Solving Y'' + Y = 4sin(t): Find The Particular Solution

Hey guys! Today, we are diving into the fascinating world of differential equations, specifically focusing on how to find a particular solution. We're going to tackle the equation y'' + y = 4sin(t). This is a classic example that combines second-order derivatives with trigonometric functions, making it a super interesting problem to solve. Understanding these concepts is crucial not only for acing your exams but also for grasping real-world applications in physics, engineering, and more. So, let's put on our math hats and get started!

Understanding the Problem: What is a Particular Solution?

Before we jump into the nitty-gritty, let's clarify what we mean by a "particular solution." In the realm of differential equations, a particular solution is a specific solution that satisfies the equation without any arbitrary constants. Think of it as finding a precise curve that fits the given differential equation perfectly. To find this, we first typically identify the general solution, which includes constants, and then use initial conditions or specific requirements to nail down the constants and arrive at our particular solution.

In our case, we have the equation y'' + y = 4sin(t). This is a second-order linear non-homogeneous differential equation. "Second-order" because it involves the second derivative (y''), "linear" because the dependent variable (y) and its derivatives appear linearly, and "non-homogeneous" because of the 4sin(t) term on the right side. This term is super important because it dictates that we'll need to find both a complementary solution (for the homogeneous part) and a particular solution (to account for the non-homogeneous part). Now, why is this important? Well, many real-world phenomena, such as the motion of a spring or the oscillations in an electrical circuit, can be modeled using this type of equation. By solving it, we can predict and understand these behaviors more effectively. Finding the particular solution is like tuning into the specific frequency of a system, allowing us to see how it responds to external forces or inputs. So, let's get our hands dirty and explore how we can find this magical particular solution.

Method of Undetermined Coefficients

Alright, let's get down to the solution! One of the most common and effective methods for finding particular solutions of non-homogeneous differential equations is the method of undetermined coefficients. This method is particularly handy when the non-homogeneous term (the 4sin(t) in our case) is a function whose derivatives have a similar form – things like polynomials, exponentials, sines, and cosines. The basic idea here is to make an educated guess about the form of the particular solution based on the non-homogeneous term, and then plug it back into the equation to solve for the unknown coefficients.

So, how does this work in practice? First, we look at our non-homogeneous term, which is 4sin(t). Since the derivative of sine is cosine and the derivative of cosine is sine (with a possible sign change), we can assume that our particular solution, y_p(t), will be a linear combination of sine and cosine functions. This means we'll guess that y_p(t) has the form:

y_p(t) = Acos(t) + Bsin(t)

Where A and B are the undetermined coefficients that we need to find. This is where the "undetermined" part of the method's name comes from! Now, this guess isn't always straightforward. Sometimes, we need to tweak it a bit to avoid issues. For instance, if the homogeneous equation (y'' + y = 0) has solutions that look like cos(t) or sin(t), our initial guess might overlap with those solutions, leading to problems. In such cases, we'd need to multiply our guess by t (or even t² if necessary) to make it linearly independent from the homogeneous solutions. This adjustment ensures that we can find a unique particular solution. But in our case, we'll see why we need an extra step to get the correct form for y_p(t). Stick with me, and we'll figure it out together!

Applying the Method to Our Equation

Okay, guys, let's roll up our sleeves and apply the method of undetermined coefficients to our equation, y'' + y = 4sin(t). Remember, we initially guessed that the particular solution, y_p(t), would look like Acos(t) + Bsin(t). However, there's a little twist in our case because cos(t) and sin(t) are already solutions to the homogeneous equation y'' + y = 0. This means that if we plug our initial guess into the original equation, we won't be able to solve for A and B properly. It's like trying to fit a puzzle piece into a space where it already fits – it just won't work!

So, what do we do? We need to modify our guess to make it linearly independent from the solutions of the homogeneous equation. The trick here is to multiply our initial guess by t. This gives us a new guess for the particular solution:

y_p(t) = t(Acos(t) + Bsin(t)) = Atcos(t) + Btsin(t)

Now, this looks more promising! To find the values of A and B, we need to plug this new guess into our original differential equation. This means we'll need to find the first and second derivatives of y_p(t). Let's start with the first derivative, y_p'(t). We'll need to use the product rule here, which states that the derivative of uv is u'v + uv'. Applying the product rule, we get:

y_p'(t) = Acos(t) - Atsin(t) + Bsin(t) + Btcos(t)

Now, let's find the second derivative, y_p''(t). We'll need to apply the product rule again. This can get a bit messy, but hang in there! We have:

y_p''(t) = -Asin(t) - Asin(t) - Atcos(t) + Bcos(t) + Bcos(t) - Btsin(t)

Simplifying, we have:

y_p''(t) = -2Asin(t) - Atcos(t) + 2Bcos(t) - Btsin(t)

Now we've got y_p(t) and y_p''(t). The next step is to plug these into the original equation y'' + y = 4sin(t). This will give us an equation that we can solve for A and B. Are you ready for the next step? Let's do it!

Solving for the Coefficients

Alright, guys, we've made it to the crucial step of solving for the coefficients A and B. We have our particular solution guess, y_p(t) = Atcos(t) + Btsin(t), and its second derivative, y_p''(t) = -2Asin(t) - Atcos(t) + 2Bcos(t) - Btsin(t). Now, let's plug these into our original differential equation, y'' + y = 4sin(t). This gives us:

(-2Asin(t) - Atcos(t) + 2Bcos(t) - Btsin(t)) + (Atcos(t) + Btsin(t)) = 4sin(t)

Notice how the Atcos(t) and Btsin(t) terms cancel out? This is exactly what we wanted! This cancellation simplifies our equation significantly, leaving us with:

-2Asin(t) + 2Bcos(t) = 4sin(t)

Now, we can equate the coefficients of the sine and cosine terms on both sides of the equation. This gives us a system of two equations:

• -2A = 4 (coefficients of sin(t))
• 2B = 0 (coefficients of cos(t))

Solving this system is pretty straightforward. From the first equation, we get A = -2. From the second equation, we get B = 0. So, we've found our coefficients! Now we can plug these values back into our particular solution guess:

y_p(t) = Atcos(t) + Btsin(t) = -2tcos(t) + 0tsin(t) = -2tcos(t)

So, there you have it! Our particular solution is y_p(t) = -2tcos(t). This is the specific solution that satisfies our differential equation without any arbitrary constants. It's a pretty cool feeling to find the solution, right? But we're not quite done yet. Let's make sure we understand what this solution means and how it fits into the bigger picture.

Verifying the Solution and Final Answer

Fantastic job, guys! We've arrived at a particular solution, y_p(t) = -2tcos(t), for the differential equation y'' + y = 4sin(t). But before we do a victory dance, it's always a good idea to verify our solution. This ensures that we haven't made any sneaky calculation errors along the way. To verify, we simply plug our solution back into the original equation and see if it holds true.

So, let's revisit our derivatives. We have:

• y_p(t) = -2tcos(t)
• y_p'(t) = -2cos(t) + 2tsin(t)
• y_p''(t) = 4sin(t) + 2tcos(t)

Now, let's plug y_p(t) and y_p''(t) into the left side of our differential equation, y'' + y:

(4sin(t) + 2tcos(t)) + (-2tcos(t)) = 4sin(t)

As you can see, the 2tcos(t) and -2tcos(t) terms cancel each other out, leaving us with 4sin(t), which is exactly the right-hand side of our original equation. Hooray! Our solution checks out. This gives us confidence that we've found the correct particular solution.

Now, let's circle back to the original question. We were asked to find a particular solution of the equation y'' + y = 4sin(t), and we've successfully found it to be y = -2tcos(t). This corresponds to option C in the choices provided. So, the final answer is C. This process not only solved the problem but also deepened our understanding of how to tackle non-homogeneous differential equations using the method of undetermined coefficients.

Conclusion: Mastering Differential Equations

Alright, guys, we've reached the end of our journey to find the particular solution of the differential equation y'' + y = 4sin(t). We've not only found the answer, which is y = -2tcos(t), but we've also walked through the process step-by-step, making sure we understand the why behind the how. We started by defining what a particular solution is and why it's important, then we dove into the method of undetermined coefficients, which is a powerful technique for solving non-homogeneous differential equations. We tackled the tricky situation where our initial guess needed modification because it overlapped with the homogeneous solution, and we learned how multiplying by t can save the day.

We then meticulously calculated the derivatives, plugged everything back into the original equation, and solved for the unknown coefficients. Finally, we verified our solution to ensure accuracy. This entire process highlights the importance of attention to detail and a systematic approach when dealing with differential equations. But beyond the specific steps, remember that the real magic happens when you grasp the underlying concepts. Understanding why we make certain guesses, why we modify them, and how each step contributes to the final solution is what truly empowers you to tackle similar problems in the future.

Differential equations might seem daunting at first, but with practice and a solid understanding of the methods, they become much less intimidating. So, keep practicing, keep exploring, and most importantly, keep asking questions. Whether you're dealing with oscillations in a mechanical system, the flow of current in an electrical circuit, or even population dynamics, the principles we've discussed today will serve you well. Keep up the great work, and happy solving!