Solving Y'' + 4y = 10e^x A Step-by-Step Guide

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Hey guys! Today, we're diving into the fascinating world of differential equations. Specifically, we're going to tackle the equation y'' + 4y = 10e^x. Don't worry if this looks intimidating; we'll break it down step by step, making it super easy to understand. This guide is designed to help you not only solve this particular problem but also grasp the underlying concepts so you can confidently tackle similar equations in the future. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into solving, let's take a moment to understand what we're dealing with. The equation y'' + 4y = 10e^x is a second-order, linear, non-homogeneous ordinary differential equation. Whew, that's a mouthful! Let's break that down:

  • Second-order: This means the highest derivative in the equation is the second derivative (y'').
  • Linear: The dependent variable (y) and its derivatives appear linearly (no squares, square roots, etc.).
  • Non-homogeneous: This is because of the 10e^x term on the right side of the equation. If it were zero, the equation would be homogeneous.

Okay, now that we've got the jargon out of the way, let's talk about why these types of equations are important. Differential equations like this pop up all over the place in physics, engineering, economics, and many other fields. They're used to model everything from the motion of a pendulum to the spread of a disease. Understanding how to solve them is a crucial skill for anyone working in these areas.

Our goal here is to find a function y(x) that satisfies the equation. This function will describe the behavior of the system being modeled. We'll achieve this by finding a general solution, which includes all possible solutions to the equation. The general solution consists of two parts: the homogeneous solution (yh) and the particular solution (yp). We will discuss these in detail later. In essence, we're going to embark on a journey of mathematical discovery, unraveling the intricacies of this equation and revealing the hidden solution within. So, stay with me, and let's get started on this exciting adventure!

Step 1: Finding the Homogeneous Solution (yh)

The first step in solving this non-homogeneous differential equation is to tackle the homogeneous part. This means we set the right-hand side of the equation to zero, giving us: y'' + 4y = 0. This simplified equation allows us to focus on the inherent behavior of the system without the influence of the external driving force represented by 10e^x.

To find the homogeneous solution (yh), we assume a solution of the form y = e^(rx), where r is a constant we need to determine. This assumption is based on the fact that exponential functions are their own derivatives, making them a natural fit for differential equations. Now, we need to find the first and second derivatives of our assumed solution:

  • y' = re^(rx)
  • y'' = r2e(rx)

Next, we substitute these derivatives back into the homogeneous equation: r2e(rx) + 4e^(rx) = 0. We can factor out e^(rx), which gives us (r^2 + 4)e^(rx) = 0. Since e^(rx) is never zero, we are left with the auxiliary equation: r^2 + 4 = 0. This quadratic equation is the key to unlocking the homogeneous solution.

Now, let's solve for r. Subtracting 4 from both sides gives us r^2 = -4. Taking the square root of both sides, we get r = ±√(-4) = ±2i, where i is the imaginary unit (√(-1)). This result tells us that the roots of the auxiliary equation are complex conjugates. In cases where we encounter complex roots (α ± βi), the general form of the homogeneous solution is given by yh = C1cos(βx) + C2sin(βx), where C1 and C2 are arbitrary constants. In our case, α = 0 and β = 2. Therefore, the homogeneous solution is yh = C1cos(2x) + C2sin(2x). This solution represents the natural oscillations of the system, and the constants C1 and C2 will be determined by initial conditions, if provided. This solution describes the system's behavior without the influence of the external force, laying the foundation for understanding the complete solution.

Step 2: Finding the Particular Solution (yp)

Now that we've nailed the homogeneous solution (yh), it's time to find the particular solution (yp). This part accounts for the non-homogeneous term in the original equation, which in our case is 10e^x. Finding yp is like figuring out how the system responds specifically to this external driving force.

To find yp, we use the method of undetermined coefficients. This method involves making an educated guess about the form of the particular solution based on the form of the non-homogeneous term. Since our non-homogeneous term is 10e^x, we guess that yp has the form yp = Ae^x, where A is a constant we need to determine. The key here is to choose a form that, when differentiated, will produce terms similar to those in the non-homogeneous part of the equation.

Next, we find the first and second derivatives of our guessed solution:

  • yp' = Ae^x
  • yp'' = Ae^x

Now, we substitute yp and its derivatives back into the original non-homogeneous equation: y'' + 4y = 10e^x. This gives us Ae^x + 4(Ae^x) = 10e^x. Simplifying, we get 5Ae^x = 10e^x. To solve for A, we divide both sides by 5e^x, which gives us A = 2. Therefore, our particular solution is yp = 2e^x. This solution represents the system's specific response to the external force 10e^x. It's the steady-state part of the solution, describing how the system behaves after any initial transient effects have died out. With both the homogeneous and particular solutions in hand, we're now ready to construct the general solution.

Step 3: Constructing the General Solution

Alright, we've done the heavy lifting! We've found both the homogeneous solution (yh) and the particular solution (yp). Now, the fun part: putting it all together to form the general solution. The general solution (y) is simply the sum of the homogeneous and particular solutions: y = yh + yp.

We found earlier that yh = C1cos(2x) + C2sin(2x) and yp = 2e^x. Therefore, the general solution to the differential equation y'' + 4y = 10e^x is: y = C1cos(2x) + C2sin(2x) + 2e^x. This equation represents the complete family of solutions to the original differential equation. It incorporates both the natural oscillatory behavior of the system (from yh) and the specific response to the external force (from yp).

The constants C1 and C2 are arbitrary constants, meaning they can take on any value. To determine their specific values, we would need additional information, such as initial conditions (e.g., y(0) and y'(0)). Initial conditions provide specific values of the solution and its derivative at a particular point, allowing us to solve for the constants and pinpoint a unique solution from the general family of solutions. Without initial conditions, the general solution represents all possible solutions to the differential equation. In essence, the general solution is a comprehensive description of the system's behavior, capturing both its inherent dynamics and its response to external influences. It's the culmination of our step-by-step journey, providing a complete and satisfying answer to the problem.

Step 4: (Optional) Applying Initial Conditions

While we've found the general solution, let's briefly discuss what happens if we have initial conditions. Initial conditions provide us with specific values of the solution and its derivative at a particular point, allowing us to determine the values of the constants C1 and C2 in our general solution. This gives us a unique solution that satisfies both the differential equation and the given initial conditions.

Let's say, for example, we have the initial conditions y(0) = 0 and y'(0) = 1. This means that at x = 0, the value of the solution is 0, and the value of its derivative is 1. To use these conditions, we first need to find the derivative of our general solution: y' = -2C1sin(2x) + 2C2cos(2x) + 2e^x.

Now, we can apply the initial conditions:

  1. y(0) = 0: Substituting x = 0 into the general solution, we get 0 = C1cos(0) + C2sin(0) + 2e^0, which simplifies to 0 = C1 + 0 + 2. This gives us C1 = -2.
  2. y'(0) = 1: Substituting x = 0 into the derivative of the general solution, we get 1 = -2C1sin(0) + 2C2cos(0) + 2e^0, which simplifies to 1 = 0 + 2C2 + 2. This gives us 2C2 = -1, so C2 = -1/2.

Therefore, with these initial conditions, the unique solution is: y = -2cos(2x) - (1/2)sin(2x) + 2e^x. This solution not only satisfies the differential equation but also passes through the specific point and has the specific slope defined by the initial conditions. In practice, initial conditions are often determined by the physical context of the problem. For instance, in a spring-mass system, initial conditions might represent the initial displacement and velocity of the mass. By applying initial conditions, we transform the general solution into a specific solution that accurately describes the behavior of the system under those particular circumstances. This step highlights the power of initial conditions in tailoring the general solution to fit real-world scenarios.

Key Takeaways

Okay, guys, we've reached the end of our journey! Let's recap the key steps we took to solve the differential equation y'' + 4y = 10e^x:

  1. Finding the Homogeneous Solution (yh): We set the right-hand side of the equation to zero and solved the resulting homogeneous equation. This involved assuming a solution of the form y = e^(rx), finding the auxiliary equation, and solving for r. The roots of the auxiliary equation determined the form of yh.
  2. Finding the Particular Solution (yp): We used the method of undetermined coefficients to guess the form of yp based on the non-homogeneous term. We substituted our guess into the original equation and solved for the unknown coefficient(s).
  3. Constructing the General Solution: We added yh and yp together to obtain the general solution, which represents the family of all possible solutions to the differential equation.
  4. (Optional) Applying Initial Conditions: If we had initial conditions, we used them to determine the values of the arbitrary constants in the general solution, giving us a unique solution.

The main idea here is that solving a non-homogeneous differential equation involves breaking it down into smaller, more manageable parts. By finding the homogeneous and particular solutions separately, we can combine them to create the general solution. This approach is a powerful tool for tackling a wide variety of differential equations.

Remember, practice makes perfect! The more you work through these types of problems, the more comfortable you'll become with the process. Don't be afraid to make mistakes – they're a natural part of learning. And most importantly, have fun with it! Differential equations might seem daunting at first, but they're actually quite fascinating once you start to understand them. They're the language of change, describing how things evolve and interact in the world around us. So keep exploring, keep learning, and keep solving!

Conclusion

We've successfully navigated the world of differential equations and solved y'' + 4y = 10e^x! We've seen how to find the homogeneous and particular solutions, construct the general solution, and even apply initial conditions. This step-by-step approach can be applied to a wide range of similar problems, empowering you to tackle more complex equations with confidence.

The beauty of differential equations lies in their ability to model real-world phenomena. From the oscillations of a pendulum to the flow of current in a circuit, these equations provide a mathematical framework for understanding and predicting the behavior of dynamic systems. Mastering the techniques for solving them opens doors to a deeper understanding of the world around us.

So, keep practicing, keep exploring, and never stop questioning. The world of mathematics is full of exciting challenges and rewarding discoveries. And remember, every complex problem can be broken down into smaller, manageable steps. With patience, persistence, and a dash of curiosity, you can conquer any mathematical hurdle that comes your way. Keep up the great work, and I'll see you in the next mathematical adventure!