Solving The Differential Equation Y' = 8 - 16x^3 - 10e^(-5x+5) With Y(1) = 4

by Axel SΓΈrensen 77 views

Hey everyone! Today, we're diving into a super interesting problem from the realm of differential equations. We've got a first-order differential equation that we need to solve, but that's not all – we also have an initial condition to satisfy. This means we're not just looking for any solution; we need the specific solution that passes through the point (1,4). So, grab your thinking caps, and let's get started!

Understanding the Problem

Okay, let's break down what we're dealing with here. We're given the differential equation yβ€²=8βˆ’16x3βˆ’10eβˆ’5x+5y^{\prime} = 8 - 16x^3 - 10e^{-5x+5}. Now, what does this actually mean? Well, yβ€²y^{\prime} represents the derivative of the function yy with respect to xx. In simpler terms, it's the rate of change of yy as xx changes. Our goal is to find the function y(x)y(x) that satisfies this equation.

But wait, there's more! We also have the initial condition (1,4). This tells us that when x=1x = 1, the value of the function yy is 4. This is crucial because it will help us pinpoint the exact solution we're looking for, eliminating any extra constants that might pop up during the solving process. Think of it like this: the differential equation gives us a family of solutions, and the initial condition tells us which specific member of that family we want.

To summarize, we need to:

  1. Find the general solution to the differential equation yβ€²=8βˆ’16x3βˆ’10eβˆ’5x+5y^{\prime} = 8 - 16x^3 - 10e^{-5x+5}.
  2. Use the initial condition (1,4) to determine the particular solution.

Step-by-Step Solution

1. Finding the General Solution

The key to solving this differential equation is recognizing that it's a first-order equation where we can directly integrate both sides. Remember, integration is essentially the reverse process of differentiation. If we have the derivative of a function, we can integrate it to find the original function (plus a constant of integration, of course!).

So, let's integrate both sides of the equation yβ€²=8βˆ’16x3βˆ’10eβˆ’5x+5y^{\prime} = 8 - 16x^3 - 10e^{-5x+5} with respect to xx:

∫yβ€²dx=∫(8βˆ’16x3βˆ’10eβˆ’5x+5)dx\int y^{\prime} dx = \int (8 - 16x^3 - 10e^{-5x+5}) dx

The left side is straightforward: the integral of yβ€²y^{\prime} with respect to xx is simply y(x)y(x). On the right side, we need to integrate each term separately.

  • ∫8dx=8x\int 8 dx = 8x

  • βˆ«βˆ’16x3dx=βˆ’16∫x3dx=βˆ’16βˆ—x44=βˆ’4x4\int -16x^3 dx = -16 \int x^3 dx = -16 * \frac{x^4}{4} = -4x^4

  • βˆ«βˆ’10eβˆ’5x+5dx=βˆ’10∫eβˆ’5x+5dx\int -10e^{-5x+5} dx = -10 \int e^{-5x+5} dx. For this one, we'll use a little u-substitution. Let u=βˆ’5x+5u = -5x + 5. Then, du=βˆ’5dxdu = -5 dx, so dx=duβˆ’5dx = \frac{du}{-5}. Substituting these into the integral, we get:

    βˆ’10∫euduβˆ’5=2∫eudu=2eu=2eβˆ’5x+5-10 \int e^u \frac{du}{-5} = 2 \int e^u du = 2e^u = 2e^{-5x+5}

Now, let's put it all together. The general solution is:

y(x)=8xβˆ’4x4+2eβˆ’5x+5+Cy(x) = 8x - 4x^4 + 2e^{-5x+5} + C

where C is the constant of integration. This constant represents the family of solutions we talked about earlier. Each different value of C gives us a different curve that satisfies the differential equation.

2. Using the Initial Condition to Find the Particular Solution

This is where the fun really begins! We have a general solution, but we want the specific solution that passes through the point (1,4). This means when x=1x = 1, y=4y = 4. Let's plug these values into our general solution and see what we get:

4=8(1)βˆ’4(1)4+2eβˆ’5(1)+5+C4 = 8(1) - 4(1)^4 + 2e^{-5(1)+5} + C

Simplify this equation:

4=8βˆ’4+2e0+C4 = 8 - 4 + 2e^0 + C

Since e0=1e^0 = 1, we have:

4=8βˆ’4+2+C4 = 8 - 4 + 2 + C

4=6+C4 = 6 + C

Now, solve for C:

C=4βˆ’6=βˆ’2C = 4 - 6 = -2

Excellent! We've found the value of the constant of integration that corresponds to our specific solution.

3. The Particular Solution

Now that we know C=βˆ’2C = -2, we can plug this value back into our general solution to get the particular solution:

y(x)=8xβˆ’4x4+2eβˆ’5x+5βˆ’2y(x) = 8x - 4x^4 + 2e^{-5x+5} - 2

And that's it! This is the solution to the differential equation that passes through the point (1,4). We've successfully navigated the world of differential equations and found our treasure!

Comparing with the Given Options

Let's take a look at the answer choices provided and see which one matches our solution:

A. y=βˆ’4x4+2eβˆ’5x+5+6y = -4x^4 + 2e^{-5x+5} + 6 B. y=8xβˆ’4x4+2eβˆ’5x+5βˆ’2y = 8x - 4x^4 + 2e^{-5x+5} - 2 C. y=8xβˆ’4x4βˆ’10eβˆ’5x+5+10y = 8x - 4x^4 - 10e^{-5x+5} + 10 D. y=8xβˆ’4x4βˆ’2eβˆ’5x+5+2y = 8x - 4x^4 - 2e^{-5x+5} + 2

Ah, Option B is a perfect match! Our solution, y(x)=8xβˆ’4x4+2eβˆ’5x+5βˆ’2y(x) = 8x - 4x^4 + 2e^{-5x+5} - 2, aligns perfectly with option B.

Key Concepts and Takeaways

Before we wrap up, let's recap the key concepts and takeaways from this problem. This will help solidify your understanding and make you even more confident when tackling similar problems in the future.

  • Differential Equations: At their heart, differential equations describe relationships between functions and their derivatives. They're used extensively in physics, engineering, economics, and many other fields to model dynamic systems.
  • First-Order Differential Equations: These are differential equations that involve the first derivative of the unknown function. They often model processes that change over time.
  • General Solution: This is the family of solutions to a differential equation, typically involving a constant of integration (like C in our problem). It represents all possible solutions to the equation.
  • Initial Condition: This is a specific point that the solution curve must pass through. It helps us narrow down the general solution to a unique particular solution.
  • Particular Solution: This is the specific solution to a differential equation that satisfies both the equation and the initial condition. It's the unique function that solves our problem.
  • Integration: This is the process of finding the antiderivative of a function. It's the reverse operation of differentiation and is crucial for solving many differential equations.
  • U-Substitution: This is a technique used to simplify integrals by substituting a new variable for a more complex expression within the integral. It's like a secret weapon for taming tricky integrals!

Real-World Applications

You might be thinking, "Okay, this is cool, but where would I actually use this stuff?" Well, differential equations are everywhere in the real world!

  • Physics: They're used to model the motion of objects, the flow of heat, the behavior of circuits, and countless other phenomena.
  • Engineering: They're essential for designing bridges, aircraft, and all sorts of structures and systems.
  • Biology: They can model population growth, the spread of diseases, and the interactions between species.
  • Economics: They're used to study economic growth, market dynamics, and financial models.

The ability to solve differential equations is a powerful tool that can unlock a deeper understanding of the world around us. So, keep practicing and exploring, and you'll be amazed at what you can achieve!

Practice Makes Perfect

Solving differential equations is a skill that improves with practice. The more problems you tackle, the more comfortable you'll become with the techniques and concepts involved. Here are a few tips to help you on your journey:

  • Start with the basics: Make sure you have a solid understanding of integration and differentiation. These are the fundamental building blocks for solving differential equations.
  • Work through examples: Find worked examples in textbooks or online resources and carefully follow each step. This will help you see how the different techniques are applied.
  • Practice, practice, practice: The best way to learn is by doing. Solve as many problems as you can, starting with simpler ones and gradually moving on to more challenging ones.
  • Don't be afraid to ask for help: If you get stuck, don't hesitate to ask your teacher, classmates, or online communities for assistance. Collaboration is a great way to learn and overcome challenges.
  • Check your answers: Whenever possible, check your solutions by plugging them back into the original differential equation and initial condition. This will help you catch any errors and build confidence in your work.

Conclusion

Alright guys, we've reached the end of our journey through this differential equation problem. We successfully found the particular solution that satisfies both the equation and the initial condition. Remember, the key to mastering differential equations is to break down the problem into smaller steps, understand the underlying concepts, and practice consistently. So keep up the great work, and you'll be solving even the trickiest differential equations in no time!

Choose the Correct Answer

Based on our calculations, the correct answer is:

B. y=8xβˆ’4x4+2eβˆ’5x+5βˆ’2y = 8x - 4x^4 + 2e^{-5x+5} - 2