Solving Y'' + 4y' + 4y = Sin(3t) A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of differential equations, specifically how to solve a non-homogeneous one. We'll be tackling the equation y'' + 4y' + 4y = sin(3t). This type of problem might seem intimidating at first, but I promise we'll break it down into manageable steps so you can conquer it with confidence. So, let's get started and find the general solution to this equation!

1. Understanding Non-Homogeneous Differential Equations

Before we jump into the nitty-gritty of solving our specific equation, let's quickly recap what non-homogeneous differential equations are all about. Think of it this way: a homogeneous differential equation is one where the right-hand side is equal to zero. Our equation, however, has sin(3t) on the right side, making it non-homogeneous. This seemingly small difference has a big impact on how we approach the solution.

The general solution to a non-homogeneous differential equation is composed of two parts: the homogeneous solution (yh) and the particular solution (yp). The homogeneous solution is the solution to the corresponding homogeneous equation (where we set the right-hand side to zero). The particular solution, on the other hand, is any solution that satisfies the original non-homogeneous equation. We find the general solution by adding these two parts together: y = yh + yp.

The reason we need both parts is that the homogeneous solution takes care of the inherent behavior of the system described by the differential equation (its natural response), while the particular solution accounts for the influence of the external forcing function (in our case, the sin(3t) term). Imagine it like this: the homogeneous solution is the sound a guitar string makes when you pluck it, while the particular solution is how the sound changes when you hold a finger on a fret. Together, they give the complete picture of the sound.

Understanding this fundamental concept is crucial because it dictates our strategy for solving the equation. We'll first find the homogeneous solution by dealing with the left-hand side of the equation. Then, we'll find a particular solution that fits the sin(3t) forcing function. Finally, we'll combine them to get the general solution. Think of it as a puzzle with two main pieces that fit together perfectly to reveal the final answer. This step-by-step approach will make the whole process much clearer and less daunting. We're essentially breaking down a complex problem into smaller, more manageable tasks, which is a valuable skill in mathematics and beyond!

2. Finding the Homogeneous Solution (yh)

Okay, let's roll up our sleeves and find the homogeneous solution (yh) first. This means we're going to tackle the equation as if it were homogeneous, meaning we'll set the right-hand side to zero. So, we're now dealing with:

y'' + 4y' + 4y = 0

The trick to solving these types of equations is to assume a solution of the form y = e^(rt), where r is a constant we need to determine. This might seem like magic, but it works because the exponential function has the special property that its derivatives are proportional to itself. This makes it a great candidate for solutions to differential equations where derivatives are related to the function itself. We're essentially making an educated guess based on the nature of exponential functions and their behavior in calculus.

Now, let's plug this assumed solution into our homogeneous equation. We need to find the first and second derivatives of y = e^(rt):

• y' = re^(rt)
• y'' = r^2e(rt)

Substituting these into our equation, we get:

r^2e(rt) + 4re^(rt) + 4e^(rt) = 0

Notice that e^(rt) is a common factor in all the terms. Since e^(rt) is never zero, we can divide both sides of the equation by it, leaving us with the characteristic equation:

r^2 + 4r + 4 = 0

This is a quadratic equation, and we can solve it using factoring, the quadratic formula, or completing the square. In this case, it factors nicely:

(r + 2)(r + 2) = 0

This gives us a repeated root of r = -2. When we have repeated roots, the form of the homogeneous solution is a little different. For a single root, we'd have e^(rt). But for a repeated root, we need to include an extra term multiplied by t. So, our homogeneous solution becomes:

yh = c1e^(-2t) + c2te^(-2t)

Where c1 and c2 are arbitrary constants. These constants will be determined by initial conditions, if we had them, but since we're looking for the general solution, we'll leave them as constants for now. We've successfully conquered the first part of our problem! We've found the homogeneous solution that describes the natural behavior of the system without the influence of the sin(3t) term. Now, let's move on to the particular solution, which will account for that external influence.

3. Finding the Particular Solution (yp)

Alright, we've nailed the homogeneous solution, so now it's time to find the particular solution (yp). This is the part where we deal with the sin(3t) on the right-hand side of our original equation: y'' + 4y' + 4y = sin(3t). To find yp, we'll use a technique called 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 forcing function (in this case, sin(3t)).

Since our forcing function is sin(3t), we'll guess that the particular solution has the form:

yp = Acos(3t) + Bsin(3t)

Where A and B are constants that we need to determine. We include both cos(3t) and sin(3t) because the derivatives of sine are cosine and vice-versa, so we need to account for both possibilities. It's like preparing for all the possible outcomes when you're making a prediction. If we only included the sine term, we might miss a crucial part of the solution that involves the cosine.

Now, let's find the first and second derivatives of our guessed solution:

• yp' = -3Asin(3t) + 3Bcos(3t)
• yp'' = -9Acos(3t) - 9Bsin(3t)

Next, we'll substitute yp, yp', and yp'' into our original non-homogeneous equation:

(-9Acos(3t) - 9Bsin(3t)) + 4(-3Asin(3t) + 3Bcos(3t)) + 4(Acos(3t) + Bsin(3t)) = sin(3t)

Now, we need to simplify and collect like terms. Let's group the cosine terms and the sine terms together:

(-9A + 12B + 4A)cos(3t) + (-9B - 12A + 4B)sin(3t) = sin(3t)

This simplifies to:

(-5A + 12B)cos(3t) + (-12A - 5B)sin(3t) = sin(3t)

For this equation to hold true for all values of t, the coefficients of the cosine and sine terms on both sides must be equal. This gives us a system of two equations:

• -5A + 12B = 0
• -12A - 5B = 1

We can solve this system of equations for A and B. Let's use substitution. From the first equation, we can express A in terms of B:

A = (12/5)B

Now, substitute this into the second equation:

-12(12/5)B - 5B = 1

Simplify and solve for B:

(-144/5)B - 5B = 1 (-144/5)B - (25/5)B = 1 (-169/5)B = 1 B = -5/169

Now, plug this value of B back into the equation for A:

A = (12/5)(-5/169) A = -12/169

We've found A and B! Now we can write our particular solution:

yp = (-12/169)cos(3t) + (-5/169)sin(3t)

We've successfully found the particular solution, which accounts for the influence of the sin(3t) forcing function. It might have seemed like a lot of steps, but we broke it down systematically, and now we have a crucial piece of our puzzle. We're almost there! Just one more step to go, and we'll have the general solution.

4. Constructing the General Solution

Okay, the moment we've been working towards! We've found both the homogeneous solution (yh) and the particular solution (yp). Now, to get the general solution (y), we simply add them together:

y = yh + yp

Remember, the homogeneous solution was:

yh = c1e^(-2t) + c2te^(-2t)

And the particular solution was:

yp = (-12/169)cos(3t) + (-5/169)sin(3t)

So, the general solution is:

y = c1e^(-2t) + c2te^(-2t) + (-12/169)cos(3t) + (-5/169)sin(3t)

And there you have it! We've successfully solved the non-homogeneous differential equation y'' + 4y' + 4y = sin(3t). The general solution includes the arbitrary constants c1 and c2, which would be determined by specific initial conditions if we had them. But since we were looking for the general solution, this is our final answer.

Let's take a moment to appreciate what we've accomplished. We started with a seemingly complex equation and, by breaking it down into smaller, manageable steps, we were able to find the complete solution. We understood the nature of non-homogeneous equations, found the homogeneous solution by solving the characteristic equation, and then used the method of undetermined coefficients to find the particular solution. Finally, we combined these pieces to arrive at the general solution. This is a testament to the power of systematic problem-solving and the beauty of differential equations!

5. Key Takeaways and Further Exploration

Wow, guys, we really went deep into solving that differential equation! Let's recap the key takeaways from our journey:

• Non-homogeneous equations have a right-hand side that's not zero, and their solutions are the sum of the homogeneous and particular solutions.
• The homogeneous solution (yh) represents the natural behavior of the system and is found by solving the equation with the right-hand side set to zero.
• The particular solution (yp) accounts for the external forcing function and can be found using methods like undetermined coefficients.
• The general solution (y) is the sum of yh and yp and represents all possible solutions to the non-homogeneous equation.
• The method of undetermined coefficients involves making an educated guess about the form of yp based on the forcing function.

If you're feeling adventurous and want to explore further, here are some ideas:

• Try solving similar non-homogeneous equations with different forcing functions, like e^(2t) or t^2. This will help you solidify your understanding of the method of undetermined coefficients.
• Investigate other methods for finding particular solutions, such as variation of parameters. This method is more general and can handle a wider range of forcing functions.
• Explore applications of differential equations in various fields, such as physics, engineering, and economics. You'll be amazed at how these equations can model real-world phenomena.
• Look into solving higher-order differential equations or systems of differential equations. These are more complex but build upon the same fundamental principles we've discussed.

Solving differential equations is a valuable skill that opens doors to many exciting areas of mathematics and its applications. Keep practicing, keep exploring, and you'll become a differential equation master in no time! Keep up the great work, and I'll see you in the next mathematical adventure!