Solving Initial Value Problem: A Step-by-Step Guide

Let's dive into solving this fascinating initial value problem! If you're tackling differential equations, you've likely encountered these types of problems. They might seem daunting at first, but with a systematic approach, they become quite manageable. In this guide, we'll break down the steps to solve the given problem: $\frac{d^2 s}{d t^2}=-25 \text{sin}(5t-\frac{\pi}{2})$, with initial conditions $s'(0) = 300$ and $s(0) = 0$. We'll make sure to cover each step in detail, so you can confidently tackle similar problems in the future. This is an important concept in various fields, including physics and engineering, where understanding motion and change is crucial. So, let's get started and unravel this mathematical puzzle together!

Understanding the Problem

Before we jump into calculations, it’s essential to understand what we’re dealing with. The problem presents us with a second-order differential equation: $\frac{d^2 s}{d t^2}=-25 \text{sin}(5t-\frac{\pi}{2})$. This equation describes the acceleration ($\frac{d^2 s}{d t^2}$) of an object as a function of time ($t$). The expression $-25 \text{sin}(5t-\frac{\pi}{2})$ tells us how the acceleration changes with time, specifically in a sinusoidal manner. Now, what makes this an initial value problem? It's the addition of initial conditions: $s'(0) = 300$ and $s(0) = 0$. These conditions give us specific values of the object's velocity ($s'(t)$) and position ($s(t)$) at time $t = 0$. Think of it like this: we’re not just finding a general solution, but a particular solution that fits these initial circumstances. Understanding these initial conditions is crucial because they allow us to pinpoint a unique solution from a family of possible solutions. Without them, we'd have a general solution with arbitrary constants, but with them, we can find the exact function that describes the object's motion. This is why initial value problems are so important in real-world applications, as they allow us to model specific scenarios with precision. For instance, in physics, this might represent the motion of a pendulum, and the initial conditions would describe its starting position and velocity. In engineering, it could represent the behavior of a system starting from a known state. So, grasping this concept thoroughly sets the stage for a successful solution.

Step 1: First Integration

The first key step in solving this initial value problem involves integrating the given differential equation once. We start with $\frac{d^2 s}{d t^2}=-25 \text{sin}(5t-\frac{\pi}{2})$. To find the velocity function, $s'(t)$, we need to integrate both sides of the equation with respect to $t$. Remember, integration is the reverse process of differentiation, so it helps us move from acceleration to velocity. When we integrate $-25 \text{sin}(5t-\frac{\pi}{2})$ with respect to $t$, we're essentially asking: “What function, when differentiated, gives us $-25 \text{sin}(5t-\frac{\pi}{2})$?” The integral of $\text{sin}(ax + b)$ is $-\frac{1}{a} \text{cos}(ax + b)$, so we need to apply this rule here. Integrating $-25 \text{sin}(5t-\frac{\pi}{2})$ gives us $5 \text{cos}(5t-\frac{\pi}{2}) + C_1$, where $C_1$ is the constant of integration. This constant is vital because the derivative of a constant is zero, meaning many functions could have the same derivative. Therefore, we write $s'(t) = 5 \text{cos}(5t-\frac{\pi}{2}) + C_1$. Now, we use the initial condition $s'(0) = 300$ to find the value of $C_1$. Plugging in $t = 0$, we get $300 = 5 \text{cos}(-\frac{\pi}{2}) + C_1$. Since $\text{cos}(-\frac{\pi}{2}) = 0$, we find that $C_1 = 300$. So, our velocity function becomes $s'(t) = 5 \text{cos}(5t-\frac{\pi}{2}) + 300$. This step is crucial because it not only gives us the velocity function but also incorporates the initial condition, narrowing down our solution. By carefully performing this integration and applying the initial condition, we're one step closer to solving the entire problem.

Step 2: Second Integration

Having found the velocity function, $s'(t) = 5 \text{cos}(5t-\frac{\pi}{2}) + 300$, the next step is to integrate again to find the position function, $s(t)$. This process is similar to the first integration, but this time we are moving from velocity to position. We need to integrate both terms of the velocity function separately with respect to $t$. Let's start with the first term: $5 \text{cos}(5t-\frac{\pi}{2})$. The integral of $\text{cos}(ax + b)$ is $\frac{1}{a} \text{sin}(ax + b)$, so we integrate $5 \text{cos}(5t-\frac{\pi}{2})$ to get $\text{sin}(5t-\frac{\pi}{2})$. Next, we integrate the constant term, 300, with respect to $t$. The integral of a constant $k$ is simply $kt$, so the integral of 300 is $300t$. Now, combining these results, we have $s(t) = \text{sin}(5t-\frac{\pi}{2}) + 300t + C_2$, where $C_2$ is another constant of integration. Just like before, this constant is essential for capturing the general solution. To find the specific solution that satisfies our initial conditions, we use the second initial condition: $s(0) = 0$. Plugging in $t = 0$, we get $0 = \text{sin}(-\frac{\pi}{2}) + 300(0) + C_2$. Since $\text{sin}(-\frac{\pi}{2}) = -1$, we have $0 = -1 + C_2$, which gives us $C_2 = 1$. Thus, the position function becomes $s(t) = \text{sin}(5t-\frac{\pi}{2}) + 300t + 1$. This second integration is just as crucial as the first. It not only provides us with the position function but also incorporates the second initial condition, pinning down the unique solution to our initial value problem. By meticulously integrating and applying the initial condition, we complete the process of finding the specific function that describes the object's position at any given time.

Step 3: The Solution

After performing the two integrations and applying the initial conditions, we have arrived at the solution to the initial value problem. We started with the differential equation $\frac{d^2 s}{d t^2}=-25 \text{sin}(5t-\frac{\pi}{2})$, and through our step-by-step process, we have found the position function, which is: $s(t) = \text{sin}(5t-\frac{\pi}{2}) + 300t + 1$. This function, $s(t)$, describes the position of the object at any time $t$, satisfying both the original differential equation and the initial conditions $s'(0) = 300$ and $s(0) = 0$. Let's recap how we got here. First, we integrated the acceleration function to find the velocity function, $s'(t)$, and used the first initial condition to determine the constant of integration. Then, we integrated the velocity function to find the position function, $s(t)$, and used the second initial condition to find the second constant of integration. The final solution, $s(t) = \text{sin}(5t-\frac{\pi}{2}) + 300t + 1$, is the unique solution that fits the given problem. It tells us exactly how the object moves over time, given its initial state and the forces acting upon it. This is the power of solving initial value problems – we move from a general description of motion (the differential equation) to a specific, concrete description (the solution function). This solution is not just a mathematical expression; it's a model that can be used to predict and understand the object's behavior. Whether it's the motion of a spring, the trajectory of a projectile, or the behavior of an electrical circuit, the ability to solve these problems is fundamental in many areas of science and engineering.

Conclusion

In conclusion, we have successfully solved the initial value problem $\frac{d^2 s}{d t^2}=-25 \text{sin}(5t-\frac{\pi}{2})$ with initial conditions $s'(0) = 300$ and $s(0) = 0$. We achieved this by systematically integrating the differential equation twice, first to find the velocity function and then to find the position function. We then used the given initial conditions to determine the constants of integration, leading us to the unique solution: $s(t) = \text{sin}(5t-\frac{\pi}{2}) + 300t + 1$. This process highlights the importance of understanding each step – from recognizing the type of problem and understanding the significance of initial conditions to performing the integrations and solving for the constants. Initial value problems are a cornerstone of differential equations and have broad applications in various fields, including physics, engineering, and economics. They allow us to model real-world phenomena with a high degree of accuracy, making them an essential tool in many disciplines. The ability to solve these problems confidently requires a solid understanding of calculus, particularly integration, and a methodical approach to problem-solving. By breaking down the problem into manageable steps and carefully applying the initial conditions, we can arrive at a precise solution that describes the behavior of the system under consideration. Remember, practice is key to mastering these concepts. The more you work through different types of initial value problems, the more comfortable and proficient you will become. So, keep practicing, keep exploring, and continue to build your mathematical skills!

For further learning about differential equations and initial value problems, you might find resources on websites like Khan Academy's Differential Equations helpful. They offer excellent explanations and practice problems to enhance your understanding.