx/dt`, `x(t)` itself, and `y(t)`. Notice how `y(t)` appears in the second equation, directly linking it to `x(t)`. This is the essence of being "coupled"—the rate of change of `x` (and its second derivative) is influenced not just by `x` itself, but also by `y`. These equations describe a dynamic relationship where `x` and `y` evolve together over time, constantly influencing each other's rates of change. Think of it like two interacting components in a machine or two populations in an ecosystem; their behavior isn't isolated. To simplify our work, we'll be using a handy notation: `$\Delta = d/dt

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. This `operator method` lets us treat differentiation almost like an algebraic operation, making the manipulation of these equations much cleaner and more intuitive. So, `dx/dt` becomes `$\Delta x

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, `dy/dt` becomes `$\Delta y

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, and `d^2x/dt^2` becomes `$\Delta^2 x

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. This notation is a total game-changer for organizing our thoughts and calculations when we start rearranging terms and eliminating variables. Understanding these components is the first crucial step in developing a robust **solution to the system**. Without a clear picture of what each part means, trying to find a solution would be like trying to navigate a maze blindfolded. So, take a moment to really soak in these equations and the power of the `$\Delta

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operator; it's going to make our lives a whole lot easier as we move forward! This foundation is truly *essential* for mastering the approach to these kinds of problems, and it’s the gateway to performing accurate and efficient algebraic manipulations with our differential operators.\n\n## The Game Plan: How to Approach Coupled Systems\n\nOkay, so we've got our challenging **system of differential equations** in front of us. Now, how do we actually *solve* it? This is where having a solid `game plan` comes into play. When dealing with *coupled linear differential equations*, one of the most powerful and common strategies is the **elimination method**, often combined with the `operator method` we just talked about. Think of it like solving a system of algebraic equations, but with derivatives! The core idea is to manipulate the equations in such a way that you can eliminate one of the dependent variables (either `x` or `y`), leaving you with a single, higher-order differential equation for the *remaining* variable. Once you've got that single equation, solving it becomes a standard process that you've probably encountered before (homogeneous and particular solutions, anyone?). Then, once you've found one variable, you can substitute it back into one of the original equations to find the other. It sounds simple, but the devil, as they say, is in the details—specifically, how you perform those `operator manipulations` cleanly and accurately. Our steps will look something like this:\n\n1. ***Rewrite the equations using the `$\Delta

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operator:*** This step transforms our calculus-heavy expressions into a more algebraic form, making them easier to manage. It's like switching from a complicated language to a simplified code.\n2. ***Manipulate the equations to eliminate one variable:*** We'll treat the `$\Delta

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operator much like a constant or a variable in algebra. We'll multiply equations by operators, add, and subtract them to get rid of either `x` or `y`. Our aim here will be to derive a single *higher-order differential equation* purely in terms of `x` (or `y`, but `x` looks a bit more straightforward here due to `d^2x/dt^2`). This is the most crucial part of the **elimination method** and requires careful attention to detail.\n3. ***Solve the resulting single differential equation:*** Once we have an equation for `x(t)` (or `y(t)`), we'll find its `general solution`. This typically involves finding both a `homogeneous solution` (the solution to the equation with zero on the right-hand side) and a `particular solution` (a solution that accounts for the non-homogeneous term, in our case `e^t`). This part leverages techniques you've likely mastered in ordinary differential equations.\n4. ***Substitute back to find the other variable:*** With `x(t)` now known, we'll plug it (and its derivatives) back into one of the original `system of differential equations` to determine `y(t)`. This ensures that our `solution to the system` satisfies *both* initial conditions simultaneously.\n\nThis `systematic approach` is what separates the pros from the newbies when it comes to solving **coupled differential equations**. It ensures we don't get lost in the algebra and calculus and provides a clear path to the correct `solution`. Each step builds upon the last, so mastering each stage is key. We're not just solving a problem; we're learning a robust methodology for tackling a whole class of similar mathematical challenges. Remember, patience and precision are your best friends here. Let’s dive into applying this `game plan` and seeing it in action! This detailed approach is not merely a suggestion but a *necessity* for navigating the complexities inherent in such interconnected systems, ensuring that every derivative and every term is accounted for precisely.\n\n### Step 1: Rewriting Our Equations with the Operator $\Delta$\n\nAlright, team, let's kick off our `solution to the system` by making our **coupled differential equations** more manageable. This is where our good friend, the `operator method`, specifically using `$\Delta = d/dt

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, comes into play. It’s seriously a _game-changer_ for simplifying the look and feel of these equations, turning intimidating calculus notation into something that feels much more like algebra. Trust me, this small step will save you a ton of headaches later on! Let’s take our original system:\n\n1. `dx/dt + dy/dt = e^t`\n2. `-d^2x/dt^2 + dx/dt = -2x - y`\n\nNow, let's apply our `$\Delta

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operator to each term. Remember, `dx/dt` simply becomes `$\Delta x

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, `dy/dt` becomes `$\Delta y

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, and the second derivative `d^2x/dt^2` transforms into `$\Delta^2 x

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. We'll also move all the `x` and `y` terms to one side of the equation, making it easier to see their relationships and prepare for elimination.\n\nEquation (1) transforms into:\n\n`$\Delta x + \Delta y = e^t

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\n\nThis is pretty straightforward, right? We just replaced the derivative notation with our operator.\n\nNow for Equation (2). This one has a few more terms:\n\n`$-\Delta^2 x + \Delta x = -2x - y

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\n\nLet’s bring all the `x` and `y` terms to the left-hand side. When `y` moves over, it becomes `+y`. When `-2x` moves over, it becomes `+2x`. So, we get:\n\n`$-\Delta^2 x + \Delta x + 2x + y = 0

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\n\nTo make it even cleaner and group the `x` terms together, we can factor out `x` using the operator:\n\n`$(-\Delta^2 + \Delta + 2)x + y = 0

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\n\nSo, our rewritten **system of differential equations** using the `$\Delta

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operator now looks like this:\n\n(A) `$\Delta x + \Delta y = e^t

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\n(B) `$(-\Delta^2 + \Delta + 2)x + y = 0

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\n\nSee how much neater that looks? The `operator method` allows us to manipulate these equations almost as if `$\Delta

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were a simple constant. This simplified notation is absolutely crucial for the next step, where we'll be performing algebraic-like operations to eliminate one of the variables. It helps prevent errors and makes the entire `solution process` far more transparent. This transformation is not just a cosmetic change; it's a fundamental shift in how we approach the problem, enabling us to leverage algebraic intuition in the realm of calculus. By clearly defining our `differential operators`, we set the stage for a much smoother `elimination method`, directly contributing to finding the accurate `solution` for our `coupled system`. Getting this step right is foundational for a successful journey through the rest of the problem, ensuring that our `differential equations` are perfectly poised for the next set of manipulations.\n\n### Step 2: Isolating Variables – The Elimination Method\n\nAlright, now that we've got our `system of differential equations` neatly expressed using the `$\Delta

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operator, it's time for the real magic: the **elimination method**! Our goal here is to get rid of one variable, say `y`, so we're left with a single, solvable *differential equation* for `x`. This is where we treat `$\Delta

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like an algebraic term, which is super cool and powerful. Let's revisit our transformed equations:\n\n(A) `$\Delta x + \Delta y = e^t

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\n(B) `$(-\Delta^2 + \Delta + 2)x + y = 0

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\n\nLooking at these, it seems easier to eliminate `y`. Why? Because Equation (B) has a simple `+y` term. If we can get `$\Delta y

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from Equation (A) and `y` from Equation (B) to match up for elimination, that'd be perfect.\n\nFrom Equation (B), we can easily express `y` in terms of `x` and `$\Delta

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operators:\n\n`y = -(-\Delta^2 + \Delta + 2)x`\n`y = (\Delta^2 - \Delta - 2)x` (Let's call this (C))\n\nNow, if we differentiate `y` with respect to `t` (i.e., apply the `$\Delta

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operator to both sides of (C)), we'll get `$\Delta y$!`: \n\n`$\Delta y = \Delta(\Delta^2 - \Delta - 2)x

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\n`$\Delta y = (\Delta^3 - \Delta^2 - 2\Delta)x

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(Let's call this (D))\n\nThis `$\Delta y

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expression is exactly what we need to substitute back into Equation (A)! So, let's plug (D) into (A):\n\n`$\Delta x + (\Delta^3 - \Delta^2 - 2\Delta)x = e^t

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\n\nNow, we can group all the `x` terms together:\n\n`$(\Delta^3 - \Delta^2 - 2\Delta + \Delta)x = e^t

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\n`$(\Delta^3 - \Delta^2 - \Delta)x = e^t

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\n\n_Boom!_ We've done it! We've successfully eliminated `y` and are left with a single, *third-order non-homogeneous differential equation* for `x(t)`:\n\n`$\Delta^3 x - \Delta^2 x - \Delta x = e^t

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\n\nOr, written in its more familiar derivative form:\n\n`$d^3x/dt^3 - d^2x/dt^2 - dx/dt = e^t

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\n\nThis is a fantastic achievement, guys! This `elimination method` using the `operator method` has transformed a complicated `coupled system` into a problem we now know how to tackle. The resulting equation for `x(t)` is a standard form for which we have established `solution techniques`. The key here was the careful application of the `$\Delta

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operator as an algebraic entity, allowing us to perform substitutions and combine terms effectively. This demonstrates the immense power of converting our original `differential equations` into the operator form, streamlining the process of isolating variables. It’s a crucial milestone in our journey to find the complete `solution to the system`, and now we're perfectly set up for the next step: solving this individual `differential equation` for `x(t)`. The precision involved in these `operator manipulations` is paramount, as any small error here would propagate through the subsequent steps, leading to an incorrect `solution`. We have successfully reduced the complexity, allowing us to focus on standard methods for solving a single, higher-order ODE, bringing us much closer to understanding the full dynamics described by our initial `coupled differential equations`.\n\n### Step 3: Solving for `x(t)` – The Homogeneous and Particular Solutions\n\nOkay, champions, we've successfully whittled down our complex **system of differential equations** to a single, more familiar one for `x(t)`:\n\n`$\Delta^3 x - \Delta^2 x - \Delta x = e^t

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\n\nOr, if you prefer the classic notation:\n\n`$d^3x/dt^3 - d^2x/dt^2 - dx/dt = e^t

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\n\nNow, our mission is to find the **general solution** for `x(t)`. Remember, the `general solution` to a non-homogeneous linear differential equation is always the sum of two parts: the `homogeneous solution` (`$x_h(t)

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) and the `particular solution` (`$x_p(t)

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). So, `x(t) = $x_h(t) + x_p(t)

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.\n\nLet's tackle the `homogeneous solution` first. This involves setting the right-hand side of our equation to zero:\n\n`$\Delta^3 x - \Delta^2 x - \Delta x = 0

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\n\nThe `characteristic equation` for this homogeneous ODE is found by replacing `$\Delta

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with a variable, say `r`:\n\n`$r^3 - r^2 - r = 0

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\n\nNow, we need to find the roots of this polynomial. We can factor out `r`:\n\n`$r(r^2 - r - 1) = 0

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\n\nSo, one root is `r = 0`. For the quadratic part, `$(r^2 - r - 1 = 0)

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, we use the quadratic formula `$(r = [-b \pm \sqrt{b^2 - 4ac}] / 2a)

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:\n\n`$r = [1 \pm \sqrt{(-1)^2 - 4(1)(-1)}] / 2(1)

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\n`$r = [1 \pm \sqrt{1 + 4}] / 2

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\n`$r = [1 \pm \sqrt{5}] / 2

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\n\nSo, our three distinct real roots are: `r1 = 0`, `r2 = $(1 + \sqrt{5}) / 2

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, and `r3 = $(1 - \sqrt{5}) / 2

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.\nTherefore, the `homogeneous solution` is:\n\n`$x_h(t) = C_1 e^{0t} + C_2 e^{[(1 + \sqrt{5}) / 2]t} + C_3 e^{[(1 - \sqrt{5}) / 2]t}

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\n`$x_h(t) = C_1 + C_2 e^{[(1 + \sqrt{5}) / 2]t} + C_3 e^{[(1 - \sqrt{5}) / 2]t}

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\n\nWhere `C1`, `C2`, and `C3` are arbitrary constants that would be determined by initial conditions (if we had them!).\n\nNext up, the `particular solution`, `$x_p(t)

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. Since our forcing function on the right-hand side is `e^t`, we might initially guess `$x_p(t) = A e^t

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. However, we need to be careful! Notice that `r=1` is *not* a root of our characteristic equation. If it were, we'd need to multiply by `t`. Since `r=1` is not a root, our simple guess `$x_p(t) = A e^t

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should work.\n\nLet's find the derivatives of `$x_p(t)

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:\n`$x_p(t) = A e^t

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\n`$dx_p/dt = A e^t

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\n`$d^2x_p/dt^2 = A e^t

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\n`$d^3x_p/dt^3 = A e^t

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\n\nSubstitute these into our non-homogeneous equation:\n\n`$A e^t - A e^t - A e^t = e^t

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\n`$-A e^t = e^t

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\n\nThis means `$-A = 1

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, so `A = -1`.\nThus, our `particular solution` is:\n\n`$x_p(t) = -e^t

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\n\nCombining the `homogeneous` and `particular solutions`, we get the `general solution` for `x(t)`:\n\n`$x(t) = C_1 + C_2 e^{[(1 + \sqrt{5}) / 2]t} + C_3 e^{[(1 - \sqrt{5}) / 2]t} - e^t

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\n\nThis is a *major milestone* in solving our `coupled differential equations`! We've successfully isolated and solved for one of the variables. This step is a testament to the power of breaking down complex problems into manageable parts: first, isolating the variable, then applying standard techniques for solving linear ODEs. The accuracy here is absolutely vital, as `x(t)` will be used to derive `y(t)`. The detailed process of finding roots and then determining the form of the particular solution highlights the importance of a strong foundation in solving single-variable `differential equations`. We are now very close to finding the complete `solution to the system`, demonstrating the efficacy of our `operator method` and `elimination strategy`.\n\n### Step 4: Finding `y(t)` – Bringing It All Together\n\nAlright, we’re in the home stretch, folks! We've done the heavy lifting and found the **general solution** for `x(t)` from our **system of differential equations**. Now, the final piece of the puzzle is to determine `y(t)`. This is where we loop back to our original `coupled equations` or one of our intermediate operator forms. The easiest way to find `y(t)` is to use an equation where `y` appears in a simple form. Remember Equation (B) in its operator form from Step 1, which we manipulated to get `y = (\Delta^2 - \Delta - 2)x` (our Equation C)? This is a perfect candidate!\n\nLet's recall our `x(t)` solution:\n\n`$x(t) = C_1 + C_2 e^{[(1 + \sqrt{5}) / 2]t} + C_3 e^{[(1 - \sqrt{5}) / 2]t} - e^t

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\n\nWe need to calculate `$\Delta x

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, `$\Delta^2 x

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, which are `dx/dt` and `d^2x/dt^2` respectively, and then substitute them into the expression for `y`.\n\nFirst, let's find `dx/dt` (which is `$\Delta x

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):\n\n`$dx/dt = 0 + C_2 [(1 + \sqrt{5}) / 2] e^{[(1 + \sqrt{5}) / 2]t} + C_3 [(1 - \sqrt{5}) / 2] e^{[(1 - \sqrt{5}) / 2]t} - e^t

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\n\nNext, `d^2x/dt^2` (which is `$\Delta^2 x

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):\n\n`$d^2x/dt^2 = C_2 [(1 + \sqrt{5}) / 2]^2 e^{[(1 + \sqrt{5}) / 2]t} + C_3 [(1 - \sqrt{5}) / 2]^2 e^{[(1 - \sqrt{5}) / 2]t} - e^t

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\n\nNow, we use our expression for `y` from earlier: `y = $(\Delta^2 - \Delta - 2)x

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.\nThis expands to: `y = $d^2x/dt^2 - dx/dt - 2x

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\n\nLet's plug in `x(t)`, `dx/dt`, and `d^2x/dt^2` into this equation. This is where the algebra can get a bit hefty, but sticking to it is key!\n\n`$y(t) = (C_2 [(1 + \sqrt{5}) / 2]^2 e^{r_2 t} + C_3 [(1 - \sqrt{5}) / 2]^2 e^{r_3 t} - e^t)`\n`$ - (C_2 [(1 + \sqrt{5}) / 2] e^{r_2 t} + C_3 [(1 - \sqrt{5}) / 2] e^{r_3 t} - e^t)`\n`$ - 2(C_1 + C_2 e^{r_2 t} + C_3 e^{r_3 t} - e^t)

Mastering Coupled Differential Equations: A Step-by-Step Guide

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\n\nWhere `r2 = $(1 + \sqrt{5}) / 2

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

and `r3 = $(1 - \sqrt{5}) / 2

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\n\nLet's group terms by `C1`, `C2`, `C3`, and the `e^t` term.\n\nFor `C1` terms:\n`$-2 C_1

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

\n\nFor `C2` terms (factor out `$C_2 e^{r_2 t}

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

):\n`$C_2 e^{r_2 t} [r_2^2 - r_2 - 2]

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

\nRecall that `r2` is a root of `$r^2 - r - 1 = 0

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

. This means `$r_2^2 - r_2 - 1 = 0

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

, so `$r_2^2 - r_2 = 1

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\nTherefore, `$r_2^2 - r_2 - 2 = (r_2^2 - r_2) - 2 = 1 - 2 = -1

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\nSo, the `C2` term simplifies to `$-C_2 e^{r_2 t}

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\n\nFor `C3` terms (factor out `$C_3 e^{r_3 t}

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

):\n`$C_3 e^{r_3 t} [r_3^2 - r_3 - 2]

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

\nSimilarly, `$r_3^2 - r_3 - 1 = 0

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

, so `$r_3^2 - r_3 = 1

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\nTherefore, `$r_3^2 - r_3 - 2 = (r_3^2 - r_3) - 2 = 1 - 2 = -1

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\nSo, the `C3` term simplifies to `$-C_3 e^{r_3 t}

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

.\n\nFor the `e^t` terms:\n`$-e^t - (-e^t) - 2(-e^t) = -e^t + e^t + 2e^t = 2e^t

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

\n\nPutting it all together, our `solution for y(t)` is:\n\n`$y(t) = -2C_1 - C_2 e^{[(1 + \sqrt{5}) / 2]t} - C_3 e^{[(1 - \sqrt{5}) / 2]t} + 2e^t

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

\n\nAnd there you have it! The complete **solution to the system** of `coupled differential equations`. This step, while algebraically intensive, is crucial for obtaining the full picture. It highlights how intrinsically linked `x(t)` and `y(t)` are, with the arbitrary constants and exponential terms carrying over and combining in specific ways. Remember, the elegance of the `operator method` really shines through in how it allows us to handle these complex relationships systematically, providing a clear path to the final `solution`. This full solution, comprising both `x(t)` and `y(t)`, paints the entire dynamic portrait of our system, showing how each component evolves and interacts over time.\n\n## Final Thoughts and What We've Learned\n\nPhew! You made it, guys! We've just navigated a pretty intricate **system of differential equations** from start to finish. We began with two *coupled linear differential equations*, one first-order and one second-order, and systematically worked our way through to find their explicit `solutions`, `x(t)` and `y(t)`. This journey wasn't just about crunching numbers; it was about understanding a powerful `methodology` that you can apply to a wide range of similar problems.\n\nLet's quickly recap the key takeaways:\n\n* ***Understanding Coupled Systems:*** We learned that *coupled differential equations* are interconnected, meaning the rate of change of one variable depends on the other. This interdependence is what makes them so fascinating and essential in modeling real-world phenomena.\n* ***The Power of the Operator Method ($\Delta$):*** Using the `$\Delta = d/dt

Mastering Coupled Differential Equations: A Step-by-Step Guide

Dec 8, 2025 by Admin 63 views

operator was a total *game-changer*. It allowed us to transform intimidating calculus expressions into more manageable algebraic forms, making the manipulation and `elimination method` much smoother and less prone to errors. It's truly a _superhero_ tool for these kinds of problems!\n* ***Systematic Elimination:*** Our `game plan` involved strategically eliminating one variable to derive a single, higher-order differential equation. This process is the heart of solving `coupled systems` and requires careful, step-by-step application of `operator algebra`.\n* ***Homogeneous and Particular Solutions:*** Once we had a single ODE for `x(t)`, we leveraged our knowledge of finding both the `homogeneous solution` (from the characteristic equation) and the `particular solution` (using the method of undetermined coefficients for `e^t`). This is a foundational skill in solving _any_ linear non-homogeneous differential equation.\n* ***Back-Substitution for the Complete Solution:*** Finally, we used one of our derived operator equations to find `y(t)` by substituting our hard-earned `x(t)` and its derivatives. This step completed our **solution to the system**, providing a pair of functions that satisfy both original equations simultaneously.\n\nThe **solution to the system** we found, involving various exponential terms and arbitrary constants, elegantly describes the long-term behavior of `x` and `y` under the influence of the `e^t` forcing function. These constants (`C1`, `C2`, `C3`) would be determined by initial conditions, which weren't provided in this problem but are crucial in real-world applications to pinpoint a unique solution.\n\nMastering **coupled differential equations** like these opens up a whole new level of understanding in various scientific and engineering fields. Whether you're modeling circuits, population dynamics, or mechanical vibrations, the ability to solve such systems is incredibly valuable. So, give yourselves a pat on the back! You've tackled a challenging problem with a robust and elegant approach. Keep practicing, keep exploring, and remember that with the right tools and a systematic approach, even the most complex mathematical puzzles can be solved. You've got this! Keep rocking those equations!" width="300" height="200"/>