Solving Y' = 6xy - 2x, Y(0) = 3: A Step-by-Step Guide
Hey guys! Today, we're diving into a fun little math problem: solving the differential equation y' = 6xy - 2x with the initial condition y(0) = 3. This might sound intimidating, but trust me, we'll break it down step by step so it's super easy to follow. This is a classic example of a first-order linear differential equation, and we're going to tackle it using a method called the integrating factor. So, grab your favorite beverage, maybe a donut or two, and let's get started!
1. Understanding the Problem
Before we jump into solving, let's make sure we understand what we're dealing with. The equation y' = 6xy - 2x is a differential equation because it involves a derivative (y', which means dy/dx). This equation tells us something about the relationship between a function y(x) and its rate of change. The initial condition, y(0) = 3, gives us a specific point that the solution must pass through. Think of it like a starting point for our solution curve. Our goal is to find the function y(x) that satisfies both the differential equation and the initial condition.
Differential equations pop up all over the place in science and engineering. They're used to model everything from the motion of a pendulum to the spread of a disease. So, understanding how to solve them is a pretty valuable skill! In this particular equation, we see that the rate of change of y depends on both y itself and the variable x. This makes it a bit more interesting than a simple derivative problem. The initial condition is crucial because there are infinitely many functions that satisfy the differential equation alone. The condition y(0) = 3 narrows it down to just one specific solution.
The reason we need an initial condition is because when we solve a differential equation, we're essentially finding a family of functions. Think of it like having a stencil – the differential equation gives us the shape of the stencil, but we can slide it up and down the y-axis. The initial condition acts like a dot that tells us exactly where to place the stencil. Without it, we'd have a whole range of possible solutions, each differing by a constant. So, the initial condition is what makes our solution unique and specific. It anchors our solution curve at a particular point, giving us a definite answer instead of a general form.
2. Rewriting the Equation
The first step in solving this type of differential equation is to rewrite it in a standard form. This makes it easier to identify the different parts of the equation and apply the correct solution method. We want to get the equation into the form:
y' + P(x)y = Q(x)
where P(x) and Q(x) are functions of x. To do this, we'll rearrange our original equation:
y' = 6xy - 2x
Subtract 6xy from both sides to get:
y' - 6xy = -2x
Now, our equation is in the standard form. We can clearly see that P(x) = -6x and Q(x) = -2x. This is a crucial step because the integrating factor method relies on having the equation in this specific format. By rearranging the terms, we've essentially set the stage for the next steps. It's like organizing your ingredients before you start cooking – it makes the whole process smoother and less chaotic.
Getting the equation into this form allows us to use a specific formula for the integrating factor, which is what we'll calculate next. Without this step, we wouldn't be able to apply the method effectively. The standard form highlights the key components of the equation, making it easier to see the structure and how the different terms relate to each other. It’s like putting on your glasses before reading a book – everything becomes much clearer and easier to understand. This standard form is a gateway to solving the equation, and we've just unlocked it!
3. Finding the Integrating Factor
The integrating factor is a special function that we multiply our differential equation by to make it easier to solve. It's like a magic ingredient that transforms a seemingly difficult problem into a manageable one. The integrating factor, often denoted by μ(x), is calculated using the following formula:
μ(x) = e^(∫P(x) dx)
where P(x) is the same function we identified in the previous step (the coefficient of y in the standard form). In our case, P(x) = -6x. So, we need to calculate the integral of -6x with respect to x:
∫-6x dx = -3x^2 + C
We can ignore the constant of integration C when calculating the integrating factor because it will eventually cancel out in the subsequent steps. So, we'll just use -3x^2. Now, we can plug this into the formula for the integrating factor:
μ(x) = e(-3x2)
This exponential function is our integrating factor. It looks a bit intimidating, but it's going to work wonders for us. The integrating factor is essentially a tool that helps us reverse the product rule of differentiation. It transforms the left-hand side of our differential equation into the derivative of a product, which we can then easily integrate. Think of it like a key that unlocks the solution – it’s the perfect fit for our equation.
The integrating factor is a unique function tailored to our specific differential equation. It’s not just a random formula; it’s carefully constructed to achieve a specific purpose. By multiplying our equation by this factor, we’re essentially setting up the equation for integration. It’s like preparing the canvas before you start painting – you need to prime it so that the paint adheres properly. The integrating factor primes our equation for the next step, making the solution process much smoother and more efficient. It’s a powerful technique that simplifies the problem significantly.
4. Multiplying by the Integrating Factor
Now that we have our integrating factor, μ(x) = e(-3x2), we're going to multiply both sides of our differential equation in standard form by it. This is where the magic really starts to happen. Our equation in standard form was:
y' - 6xy = -2x
Multiplying both sides by e(-3x2), we get:
e(-3x2)y' - 6xe(-3x2)y = -2xe(-3x2)
This might look even more complicated than before, but hold on! The left-hand side of this equation is now the derivative of a product. Specifically, it's the derivative of y(x)e(-3x2). This is the crucial step where the integrating factor shows its power. We've transformed a complex expression into a simple derivative, which we can easily integrate. It’s like turning a tangled mess of wires into a neat, organized cable – it might look like you've just added more wires, but you've actually made things much easier to manage.
The reason this works is due to the product rule of differentiation. If you differentiate y(x)e(-3x2), you get exactly the left-hand side of our equation. The integrating factor is designed to make this happen. It’s not just a coincidence; it’s a carefully engineered trick that simplifies the problem. This step is the heart of the integrating factor method. By multiplying by this special factor, we’ve transformed the equation into a form that we can easily integrate. It’s like finding the right tool for the job – once you have it, the task becomes much easier and more straightforward. The complexity has been cleverly hidden, and we’re now on the home stretch.
5. Integrating Both Sides
As we noted in the previous step, the left-hand side of our equation is now the derivative of a product. So, we can rewrite the equation as:
d/dx [y(x)e(-3x2)] = -2xe(-3x2)
Now, to get rid of the derivative, we simply integrate both sides with respect to x:
∫ d/dx [y(x)e(-3x2)] dx = ∫ -2xe(-3x2) dx
The integral on the left side cancels out the derivative, leaving us with:
y(x)e(-3x2) = ∫ -2xe(-3x2) dx
Now, we need to evaluate the integral on the right-hand side. This might look a bit tricky, but we can use a simple substitution. Let u = -3x^2. Then, du = -6x dx. We can rewrite our integral as:
∫ -2xe(-3x2) dx = (1/3) ∫ e^u du
This integral is much easier to solve:
(1/3) ∫ e^u du = (1/3)e^u + C
Substituting back for u, we get:
(1/3)e(-3x2) + C
So, our equation becomes:
y(x)e(-3x2) = (1/3)e(-3x2) + C
We've successfully integrated both sides, and we're one step closer to finding our solution. Integrating both sides is like undoing the differentiation process. We're essentially reversing the operation that created the derivative, allowing us to isolate the function y(x). This step is a crucial turning point in the solution process. We've moved from dealing with derivatives to dealing with the function itself. It’s like unwrapping a gift – we’re finally getting to the good stuff!
The substitution we used to evaluate the integral is a common technique in calculus. It allows us to simplify complex integrals by changing the variable of integration. This makes the integral more recognizable and easier to solve. The constant of integration, C, appears because there are infinitely many functions whose derivative is -2xe(-3x2). This is where our initial condition will come into play later – it will help us determine the specific value of C that corresponds to our particular solution. For now, we’ve successfully performed the integration, and we’re well on our way to finding the function y(x).
6. Solving for y(x)
To isolate y(x), we need to divide both sides of our equation by e(-3x2):
y(x) = [(1/3)e(-3x2) + C] / e(-3x2)
We can simplify this by dividing each term in the numerator by e(-3x2):
y(x) = (1/3) + Ce(3x2)
This is the general solution to our differential equation. It represents a family of functions that satisfy the equation. The constant C can take on any value, resulting in different solutions. This is where our initial condition comes in – it will help us find the specific value of C that gives us the particular solution we're looking for. Solving for y(x) is like finding the main ingredient in a recipe. We’ve isolated the function we’re interested in, and we’re now ready to fine-tune it using the initial condition.
The general solution represents a whole range of possibilities. It’s like having a map of the area but not knowing exactly where you are. The initial condition acts as a landmark, pinpointing our exact location on the map. The term Ce(3x2) represents the family of solutions, and the constant C determines the specific member of that family that we’re dealing with. This is why the initial condition is so important – it narrows down the possibilities and gives us a unique solution. We’re now in the final stages of solving the problem, and we’re about to bring everything together to get our final answer.
7. Applying the Initial Condition
Now, we'll use the initial condition y(0) = 3 to find the value of C. This means that when x = 0, y(x) = 3. Plug these values into our general solution:
3 = (1/3) + Ce(3(0)2)
Since e^(0) = 1, this simplifies to:
3 = (1/3) + C
Subtract (1/3) from both sides to solve for C:
C = 3 - (1/3) = 8/3
Now that we have the value of C, we can plug it back into our general solution to get the particular solution:
y(x) = (1/3) + (8/3)e(3x2)
This is the solution to our differential equation that satisfies the initial condition y(0) = 3. Applying the initial condition is like adding the final touch to a masterpiece. We’ve taken the general solution and tailored it to fit our specific problem. The initial condition acts as a constraint, forcing the solution to pass through a particular point. This gives us a unique solution, a function that perfectly describes the behavior we’re interested in. It’s like finding the missing piece of a puzzle – everything clicks into place, and the picture is complete.
The initial condition is what makes our solution concrete and meaningful. Without it, we would have an infinite number of possible solutions. But with it, we have a single, definite answer. This is the power of initial conditions in differential equations – they allow us to pinpoint the specific solution that matches our real-world situation. We’ve now found the function y(x) that not only satisfies the differential equation but also starts at the point specified by the initial condition. This is the ultimate goal of solving an initial value problem, and we’ve successfully achieved it!
8. The Final Solution
So, the solution to the differential equation y' = 6xy - 2x with the initial condition y(0) = 3 is:
y(x) = (1/3) + (8/3)e(3x2)
That's it! We've successfully navigated through the steps and found the solution. This function describes the behavior of y as a function of x, and it satisfies both the differential equation and the initial condition. We started with a seemingly complex problem, but by breaking it down into smaller, manageable steps, we were able to solve it. Remember, differential equations might seem daunting at first, but with the right techniques and a bit of practice, you can conquer them! The final solution is the culmination of all our efforts. It’s the answer we’ve been searching for, the function that perfectly fits the problem we set out to solve. This function tells us everything we need to know about the relationship between y and x, given the conditions we started with. It’s like reaching the summit of a mountain after a long climb – the view is spectacular, and the sense of accomplishment is immense.
Congratulations, guys! You made it through a differential equation problem! Remember, practice makes perfect, so keep tackling those problems and you'll become a differential equation master in no time. Keep exploring, keep learning, and keep having fun with math!