Solving Dy/dx = 2x³√(y - 4xy) Find Y(1)

Hey guys! Today, we're diving into a fun and challenging differential equation problem. We've got a differential equation that looks a bit intimidating at first glance, but don't worry, we'll break it down step by step and conquer it together. Our mission is to find the value of y(1) given the differential equation and an initial condition. Let's jump right in!

Understanding the Problem

Our main task here is to solve the differential equation:

dy/dx = 2x³√(y - 4xy)

with the initial condition y(0) = 0. This means we need to find a function y that satisfies this equation and also passes through the point (0, 0). Once we find that function, we can simply plug in x = 1 to find y(1). These types of problems are common in calculus and differential equations courses, and they test our ability to manipulate equations, integrate, and apply initial conditions. So, let’s get our hands dirty and work through the solution!

When approaching a differential equation, the first step is always to identify its type. In this case, we have a first-order differential equation, meaning it involves the first derivative dy/dx. The presence of the square root term, √(y - 4xy), makes it a bit tricky. Our goal will be to simplify this equation into a form that we can solve using standard techniques. Often, this involves separation of variables, where we try to get all the y terms on one side and all the x terms on the other. We might also consider substitutions to simplify the equation further. Let's see how we can manipulate the given equation to make it more manageable.

Let's rewrite the square root term to see if it sparks any ideas. We have √(y - 4xy), which can be factored as √[y(1 - 4x)]. This might suggest that separating variables could be a viable strategy. We want to isolate y terms on one side and x terms on the other. Remember, our ultimate goal is to integrate both sides of the equation to find y as a function of x. So, keeping this in mind, let's proceed with the separation of variables and see where it leads us. It’s like solving a puzzle, guys, where each step gets us closer to the final solution!

Separating Variables

The first key step in solving this differential equation is separating the variables. This means we want to rearrange the equation so that all the y terms are on one side and all the x terms are on the other. Our equation is:

dy/dx = 2x³√(y - 4xy)

First, let’s rewrite the term inside the square root to make it clearer:

dy/dx = 2x³√[y(1 - 4x)]

Now, we can separate the variables by dividing both sides by √[y(1 - 4x)] and multiplying both sides by dx. This gives us:

dy / √[y(1 - 4x)] = 2x³ dx

However, notice that we have a slight issue here. The term (1 - 4x) is still inside the square root and is mixed with y. To fully separate the variables, we need to isolate y completely. This requires us to make a small adjustment. Instead of directly separating, let's rewrite the equation in a slightly different form to make the separation cleaner. Think of this as choosing the right tool for the job – sometimes, a small tweak can make a big difference!

Let's go back to the original form of the equation:

dy/dx = 2x³√(y - 4xy)

We can factor out y from the square root:

dy/dx = 2x³√[y(1 - 4x)] = 2x³√y √(1 - 4x)

Now, the separation becomes more straightforward. We divide both sides by √y and multiply both sides by dx:

(1 / √y) dy = 2x³√(1 - 4x) dx

Great! Now we have all the y terms on the left side and all the x terms on the right side. This is a crucial step because now we can integrate both sides independently. Separating variables allows us to treat the differential equation as two separate integrals, which is a much simpler task. It’s like breaking a big problem into smaller, manageable pieces. We’re making progress, guys! Next, we’ll integrate both sides and see what we get.

Integrating Both Sides

Now that we’ve separated the variables, the next step is to integrate both sides of the equation. We have:

(1 / √y) dy = 2x³√(1 - 4x) dx

Let’s integrate the left side first. The integral of (1 / √y) with respect to y can be rewritten as the integral of y^(-1/2). This is a simple power rule integration:

∫ y^(-1/2) dy = 2y^(1/2) + C₁

where C₁ is the constant of integration. Remember, when we integrate, we always need to include a constant of integration because the derivative of a constant is zero. So, when we reverse the process, we need to account for any possible constant term.

Now, let’s tackle the right side of the equation:

∫ 2x³√(1 - 4x) dx

This integral looks a bit more complicated. We might need to use a substitution to solve it. Let’s try substituting u = 1 - 4x. Then, du = -4 dx, or dx = -du/4. Also, we need to express x in terms of u. From u = 1 - 4x, we get x = (1 - u)/4. Now, we can substitute these into the integral:

∫ 2x³√(1 - 4x) dx = ∫ 2[(1 - u)/4]³√u (-du/4)

This simplifies to:

∫ 2[(1 - u)³/64]√u (-du/4) = -1/128 ∫ (1 - u)³√u du

Expanding (1 - u)³ gives us 1 - 3u + 3u² - u³. So, the integral becomes:

-1/128 ∫ (1 - 3u + 3u² - u³)√u du = -1/128 ∫ (u^(1/2) - 3u^(3/2) + 3u^(5/2) - u^(7/2)) du

Now, we can integrate term by term using the power rule:

-1/128 [2/3 u^(3/2) - 3(2/5)u^(5/2) + 3(2/7)u^(7/2) - 2/9 u^(9/2)] + C₂

where C₂ is another constant of integration. This looks a bit messy, but we’re almost there! Remember, we're just applying the rules of integration step by step. It’s like building a complex structure – each component has its place, and together they form the whole solution.

Now, we need to substitute back u = 1 - 4x and simplify the expression. This will give us the integral in terms of x. Once we have both integrals, we’ll combine them and use the initial condition to find the constant of integration. So, let’s keep going, guys! We’re on the right track.

Solving for y

Alright, we've integrated both sides of our separated differential equation. Now, let's put the pieces together and solve for y. We had:

∫ (1 / √y) dy = ∫ 2x³√(1 - 4x) dx

We found the left side integral to be:

2√y + C₁

And the right side integral, after substituting u = 1 - 4x, was:

-1/128 [2/3 u^(3/2) - 6/5 u^(5/2) + 6/7 u^(7/2) - 2/9 u^(9/2)] + C₂

Substituting back u = 1 - 4x, we get a rather lengthy expression. Let’s call it F(x) for simplicity:

F(x) = -1/128 [2/3 (1 - 4x)^(3/2) - 6/5 (1 - 4x)^(5/2) + 6/7 (1 - 4x)^(7/2) - 2/9 (1 - 4x)^(9/2)] + C₂

So, our equation now looks like:

2√y = F(x) + C

where C = C₂ - C₁ is a combined constant of integration. This is a crucial step – we’ve related y and x through this equation. Now, we need to use the initial condition y(0) = 0 to find the value of C. This is where the initial condition comes into play, giving us a specific solution rather than a general one. It’s like finding the exact path on a map rather than just a general direction.

Plugging in x = 0 and y = 0, we get:

2√0 = F(0) + C

So, 0 = F(0) + C, which means C = -F(0). Let’s calculate F(0):

F(0) = -1/128 [2/3 (1)^(3/2) - 6/5 (1)^(5/2) + 6/7 (1)^(7/2) - 2/9 (1)^(9/2)]

F(0) = -1/128 [2/3 - 6/5 + 6/7 - 2/9]

Now, we need to find a common denominator for these fractions, which is 315. So:

F(0) = -1/128 [(210 - 378 + 270 - 70) / 315]

F(0) = -1/128 [32 / 315] = -1/1260

Thus, C = -F(0) = 1/1260. Now we have the specific solution:

2√y = -1/128 [2/3 (1 - 4x)^(3/2) - 6/5 (1 - 4x)^(5/2) + 6/7 (1 - 4x)^(7/2) - 2/9 (1 - 4x)^(9/2)] + 1/1260

We're getting closer to our final answer! Now we need to find y(1), which means plugging in x = 1 into this equation and solving for y. This might seem like a daunting task with all these fractions and exponents, but we're in the home stretch now. Let’s do it!

Finding y(1)

Okay, guys, we’re in the final stretch! We've got the equation:

2√y = -1/128 [2/3 (1 - 4x)^(3/2) - 6/5 (1 - 4x)^(5/2) + 6/7 (1 - 4x)^(7/2) - 2/9 (1 - 4x)^(9/2)] + 1/1260

We need to find y(1), so we plug in x = 1:

2√y(1) = -1/128 [2/3 (1 - 4)^(3/2) - 6/5 (1 - 4)^(5/2) + 6/7 (1 - 4)^(7/2) - 2/9 (1 - 4)^(9/2)] + 1/1260

Notice that (1 - 4) = -3. Since we have fractional exponents with a denominator of 2, we are taking square roots of negative numbers, which will result in imaginary numbers. This indicates that there might be a domain issue with our solution, or perhaps we made a mistake in our calculations. Let’s go back and review our steps to make sure everything is correct. It’s always a good idea to double-check our work, especially when dealing with complex calculations!

Upon reviewing, we notice that the original differential equation has a term √(y - 4xy). For this to be real, we need y - 4xy ≥ 0, which means y(1 - 4x) ≥ 0. When x = 1, this becomes y(1 - 4) ≥ 0, or -3y ≥ 0, which implies y ≤ 0. Since our initial condition is y(0) = 0, and we are looking for y(1), we need to consider the behavior of the function in this domain.

However, the presence of imaginary numbers when plugging in x = 1 suggests that our current approach might have led us to a solution that is not valid for x = 1. We need to reconsider our steps, especially the integration and substitution parts, to see if we can find a different approach or identify any errors.

Sometimes, in problems like these, a seemingly straightforward approach can lead to complications. It's essential to stay flexible and be willing to try different methods. So, let's take a step back and see if there’s another way to tackle this differential equation.

Let's rewrite the original differential equation again:

dy/dx = 2x³√(y - 4xy)

We factored out y inside the square root:

dy/dx = 2x³√[y(1 - 4x)]

We separated variables and integrated. However, the complexity of the right-side integral led us to a complicated expression. Maybe there’s a simpler substitution we could have used, or perhaps another approach altogether.

One thing we haven't explicitly tried is to check if the differential equation is separable in a different way, or if there’s a substitution that simplifies the entire equation rather than just the integral. Let’s think about that for a moment. What if we tried to make the term inside the square root a perfect square? This might lead to a more manageable integral. We’re brainstorming here, guys, and that’s a crucial part of problem-solving!

Given the complexity we encountered, it's possible that the solution involves a special function or a more advanced technique that we haven't considered yet. Differential equations can be tricky, and sometimes the “obvious” path isn't the most efficient one. We might need to consult additional resources or explore alternative methods to solve this particular problem.

Let’s pause here and recap what we’ve done. We identified the differential equation, separated variables, integrated both sides (which led to a complicated integral), applied the initial condition, and then encountered imaginary numbers when trying to find y(1). This suggests that we need to re-evaluate our approach. It’s like realizing you’ve taken a wrong turn – sometimes, the best thing to do is to go back and find a new route.

To continue, let's consider other techniques for solving differential equations, such as Bernoulli's equation or other types of substitutions. We might also look for resources that discuss similar problems or offer alternative solution methods. Solving complex problems often involves a combination of techniques and a willingness to explore different avenues. We'll keep at it, guys, and figure this out!

Conclusion

In this article, we tackled the differential equation dy/dx = 2x³√(y - 4xy) with the initial condition y(0) = 0, aiming to find y(1). We walked through the process of separating variables and integrating, but we encountered complexities when evaluating the integrals and ended up with an expression that led to imaginary numbers when trying to find y(1). This indicates that we need to re-evaluate our approach and possibly explore alternative methods or techniques to solve this problem. Differential equations can be challenging, and it's important to be persistent and adaptable in our problem-solving strategies. We'll continue to explore different avenues and resources to find the correct solution. Stay tuned for more updates as we delve deeper into this problem! Remember, guys, problem-solving is a journey, and sometimes the most valuable lessons are learned when we encounter obstacles and find new ways to overcome them.