Solving Y' = Y + E^x With Y(2) = 0: Find Y(1)
Introduction
Hey guys! Today, we're diving deep into the fascinating world of differential equations. Specifically, we're going to tackle the problem of solving the first-order linear ordinary differential equation y' = y + e^x with the initial condition y(2) = 0. This means we're looking for a function y(x) that satisfies this equation and passes through the point (2, 0). And, our ultimate goal? To find the value of y(1). This is a classic problem that pops up in various fields, from physics to engineering, so understanding how to solve it is super important. Trust me, once you grasp the concepts, you'll feel like a total math whiz!
Solving differential equations might seem daunting at first, but don't worry, we'll break it down step by step. We'll explore the different methods, discuss why they work, and highlight the key concepts. Think of it like assembling a puzzle – each piece of knowledge fits together to form the complete solution. We will begin by understanding the type of differential equation we're dealing with. The given equation, y' = y + e^x, is a first-order linear ordinary differential equation. This classification helps us choose the appropriate method for solving it. Next, we'll explore the method of integrating factors, a powerful technique for solving linear differential equations. This method involves multiplying both sides of the equation by a carefully chosen function, called the integrating factor, to make the left-hand side a perfect derivative. We'll go through the process of finding this integrating factor and applying it to our equation. After applying the integrating factor, we'll integrate both sides of the equation to obtain the general solution. This solution will contain an arbitrary constant of integration, which we'll need to determine using the initial condition y(2) = 0. Once we've found the particular solution that satisfies the initial condition, we'll be able to evaluate it at x = 1 to find y(1). This final step will give us the answer we're looking for. So, buckle up, grab your pencils, and let's get started on this mathematical adventure! We're going to unravel this differential equation and emerge victorious with a clear understanding of the solution process. Remember, math is like a muscle – the more you exercise it, the stronger it gets. And solving problems like this is the perfect workout for your mathematical muscles!
Method of Integrating Factors
The method of integrating factors is our go-to technique here. It's like a secret weapon for tackling first-order linear differential equations. These equations have a specific form: y' + P(x)y = Q(x). Notice how our equation, y' = y + e^x, neatly fits this mold if we rewrite it as y' - y = e^x. So, P(x) is -1, and Q(x) is e^x. The magic of the integrating factor lies in transforming the left side of the equation into the derivative of a product. This makes integration a breeze! Now, how do we find this magical integrating factor? It's given by the formula μ(x) = e^(∫P(x) dx). This formula might look a bit intimidating, but don't worry, we'll break it down. We take the exponential of the integral of P(x). In our case, P(x) = -1, so we need to find the integral of -1 with respect to x. This integral is simply -x (plus a constant, which we can ignore for now since we just need one integrating factor). Therefore, our integrating factor is μ(x) = e^(-x). Think of this integrating factor as a special key that unlocks the solution to our differential equation. It's a function that, when multiplied by our equation, simplifies it in a way that makes it easy to integrate. It's like adding a catalyst to a chemical reaction, speeding up the process and making it easier to obtain the desired product. Once we have the integrating factor, we multiply both sides of our differential equation (y' - y = e^x) by it. This gives us e^(-x)y' - e^(-x)y = e(-x)ex. Now, here's the crucial part: the left side of this equation is the derivative of the product e^(-x)y. This is the whole point of using the integrating factor! We've transformed the left side into something we can easily integrate. The right side simplifies to 1, which is also super easy to integrate. So, we now have d/dx (e^(-x)y) = 1. This equation is much simpler to solve than the original differential equation. We've effectively transformed a complex problem into a more manageable one. It's like taking a complicated machine and breaking it down into its individual components, making it easier to understand and repair. The next step is to integrate both sides with respect to x. This will give us the general solution to the differential equation. Remember, the integrating factor is a powerful tool in our mathematical arsenal. It allows us to solve a wide range of first-order linear differential equations, making it an essential technique for anyone studying calculus and differential equations.
Integrating Both Sides
Alright, we've got our equation in a beautifully simplified form: d/dx (e^(-x)y) = 1. Now comes the fun part: integrating both sides! This is where we undo the differentiation and get closer to our solution. Integrating the left side, ∫ d/dx (e^(-x)y) dx, simply gives us e^(-x)y. It's like the integral and derivative cancel each other out, leaving us with the original expression. On the right side, we need to find the integral of 1 with respect to x, ∫ 1 dx. This is a basic integral, and the result is x + C, where C is the constant of integration. Remember, the constant of integration is crucial! It represents the family of solutions to the differential equation. Think of it as a degree of freedom in the solution – there are infinitely many solutions that differ only by a constant. So, we now have e^(-x)y = x + C. We're almost there! We need to isolate y to get the general solution in the form y = f(x). To do this, we multiply both sides of the equation by e^x. This gives us y = (x + C)e^x. This is the general solution to our differential equation. It represents a family of functions that satisfy the equation y' = y + e^x. Each value of C corresponds to a different member of this family. This general solution is like a blueprint for all possible solutions to the differential equation. It tells us the overall shape and behavior of the solutions, but it doesn't give us a specific solution. To find a specific solution, we need to use the initial condition, which we'll do in the next step. Before we move on, let's take a moment to appreciate what we've accomplished. We've successfully integrated both sides of the equation and obtained the general solution. This is a significant milestone in solving the differential equation. We've gone from a differential equation, which relates a function to its derivative, to an algebraic equation that expresses the function itself. It's like translating a complex sentence into a simpler one, making it easier to understand. Now, we're ready to use the initial condition to find the particular solution that satisfies our specific problem. This will be the final piece of the puzzle, allowing us to find the value of y(1).
Applying the Initial Condition y(2) = 0
We've arrived at a crucial step: using the initial condition y(2) = 0. This is like having a specific landmark on a map that helps us pinpoint our exact location. The initial condition tells us that when x = 2, y = 0. We can plug these values into our general solution, y = (x + C)e^x, to solve for the constant C. Substituting x = 2 and y = 0 into the general solution, we get 0 = (2 + C)e^2. Now, we need to solve this equation for C. Since e^2 is not zero, the only way for the product (2 + C)e^2 to be zero is if 2 + C = 0. Solving for C, we find that C = -2. This is a key moment! We've found the value of C that corresponds to the particular solution we're looking for. It's like finding the missing piece of the puzzle that completes the picture. Now that we know C = -2, we can plug it back into our general solution to get the particular solution: y = (x - 2)e^x. This is the specific function that satisfies both the differential equation y' = y + e^x and the initial condition y(2) = 0. This particular solution is like a unique fingerprint of our problem. It's the one and only solution that passes through the point (2, 0) and satisfies the given differential equation. We've successfully narrowed down the infinite family of solutions represented by the general solution to a single, specific solution. Think of it like choosing a specific route from a network of roads. The general solution is like the entire road network, while the particular solution is like the specific route we choose based on our destination. Now that we have the particular solution, we're ready to find y(1). This is the final step in our journey, and it's a simple one. We just need to plug in x = 1 into our particular solution.
Finding y(1)
We're in the home stretch now! We have the particular solution, y = (x - 2)e^x, and we want to find the value of y when x = 1, which is y(1). This is like the final piece of the puzzle clicking into place. To find y(1), we simply substitute x = 1 into our particular solution: y(1) = (1 - 2)e^1. This simplifies to y(1) = (-1)e, or simply y(1) = -e. And there we have it! We've successfully solved the differential equation and found that y(1) = -e. This is the answer we've been working towards, and it feels pretty awesome to have reached the finish line. Let's take a moment to recap our journey. We started with the differential equation y' = y + e^x and the initial condition y(2) = 0. We used the method of integrating factors to find the general solution, which was y = (x + C)e^x. Then, we applied the initial condition to find the value of C, which gave us the particular solution y = (x - 2)e^x. Finally, we substituted x = 1 into the particular solution to find y(1) = -e. This problem demonstrates the power of differential equations in modeling real-world phenomena. Differential equations are used to describe a wide variety of systems, from the motion of objects to the flow of electricity to the spread of diseases. By understanding how to solve differential equations, we can gain valuable insights into these systems and make predictions about their behavior. So, congratulations on making it this far! You've successfully navigated a challenging differential equation problem. You've learned about integrating factors, general solutions, particular solutions, and initial conditions. You've honed your mathematical skills and gained a deeper understanding of differential equations. Remember, the key to mastering math is practice, practice, practice. So, keep tackling problems, keep exploring new concepts, and keep pushing your mathematical boundaries. You've got this!
Conclusion
In conclusion, we successfully navigated the solution to the differential equation y' = y + e^x with the initial condition y(2) = 0, and we found that y(1) = -e. This journey highlighted the power and elegance of the method of integrating factors, a fundamental technique in solving first-order linear ordinary differential equations. We started by identifying the type of differential equation and recognizing that the method of integrating factors was the most appropriate approach. We then carefully calculated the integrating factor, multiplied it by the equation, and integrated both sides. This process allowed us to transform the differential equation into a more manageable form, ultimately leading to the general solution. The application of the initial condition was crucial in narrowing down the infinite family of solutions represented by the general solution to a single, particular solution that satisfied our specific problem. This step emphasized the importance of initial conditions in determining the unique behavior of a system modeled by a differential equation. Finally, by substituting x = 1 into the particular solution, we were able to find the value of y(1), which was our ultimate goal. This final step demonstrated the practical application of solving differential equations – finding specific values of the solution at particular points. The process we followed in solving this problem is a testament to the systematic and logical nature of mathematics. Each step built upon the previous one, and each concept played a crucial role in reaching the final solution. This step-by-step approach is a valuable skill that can be applied to a wide range of problems, both in mathematics and in other fields. Solving differential equations is not just an academic exercise; it's a powerful tool for understanding and modeling the world around us. Differential equations are used in physics, engineering, biology, economics, and many other disciplines. They allow us to describe the relationships between quantities that change over time and to make predictions about the future behavior of systems. The skills and knowledge we've gained in solving this problem will serve us well in tackling more complex problems in the future. Whether you're studying advanced mathematics, pursuing a career in science or engineering, or simply interested in understanding the world around you, the ability to solve differential equations is a valuable asset. So, keep practicing, keep exploring, and keep challenging yourself. The world of mathematics is vast and fascinating, and there's always something new to learn.
