Finding The Integrating Factor For Y' = 8x³y + 6 With Μ(0) = 1

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Hey guys! Let's dive into solving a differential equation problem where we need to find the integrating factor μ(x){\mu(x)} for the equation y′=8x3y+6{y' = 8x^3y + 6}, given that μ(0)=1{\mu(0) = 1}. This type of problem is super common in differential equations, and understanding how to find integrating factors is a crucial skill. We're going to break it down step by step so it’s easy to follow. Our goal here is to not only find the integrating factor but also to really understand why this method works. So, grab your favorite beverage, and let's get started!

Understanding Integrating Factors

First off, what exactly is an integrating factor? Well, in the context of first-order linear differential equations, an integrating factor is a function that we multiply through the equation to make it easier to solve. Think of it like a magic ingredient that transforms a tricky equation into something much more manageable. Specifically, for a first-order linear differential equation in the form:

y′+P(x)y=Q(x){ y' + P(x)y = Q(x) }

The integrating factor μ(x){\mu(x)} is given by:

μ(x)=e∫P(x)dx{ \mu(x) = e^{\int P(x) dx} }

Why does this work? The key is that multiplying by μ(x){\mu(x)} turns the left-hand side of the equation into the derivative of a product, which we can then easily integrate. This method hinges on rewriting the differential equation into a form where the left side is the result of the product rule. By identifying and using the integrating factor, we essentially reverse-engineer the product rule to simplify the equation. The integrating factor allows us to combine the terms involving y′{y'} and y{y} into a single derivative, specifically ddx(μ(x)y){\frac{d}{dx}(\mu(x)y)}. Once the equation is in this form, we can integrate both sides with respect to x{x}, which significantly simplifies the process of finding the solution.

Rewriting the Differential Equation

Our given equation is:

y′=8x3y+6{ y' = 8x^3y + 6 }

To use the integrating factor method, we need to rewrite this in the standard form:

y′+P(x)y=Q(x){ y' + P(x)y = Q(x) }

Subtract 8x3y{8x^3y} from both sides to get:

y′−8x3y=6{ y' - 8x^3y = 6 }

Now we can clearly identify P(x){P(x)} and Q(x){Q(x)}. In this case, P(x)=−8x3{P(x) = -8x^3} and Q(x)=6{Q(x) = 6}. This step is crucial because the integrating factor depends directly on P(x){P(x)}. If we misidentify P(x){P(x)}, we'll end up with the wrong integrating factor and won't be able to solve the differential equation correctly. Ensuring the equation is in the standard form allows us to apply the formula for the integrating factor directly and accurately. By isolating y′{y'} and grouping the terms with y{y} and the constant term, we set the stage for the next step, which is calculating the integrating factor.

Calculating the Integrating Factor μ(x)

Now that we have P(x)=−8x3{P(x) = -8x^3}, we can calculate the integrating factor μ(x){\mu(x)} using the formula:

μ(x)=e∫P(x)dx{ \mu(x) = e^{\int P(x) dx} }

Plug in P(x){P(x)}:

μ(x)=e∫−8x3dx{ \mu(x) = e^{\int -8x^3 dx} }

Let's evaluate the integral:

∫−8x3dx=−8∫x3dx=−8⋅x44+C=−2x4+C{ \int -8x^3 dx = -8 \int x^3 dx = -8 \cdot \frac{x^4}{4} + C = -2x^4 + C }

So, our integrating factor is:

μ(x)=e−2x4+C{ \mu(x) = e^{-2x^4 + C} }

Since we only need one integrating factor, we can ignore the constant of integration C{C} (we'll deal with constants later when solving the differential equation). Thus:

μ(x)=e−2x4{ \mu(x) = e^{-2x^4} }

This is our integrating factor! The exponential function makes this integrating factor particularly effective because it will neatly combine with the other terms in the differential equation when we multiply through. This simplification is what makes the integrating factor method so powerful. By choosing μ(x)=e−2x4{\mu(x) = e^{-2x^4}}, we're setting up the equation to be in a form that we can easily integrate, which is the key to finding the solution.

Applying the Initial Condition μ(0) = 1

Okay, we've found μ(x)=e−2x4{\mu(x) = e^{-2x^4}}, but we also have the condition that μ(0)=1{\mu(0) = 1}. Let's check if our μ(x){\mu(x)} satisfies this condition:

μ(0)=e−2(0)4=e0=1{ \mu(0) = e^{-2(0)^4} = e^0 = 1 }

Great! Our integrating factor already satisfies the condition μ(0)=1{\mu(0) = 1}. Sometimes, we might need to adjust our integrating factor using the constant of integration from the integral, but in this case, we don't need to. This is a nice bonus! The fact that our calculated integrating factor already meets the initial condition simplifies our work significantly. It means we can proceed directly to using μ(x)=e−2x4{\mu(x) = e^{-2x^4}} to solve the differential equation without any further modifications. This check is a crucial step to ensure we're on the right track and haven't made any mistakes in our calculations.

Multiplying the Equation by the Integrating Factor

Now, we multiply our differential equation

y′−8x3y=6{ y' - 8x^3y = 6 }

by the integrating factor μ(x)=e−2x4{\mu(x) = e^{-2x^4}}:

e−2x4y′−8x3e−2x4y=6e−2x4{ e^{-2x^4}y' - 8x^3e^{-2x^4}y = 6e^{-2x^4} }

The left side should now be the derivative of μ(x)y{\mu(x)y}. Let's verify this by computing the derivative of e−2x4y{e^{-2x^4}y} using the product rule:

ddx(e−2x4y)=e−2x4y′+y(−8x3e−2x4)=e−2x4y′−8x3e−2x4y{ \frac{d}{dx}(e^{-2x^4}y) = e^{-2x^4}y' + y(-8x^3e^{-2x^4}) = e^{-2x^4}y' - 8x^3e^{-2x^4}y }

This matches the left side of our equation, which confirms that we've chosen the correct integrating factor. This is a critical step in the process. The goal of using an integrating factor is to transform the left side of the equation into a recognizable derivative. By multiplying through by e−2x4{e^{-2x^4}}, we've successfully achieved this, making the equation much easier to handle. The verification step ensures that our calculations are correct and that the integrating factor has indeed done its job.

Integrating Both Sides

Since the left side is the derivative of e−2x4y{e^{-2x^4}y}, we can rewrite our equation as:

ddx(e−2x4y)=6e−2x4{ \frac{d}{dx}(e^{-2x^4}y) = 6e^{-2x^4} }

Now, integrate both sides with respect to x{x}:

∫ddx(e−2x4y)dx=∫6e−2x4dx{ \int \frac{d}{dx}(e^{-2x^4}y) dx = \int 6e^{-2x^4} dx }

The left side simply becomes e−2x4y{e^{-2x^4}y}, so we have:

e−2x4y=∫6e−2x4dx{ e^{-2x^4}y = \int 6e^{-2x^4} dx }

The integral on the right side, ∫6e−2x4dx{\int 6e^{-2x^4} dx}, doesn't have a simple closed-form solution in terms of elementary functions. We can express it using a special function or leave it in integral form. For the purpose of finding the integrating factor, we've done our job. If we needed to find an explicit solution for y{y}, we might use numerical methods or express the solution in terms of a special function. However, for this problem, we're primarily focused on finding μ(x){\mu(x)}, so we'll proceed with that in mind. Integrating both sides is a key step because it undoes the derivative we created by multiplying by the integrating factor. This allows us to isolate the product e−2x4y{e^{-2x^4}y}, bringing us closer to solving for y{y}.

Finalizing the Integrating Factor

From our calculations, the integrating factor is:

μ(x)=e−2x4{ \mu(x) = e^{-2x^4} }

And it satisfies the condition μ(0)=1{\mu(0) = 1}. So, we've found our answer! This integrating factor is what makes the original differential equation solvable using the techniques we've discussed. We successfully navigated through the process of identifying the correct form of the equation, calculating the integrating factor, and verifying that it meets the given condition. This final step confirms that we've achieved our goal. Knowing the integrating factor allows us to rewrite the differential equation in a way that can be readily integrated, leading us to the general solution. This is a powerful method for solving a wide range of first-order linear differential equations.

Conclusion

So, there you have it! We successfully found the integrating factor μ(x)=e−2x4{\mu(x) = e^{-2x^4}} for the differential equation y′=8x3y+6{y' = 8x^3y + 6}, with the condition μ(0)=1{\mu(0) = 1}. This problem illustrates the power of integrating factors in solving linear differential equations. I hope this breakdown has been helpful and made the process clear. Keep practicing, and you'll become a pro at these in no time! Understanding the concept of integrating factors and how to apply them is a valuable skill in differential equations. By mastering this technique, you'll be well-equipped to tackle a variety of problems and gain a deeper understanding of the behavior of solutions to differential equations. Remember, the key is to break down the problem into manageable steps and to understand the underlying principles behind each step. With practice and perseverance, you can conquer even the most challenging differential equations.