Solve The Differential Equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0

This article provides a step-by-step guide to solving the differential equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0. This type of equation falls under the category of first-order ordinary differential equations, and we will explore techniques to find its solution. Our main goal is to transform the given equation into a form that we can readily integrate. Let's dive in and unravel the solution to this fascinating problem. Understanding differential equations is crucial in many fields such as physics, engineering, and economics, as they often model dynamic systems and processes. This particular equation, while seemingly complex, showcases the beauty and elegance of mathematical problem-solving. Through careful manipulation and application of relevant techniques, we will demonstrate how to arrive at a general solution, and in doing so, enhance your understanding of differential equations and their applications.

Identifying the Type of Differential Equation

Before attempting to solve the differential equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0, it's crucial to identify its type. This equation is a first-order ordinary differential equation, which means it involves only first derivatives and a single independent variable. Specifically, we need to check if it's an exact differential equation. An exact differential equation is one that can be expressed as the total differential of some function, say, ${F(x, y)}$. If the equation is in the form ${M(x, y)dx + N(x, y)dy = 0}$, it is exact if and only if ${\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}}$. This criterion is a cornerstone in determining the solvability of a differential equation in this manner. In our case, ${M(x, y) = x^4 + 2y}$ and ${N(x, y) = xy^3 + 2y^4 - 4x}$. Therefore, we need to compute the partial derivatives of ${M}$ with respect to ${y}$ and ${N}$ with respect to ${x}$. By verifying the equality of these partial derivatives, we confirm whether the equation is exact and thus amenable to a specific solution method. This preliminary step is vital as it dictates the approach we will take to find the general solution.

Checking for Exactness

To check if the given differential equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0 is exact, we need to verify the condition ${\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}}$. Let's calculate the partial derivatives. We have ${M(x, y) = x^4 + 2y}$ and ${N(x, y) = xy^3 + 2y^4 - 4x}$. The partial derivative of ${M}$ with respect to ${y}$ is:

${ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^4 + 2y) = 2 }$

And the partial derivative of ${N}$ with respect to ${x}$ is:

${ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(xy^3 + 2y^4 - 4x) = y^3 - 4 }$

Comparing these partial derivatives, we see that ${\frac{\partial M}{\partial y} = 2}$ and ${\frac{\partial N}{\partial x} = y^3 - 4}$. Since ${2 \neq y^3 - 4}$, the given differential equation is not exact. This means we cannot directly find a function ${F(x, y)}$ such that ${dF = Mdx + Ndy}$. The non-exactness necessitates the use of alternative methods, such as finding an integrating factor, to transform the equation into an exact form. Understanding this distinction is vital in selecting the correct solution technique.

Finding an Integrating Factor

Since the differential equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0 is not exact, we need to find an integrating factor to make it exact. An integrating factor, denoted by ${\mu(x, y)}$, is a function that, when multiplied by the original equation, transforms it into an exact equation. There are several ways to find an integrating factor, but a common approach is to check for integrating factors that depend only on one variable, either ${x}$ or ${y}$. We can use the following formulas:

1. If ${\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}}$ is a function of ${x}$ only, say ${f(x)}$, then the integrating factor ${\mu(x) = e^{\int f(x) dx}}$.
2. If ${\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}}$ is a function of ${y}$ only, say ${g(y)}$, then the integrating factor ${\mu(y) = e^{\int g(y) dy}}$.

Let's calculate these expressions for our equation. We already found that ${\frac{\partial M}{\partial y} = 2}$ and ${\frac{\partial N}{\partial x} = y^3 - 4}$. Thus,

${ \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{2 - (y^3 - 4)}{xy^3 + 2y^4 - 4x} = \frac{6 - y^3}{xy^3 + 2y^4 - 4x} }$

This expression does not appear to be a function of ${x}$ only. Now, let's try the second formula:

${ \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{(y^3 - 4) - 2}{x^4 + 2y} = \frac{y^3 - 6}{x^4 + 2y} }$

This expression also does not appear to be a function of ${y}$ only. However, upon closer inspection and potentially through trial and error or recognition of a pattern, we might consider an integrating factor of the form ${\mu(y) = y^k}$. Multiplying the original equation by ${y^k}$ and re-evaluating the exactness condition can lead to a suitable integrating factor. This approach showcases the ingenuity sometimes required in solving differential equations.

Multiplying by the Integrating Factor

After exploring various possibilities, let's consider an integrating factor of the form ${\mu(y) = y^2}$. Multiplying the original differential equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0 by this integrating factor, we get:

${ y^2(x^4 + 2y)dx + y^2(xy^3 + 2y^4 - 4x)dy = 0 }$

${ (x^4y^2 + 2y^3)dx + (xy^5 + 2y^6 - 4xy^2)dy = 0 }$

Now, we have a new equation with ${M_1(x, y) = x^4y^2 + 2y^3}$ and ${N_1(x, y) = xy^5 + 2y^6 - 4xy^2}$. We need to check if this new equation is exact by computing the partial derivatives again:

${ \frac{\partial M_1}{\partial y} = \frac{\partial}{\partial y}(x^4y^2 + 2y^3) = 2x^4y + 6y^2 }$

${ \frac{\partial N_1}{\partial x} = \frac{\partial}{\partial x}(xy^5 + 2y^6 - 4xy^2) = y^5 - 4y^2 }$

These partial derivatives are still not equal, indicating that ${\mu(y) = y^2}$ is not the correct integrating factor. This outcome is a reminder that finding an integrating factor can sometimes involve trying different forms and carefully re-evaluating the exactness condition. It also highlights the importance of precision in calculations. Let's try another integrating factor, ${\mu(y) = y}$. Multiplying the original equation by ${y}$ we obtain:

${ (x^4y + 2y^2)dx + (xy^4 + 2y^5 - 4xy)dy = 0 }$

Now, ${M_1(x, y) = x^4y + 2y^2}$ and ${N_1(x, y) = xy^4 + 2y^5 - 4xy}$. Let's check for exactness:

${ \frac{\partial M_1}{\partial y} = x^4 + 4y }$

${ \frac{\partial N_1}{\partial x} = y^4 - 4y }$

Still not exact. We need to re-evaluate our approach for finding the integrating factor. A more systematic method might involve looking for a function of the form ${\mu(x,y)}$ that satisfies the exactness condition directly. This process often involves more advanced techniques or potentially recognizing patterns that emerge from the structure of the differential equation itself.

Solving the Exact Equation

Let's go back and analyze the expressions we derived for the potential integrating factors. We had:

${ \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{2 - (y^3 - 4)}{xy^3 + 2y^4 - 4x} = \frac{6 - y^3}{xy^3 + 2y^4 - 4x} }$

${ \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{(y^3 - 4) - 2}{x^4 + 2y} = \frac{y^3 - 6}{x^4 + 2y} }$

It seems we haven't found a simple integrating factor that depends solely on ${x}$ or ${y}$. This suggests that the integrating factor might be a more complex function of both ${x}$ and ${y}$, or that we might need a different approach altogether. However, let's reconsider our calculations and the structure of the original equation:

${ (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0 }$

Upon careful observation, we can rearrange the terms to try to identify a potential integrating factor. Notice that the equation can be rewritten as:

${ x^4 dx - 4x dy + 2y dx + xy^3 dy + 2y^4 dy = 0 }$

Now, let's try to group terms and see if we can identify a pattern:

${ (x^4 dx - 4x dy) + (2y dx + xy^3 dy + 2y^4 dy) = 0 }$

We are looking for an integrating factor that will make the equation exact. Let's try an integrating factor of the form ${\mu(x, y) = x}$. Multiplying the original equation by ${x}$, we get:

${ (x^5 + 2xy)dx + (x^2y^3 + 2xy^4 - 4x^2)dy = 0 }$

Now, ${M_1(x, y) = x^5 + 2xy}$ and ${N_1(x, y) = x^2y^3 + 2xy^4 - 4x^2}$. Let's check for exactness:

${ \frac{\partial M_1}{\partial y} = 2x }$

${ \frac{\partial N_1}{\partial x} = 2xy^3 + 2y^4 - 8x }$

This is still not exact. This iterative process highlights the complexity of solving non-exact differential equations. We need to find a function ${F(x, y)}$ such that ${\frac{\partial F}{\partial x} = M_1}$ and ${\frac{\partial F}{\partial y} = N_1}$. If we can find such a function, the solution to the differential equation is given by ${F(x, y) = C}$, where ${C}$ is a constant.

Finding the General Solution

Given the challenges we've encountered in finding a simple integrating factor, let's consider another approach. Sometimes, recognizing a specific form within the equation can lead to a solution. Let's revisit the original equation:

${ (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0 }$

Let's rearrange the terms again:

${ x^4 dx + 2y dx + xy^3 dy + 2y^4 dy - 4x dy = 0 }$

This can be rewritten as:

${ x^4 dx + 2y^4 dy + (2y dx + xy^3 dy - 4x dy) = 0 }$

Notice that we can integrate the first two terms directly:

${ \int x^4 dx = \frac{1}{5}x^5 }$

${ \int 2y^4 dy = \frac{2}{5}y^5 }$

So, if we can find a function whose differential matches the remaining terms, we can piece together the general solution. Let's focus on the term ${(2y dx + xy^3 dy - 4x dy)}$. We are looking for a function ${F(x, y)}$ such that:

${ dF = 2y dx + xy^3 dy - 4x dy }$

This suggests that ${\frac{\partial F}{\partial x} = 2y}$ and ${\frac{\partial F}{\partial y} = xy^3 - 4x}$. Integrating ${\frac{\partial F}{\partial x} = 2y}$ with respect to ${x}$, we get:

${ F(x, y) = 2xy + g(y) }$

where ${g(y)}$ is an arbitrary function of ${y}$. Now, we differentiate this with respect to ${y}$:

${ \frac{\partial F}{\partial y} = 2x + g'(y) }$

We want this to be equal to ${xy^3 - 4x}$, so:

${ 2x + g'(y) = xy^3 - 4x }$

${ g'(y) = xy^3 - 6x }$

This equation is problematic because ${g'(y)}$ should be a function of ${y}$ only, but it contains an ${x}$ term. This indicates that there might be an error in our approach or that a simple integrating factor or direct integration isn't sufficient for this particular differential equation. This highlights the complexity of differential equations and the need for a variety of techniques to tackle them.

Conclusion

In conclusion, solving the differential equation (x^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0 proved to be a complex task. We started by identifying the type of equation and checking for exactness, which revealed that the equation was not exact. We attempted to find an integrating factor using standard methods, but none of the initial attempts were successful. Through rearranging terms and trying different integrating factors, we encountered challenges in making the equation exact and finding a suitable function to integrate. This process illustrates that not all differential equations can be solved using straightforward methods and may require advanced techniques or numerical solutions. The journey through this problem highlights the importance of careful calculation, persistence, and the application of various strategies in solving differential equations. While we did not arrive at a final solution in this exploration, the process provided valuable insights into the complexities of differential equations and the methods used to approach them.