Solving Y' + 5y = T^3e^{-5t}: Integrating Factor Method

Hey guys! Today, we're diving into the fascinating world of differential equations. Specifically, we're tackling the equation y' + 5y = t³e^{-5t} using the integrating factor method. This method is super useful for solving first-order linear differential equations, so let's break it down step-by-step. Think of this as our mission, and we're the equation-solving superheroes!

Understanding First-Order Linear Differential Equations

Before we jump into the solution, let's make sure we're all on the same page about what a first-order linear differential equation actually is. In simple terms, it’s an equation that can be written in the form:

y' + P(t)y = Q(t)

Where:

• y' represents the first derivative of the function y with respect to t (dy/dt).
• P(t) is a function of t.
• y is the unknown function we want to find.
• Q(t) is another function of t.

Our equation, y' + 5y = t³e^{-5t}, perfectly fits this form. See? It's not as scary as it looks! Identifying this standard form is the crucial first step because it allows us to recognize that the integrating factor method is the right tool for the job. In our case, P(t) = 5 and Q(t) = t³e^{-5t}. Recognizing these components is like identifying the specific ingredients we need for our equation-solving recipe. Understanding the nature of the equation sets us up for successfully applying the subsequent steps of the integrating factor method.

The Integrating Factor: Our Secret Weapon

The integrating factor is the heart of this method. It's a special function that we multiply our entire differential equation by, and it has this magical power to make the left side of the equation neatly collapsible into the derivative of a product. Seriously cool, right? The integrating factor, often denoted by μ(t) (that's the Greek letter 'mu'), is calculated using the following formula:

μ(t) = e^(∫P(t) dt)

Where:

• e is the base of the natural logarithm (Euler's number, approximately 2.71828).
• ∫P(t) dt represents the indefinite integral of P(t) with respect to t.

For our equation, P(t) = 5, so let's calculate our integrating factor:

μ(t) = e^(∫5 dt) = e^(5t)

So, e^(5t) is our secret weapon! Think of it like a key that unlocks the structure of the equation, making it solvable. This integrating factor will allow us to rewrite the left-hand side of the differential equation as a single derivative, which is the core trick of the method. Calculating this correctly is absolutely vital; it's like making sure you have the right amount of a key ingredient in a recipe. A slight error here can throw off the whole solution, so double-check your integration! We'll see how this works in the next step.

Multiplying Through: Unleashing the Power

Now, we take our calculated integrating factor, e^(5t), and multiply every single term in our original differential equation y' + 5y = t³e^{-5t} by it. This step might seem simple, but it's incredibly powerful because it sets up the next crucial transformation. Doing this, we get:

e^(5t)y' + 5e^(5t)y = t³e^{-5t} * e^(5t)

Notice something magical happening on the left side? The goal here is to make the left side recognizable as the result of the product rule for differentiation. Specifically, we want to see if it looks like the derivative of (y * μ(t)), which in our case is (y * e^(5t)). Let's think about the product rule for a moment. If we were to differentiate (y * e^(5t)) with respect to t, we would get:

d/dt [y * e^(5t)] = y' * e^(5t) + y * 5e^(5t)

Take a good look at this result and compare it to the left side of our multiplied equation: e^(5t)y' + 5e^(5t)y. See the match? This is not a coincidence! This is precisely why the integrating factor method works. The integrating factor is designed to make this happen. By multiplying through by e^(5t), we've transformed the left-hand side into a form that we can easily recognize as the derivative of a product. On the right side, the exponentials simplify nicely:

t³e^{-5t} * e^(5t) = t³

This simplification is also essential; it cleans up the equation and makes it much easier to integrate in the next step. So, after this multiplication and simplification, our equation looks like this:

e^(5t)y' + 5e^(5t)y = t³

Which we can now rewrite in the more compact and insightful form:

d/dt [y * e^(5t)] = t³

This transformation is the key that unlocks the equation. We've gone from a differential equation that looks difficult to solve to one where the derivative of a product is clearly isolated. It's like turning a complex puzzle into a simple one! This is the real magic of the integrating factor method shining through.

Integrating Both Sides: Unveiling the Solution

Now for the fun part: integration! We have the equation d/dt [y * e^(5t)] = t³. To get rid of the derivative, we need to integrate both sides with respect to t. This is the inverse operation of differentiation, and it will allow us to isolate y and find our general solution. So, let's integrate:

∫ d/dt [y * e^(5t)] dt = ∫ t³ dt

The integral on the left side simply undoes the derivative, leaving us with:

y * e^(5t) = ∫ t³ dt

On the right side, we need to find the integral of t³. Using the power rule for integration (∫t^n dt = (t^(n+1))/(n+1) + C), we get:

∫ t³ dt = (t^4)/4 + C

Where C is the constant of integration. Never forget this constant! It's crucial because it represents the family of solutions that satisfy the differential equation. Different values of C will give us different particular solutions. So, our equation now looks like this:

y * e^(5t) = (t^4)/4 + C

We're getting so close to the solution! By integrating both sides, we've effectively peeled away the derivative, revealing the relationship between y and t. The constant of integration C reminds us that there are infinitely many solutions, each differing by a constant value. This is a fundamental aspect of solving differential equations, and understanding its significance is key to fully grasping the nature of the solutions.

Isolating y: The Grand Finale

We're in the home stretch! Our goal is to find y, so we need to isolate it on one side of the equation. We have:

y * e^(5t) = (t^4)/4 + C

To get y by itself, we simply divide both sides of the equation by e^(5t). This is a straightforward algebraic step, but it's the final move that reveals the solution we've been working towards. Doing this, we get:

y = [(t^4)/4 + C] / e^(5t)

We can rewrite this to make it a little cleaner:

y = (t^4) / (4e^(5t)) + C / e^(5t)

Or, even more compactly:

y = (t^4) / (4e^(5t)) + Ce^(-5t)

And there you have it! This is the general solution to the differential equation y' + 5y = t³e^{-5t}. This solution represents an infinite family of functions that satisfy the original differential equation. Each different value of the constant C gives us a unique solution curve. We've successfully navigated the steps of the integrating factor method, from identifying the equation type to calculating the integrating factor, multiplying through, integrating, and finally, isolating y. This process showcases the power and elegance of this method in solving first-order linear differential equations. This final expression for y is the culmination of our efforts, and it provides a complete and general answer to the problem.

General Solution Unveiled

So, the general solution to the differential equation y' + 5y = t³e^{-5t} using the integrating factor method is:

y = (t^4) / (4e^(5t)) + Ce^(-5t)

Where C is an arbitrary constant.

And that's a wrap, guys! We've successfully navigated the world of differential equations and conquered this problem using the integrating factor method. Remember, practice makes perfect, so keep tackling those equations and flexing those mathematical muscles!