Solving Differential Equations A Comprehensive Guide

JU07/16/2025 08, 2025 by THE IDEN 53 views

In the realm of differential equations, finding solutions can often feel like navigating a complex labyrinth. This article aims to serve as your guide, providing a comprehensive exploration of three distinct differential equations: 4xy' + y = x⁴y³, xy² + y³ = (cos x)/x, and x²y' - y³ + 2xy = 0. We will delve into the methods required to solve each equation, offering step-by-step explanations and insightful discussions to enhance your understanding. This exploration is crucial for anyone studying mathematics, physics, engineering, or any field where differential equations play a pivotal role. Understanding how to approach and solve these equations not only sharpens your analytical skills but also provides a foundational knowledge base for tackling more complex problems in diverse scientific and engineering applications. Our journey through these equations will highlight various techniques and problem-solving strategies, making this a valuable resource for both students and professionals alike. By the end of this discussion, you will have a clearer understanding of the nuances involved in solving different types of differential equations and be better equipped to tackle similar challenges.

1. Solving the Bernoulli Equation: 4xy' + y = x⁴y³

The first equation we'll tackle is 4xy' + y = x⁴y³, which is a classic example of a Bernoulli differential equation. Bernoulli equations are nonlinear differential equations that can be transformed into linear equations through a clever substitution. This transformation is a powerful technique that simplifies the solution process considerably. Understanding how to identify and solve Bernoulli equations is a crucial skill in the study of differential equations, as they frequently appear in various applications, including fluid dynamics, population growth models, and electrical circuits. The general form of a Bernoulli equation is y' + P(x)y = Q(x)yⁿ, where n is a real number not equal to 0 or 1. Our equation fits this form, making the Bernoulli method an ideal approach. We will methodically apply the necessary steps to reduce it to a linear form and then solve it using standard techniques for linear differential equations. This process involves identifying the appropriate substitution, performing the substitution, solving the resulting linear equation, and then transforming the solution back in terms of the original variables. By mastering this technique, you'll be able to handle a wide range of nonlinear equations more effectively. Let's embark on this process by first identifying the key components of our equation and preparing it for the substitution step.

Step-by-step Solution

Divide by 4x: The first step in solving the Bernoulli equation 4xy' + y = x⁴y³ is to divide the entire equation by 4x. This prepares the equation for the subsequent substitution by isolating the y' term and bringing it into a more manageable form. Dividing by 4x gives us: y' + (1/4x)y = (x³/4)y³. This form now clearly reveals the structure of a Bernoulli equation, making it easier to apply the appropriate techniques for solving it. Recognizing the importance of this initial step can significantly simplify the overall solution process. It sets the stage for the subsequent steps, which involve substitution and solving a linear differential equation. This transformation is a fundamental aspect of solving Bernoulli equations, and mastering this step is crucial for handling more complex problems. By isolating the derivative term and arranging the equation in a standard form, we pave the way for a systematic approach that will lead us to the solution.
Divide by y³: Continuing our journey towards solving the Bernoulli equation, the next crucial step involves dividing the entire equation y' + (1/4x)y = (x³/4)y³ by y³. This step is essential for transforming the equation into a linear form, which is a key characteristic of the Bernoulli method. Dividing by y³ yields: y⁻³y' + (1/4x)y⁻² = x³/4. By performing this division, we effectively separate the terms involving the dependent variable y, making it easier to apply a substitution that will linearize the equation. This step is a cornerstone of the Bernoulli technique and is often necessary to simplify complex nonlinear differential equations. The resulting form of the equation now allows us to introduce a new variable that will transform the equation into a linear first-order differential equation, which is significantly easier to solve. Understanding this division step is critical for anyone working with Bernoulli equations, as it lays the foundation for the subsequent substitution and solution process. This strategic manipulation is a prime example of how mathematical transformations can simplify seemingly intractable problems.
Substitution: Now, we introduce a substitution to simplify the equation further. Let v = y⁻², then v' = -2y⁻³y'. This substitution is the heart of the method for solving Bernoulli equations. By introducing a new variable v related to y, we aim to transform the nonlinear equation into a linear one. The derivative v' is carefully chosen to eliminate the nonlinear terms involving y in the original equation. Substituting these expressions into our modified equation, we get: (-1/2)v' + (1/4x)v = x³/4. This substitution is a pivotal step, as it effectively linearizes the differential equation, making it amenable to standard solution techniques for linear equations. The careful selection of the substitution variable is based on the structure of the Bernoulli equation, and understanding this selection process is crucial for mastering the method. By replacing the nonlinear terms with linear ones, we can now apply familiar methods for solving linear differential equations, such as the integrating factor method. This transformation demonstrates the power of mathematical substitutions in simplifying complex problems and making them more accessible to analytical solutions. The substitution not only simplifies the equation but also provides a clear pathway to finding the solution in terms of the original variables.
Simplify and Solve Linear Equation: Multiply through by -2 to get v' - (1/2x)v = -x³/2. This is now a linear first-order differential equation. To solve it, we find the integrating factor. The integrating factor is given by e^(∫(-1/2x) dx) = e^(-(1/2)ln|x|) = x^(-1/2). Multiply the equation by the integrating factor: x^(-1/2)v' - (1/2)x^(-3/2)v = -x^(5/2)/2. The left side is the derivative of x^(-1/2)v, so we have (x^(-1/2)v)' = -x^(5/2)/2. Integrate both sides with respect to x: ∫(x^(-1/2)v)' dx = ∫(-x^(5/2)/2) dx. This gives x^(-1/2)v = -(1/2)(2/7)x^(7/2) + C, or x^(-1/2)v = -x^(7/2)/7 + C, where C is the constant of integration. Solving this linear equation is a crucial step, as it provides the solution in terms of the substituted variable v. By carefully applying the integrating factor method, we transform the equation into a form that can be easily integrated. The integrating factor is strategically chosen to make the left-hand side of the equation a perfect derivative, which simplifies the integration process significantly. This step demonstrates the power of linear differential equation techniques in solving more complex problems. The constant of integration, C, arises from the indefinite integral and represents a family of solutions. Determining the specific value of C often requires additional information, such as initial conditions. The integration process is a key step in solving differential equations, and mastering this technique is essential for finding general solutions.
Solve for v: Multiply by x^(1/2) to get v = -x⁴/7 + Cx^(1/2). This step involves isolating the variable v to express it in terms of x and the constant of integration C. By multiplying both sides of the equation by x^(1/2), we undo the effect of the integrating factor and obtain an explicit expression for v. This algebraic manipulation is a straightforward but crucial step in the solution process. It sets the stage for the final substitution, where we replace v with its original expression in terms of y. The constant of integration C remains in the equation, representing the general solution's family of curves. Different values of C correspond to different solutions of the differential equation. Solving for v is a necessary intermediate step that bridges the gap between the linear equation and the original Bernoulli equation. It showcases the importance of algebraic manipulation in simplifying and expressing solutions in a clear and concise form. This step prepares us for the final substitution that will give us the solution in terms of the original dependent variable y.
Substitute back: Recall v = y⁻², so y⁻² = -x⁴/7 + Cx^(1/2). To find y, we invert both sides and take the square root. This is the final step in solving the Bernoulli equation, where we revert back to the original variable y to express the solution. By substituting y⁻² back for v, we obtain an equation involving y and x. To isolate y, we take the reciprocal of both sides and then take the square root. This process yields the solution in terms of y, which represents the general solution to the original Bernoulli equation. The solution may involve a constant of integration, C, which arises from the integration step. Different values of C give different particular solutions. This final step underscores the importance of reversing the initial substitution to obtain the solution in the original variables. It demonstrates the elegance of the Bernoulli method, where a nonlinear equation is transformed into a linear one, solved, and then transformed back to the original variables. The resulting solution provides valuable insights into the behavior of the system described by the differential equation, and it can be used for various applications, such as modeling physical phenomena or making predictions.
Final Solution: Thus, y² = 1/(-x⁴/7 + Cx^(1/2)). This gives us y = ±1/sqrt(-x⁴/7 + Cx^(1/2)). This expression represents the general solution to the Bernoulli equation 4xy' + y = x⁴y³. The ± sign indicates that there are two families of solutions, one positive and one negative. The solution involves a constant of integration, C, which can be determined if initial conditions are given. This final solution encapsulates the entire process of solving the Bernoulli equation, from the initial identification of the equation type to the final expression for y in terms of x. The solution is a function that satisfies the original differential equation, and it describes the relationship between the dependent variable y and the independent variable x. The complexity of the solution reflects the nonlinear nature of the original equation, but the Bernoulli method provides a systematic way to find this solution. Understanding and interpreting the solution is crucial for applying it to real-world problems, such as modeling population growth or analyzing electrical circuits. The solution not only gives the functional relationship but also provides insights into the qualitative behavior of the system being modeled.

2. Solving an Exact Differential Equation: xy² + y³ = (cos x)/x

The second equation, xy² + y³ = (cos x)/x, presents a different challenge. This equation is not immediately recognizable as a standard type, such as separable or linear. However, it is structured in a way that suggests we consider whether it might be an exact differential equation. Exact differential equations are a special class of first-order differential equations that can be solved by finding a potential function. The key characteristic of an exact equation is that it can be expressed as the total differential of some function. Recognizing and solving exact differential equations is a valuable skill in the study of differential equations, as it provides a direct and elegant method for finding solutions. These equations arise in various contexts, including thermodynamics, fluid mechanics, and electromagnetism. The general form of an exact differential equation is M(x, y) dx + N(x, y) dy = 0, where the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x. This condition ensures that there exists a function ψ(x, y) such that ∂ψ/∂x = M and ∂ψ/∂y = N. The solution to the exact differential equation is then given by ψ(x, y) = C, where C is a constant. Our strategy for this equation involves rearranging it into the standard form, checking the exactness condition, finding the potential function, and then expressing the solution. This process showcases the interplay between differential calculus and differential equations, highlighting the power of mathematical structures in solving problems. Let's begin by rearranging the given equation to match the standard form of an exact differential equation.

Step-by-step Solution

Rearrange the Equation: The first step in solving the differential equation xy² + y³ = (cos x)/x as an exact equation is to rearrange it into the standard form M(x, y) dx + N(x, y) dy = 0. To do this, we rewrite the equation as (xy² + y³)dx - (cos x / x)dx = 0. Then, to truly fit the exact equation form, we must ensure all terms are on the same side and properly grouped with their respective differentials. However, a closer look reveals that we've made an error in treating the original equation. It's not in a form directly suitable for exact equation methods as it stands. Instead, we must recognize the implicit form and the need to manipulate it differently. The correct approach involves recognizing that the equation might be better suited for other methods or requires further manipulation before considering exactness. This initial misstep highlights a crucial aspect of problem-solving in differential equations: the importance of carefully assessing the equation's structure before applying a specific method. In this case, the original equation doesn't readily fit the exact equation mold, and forcing it into that form leads to an incorrect setup. We need to backtrack and reconsider our strategy. Perhaps separation of variables, or another technique, might be more appropriate. This reflective step is vital in mathematical problem-solving, allowing us to correct course and pursue a more fruitful path towards a solution.
Recognize as non-Exact and Consider Other Methods: The realization that our initial approach to treating xy² + y³ = (cos x)/x as an exact equation was misguided leads us to a critical juncture in the problem-solving process. This moment highlights the importance of adaptability and critical assessment in tackling differential equations. Not every equation fits neatly into a predefined category, and recognizing when a particular method is unsuitable is as crucial as knowing the methods themselves. In this instance, the equation's structure, particularly the presence of terms that don't cleanly separate or conform to the exactness condition, signals the need for an alternative strategy. This could involve exploring whether the equation can be transformed into a more manageable form, perhaps through a substitution or by recognizing an integrating factor that might make it exact. Alternatively, we might consider if the equation is separable, homogeneous, or belongs to another class of differential equations with known solution techniques. The process of elimination and strategic reconsideration is a hallmark of mathematical problem-solving. It requires a deep understanding of the characteristics of different types of equations and the ability to apply appropriate techniques based on those characteristics. This reflective step is not a setback but an opportunity to refine our approach and deepen our understanding of the equation's nature. It underscores the iterative nature of mathematical inquiry, where initial attempts may lead to new insights and pathways towards a solution. Let's take a moment to further analyze the equation and explore potential alternative strategies.
Analysis and Possible Separable Form Consideration: Upon further scrutiny, the equation xy² + y³ = (cos x)/x does not readily present itself as an exact differential equation nor easily transform into one. Our previous attempt to force it into the exact form revealed its unsuitability for that method. However, this does not mean the equation is unsolvable; rather, it necessitates a different approach. One avenue worth exploring is whether the equation can be rearranged into a separable form. Separable equations are those that can be written in the form f(y) dy = g(x) dx, where the variables are completely separated on either side of the equation. This form is particularly amenable to direct integration. To investigate this possibility, we must manipulate the given equation algebraically to see if we can isolate terms involving y on one side and terms involving x on the other. This might involve factoring, dividing, or other algebraic manipulations. The challenge lies in the presence of both x and y terms intertwined in a way that doesn't immediately suggest separation. If a separable form can be achieved, the solution process becomes significantly simplified, as we can integrate both sides independently. However, if separation proves elusive, we may need to consider other techniques, such as looking for an integrating factor or exploring other types of transformations. The key is to remain flexible and adaptable in our approach, allowing the structure of the equation to guide our choice of method. This iterative process of analysis, experimentation, and refinement is central to the art of solving differential equations. Let's now delve into the algebraic manipulations to assess the separability of the equation.
Attempting Separation and Encountering Difficulties: Let's embark on the endeavor to separate the variables in the differential equation xy² + y³ = (cos x)/x. The goal is to manipulate the equation algebraically to isolate terms involving y on one side and terms involving x on the other, thereby transforming it into the separable form f(y) dy = g(x) dx. However, upon closer examination and attempted manipulations, we encounter significant challenges. The presence of the sum xy² + y³ on the left-hand side and the complex expression (cos x)/x on the right-hand side makes direct separation a daunting task. Factoring out common terms or applying other algebraic techniques does not readily lead to a clear separation of variables. The intertwined nature of x and y terms, combined with the transcendental function cos x, presents a formidable obstacle to separation. This impasse highlights the inherent difficulty in solving certain differential equations. Not all equations can be neatly categorized and solved using standard techniques. Some equations require more sophisticated methods, while others may not have closed-form solutions at all. In this instance, the failure to separate variables serves as a valuable learning experience. It underscores the importance of recognizing the limitations of certain methods and the need to explore alternative approaches. This realization prompts us to reassess our strategy and consider other avenues for solving the equation. Perhaps a substitution, a different type of transformation, or even numerical methods might be more appropriate. The journey through this equation is not just about finding a solution but also about developing a deeper understanding of the landscape of differential equations and the challenges they present.
Conclusion on the Equation xy² + y³ = (cos x)/x: In conclusion, our exploration of the differential equation xy² + y³ = (cos x)/x has led us to the realization that neither the exact equation method nor the separation of variables technique provides a straightforward solution. This outcome is not uncommon in the realm of differential equations, where many equations defy simple analytical solutions. The complex interplay of terms involving both x and y, coupled with the presence of the transcendental function cos x, creates a structure that resists easy manipulation. While we might consider more advanced techniques, such as searching for an integrating factor or exploring other transformations, these methods often lead to intricate calculations and may not guarantee a closed-form solution. In some cases, numerical methods might be the most viable approach for approximating solutions. This journey through a challenging equation underscores the diversity and complexity of differential equations. It highlights the importance of having a broad toolkit of solution techniques and the ability to adapt one's approach based on the specific characteristics of the equation at hand. The absence of a simple solution does not diminish the value of the exploration. Rather, it deepens our understanding of the nuances of differential equations and the limitations of analytical methods. The equation xy² + y³ = (cos x)/x serves as a reminder that mathematical problem-solving often involves navigating uncertainty and embracing the challenge of complexity. Moving forward, we acknowledge the intricacies of this equation and shift our focus to the next problem, armed with the insights gained from this exploration.

3. Solving a Homogeneous Equation: x²y' - y³ + 2xy = 0

Now, let's turn our attention to the third equation: x²y' - y³ + 2xy = 0. This equation, at first glance, might not immediately reveal its nature. However, a closer inspection suggests that it could be a homogeneous differential equation. Recognizing this is the first step toward applying the appropriate solution technique. Homogeneous equations are a special class of first-order differential equations that exhibit a particular type of symmetry. They can be written in the form y' = f(y/x), where the function f depends only on the ratio y/x. This form is crucial because it allows us to employ a substitution that simplifies the equation, making it solvable. The technique for solving homogeneous equations involves introducing a new variable, typically v = y/x, which transforms the original equation into a separable equation in terms of v and x. Solving this separable equation and then substituting back in terms of y yields the solution to the homogeneous equation. Understanding homogeneous equations is essential in the study of differential equations, as they appear in various applications, including geometry, physics, and engineering. They often arise in problems involving scaling or similarity, where the equation's form remains unchanged under certain transformations. Our strategy for this equation involves rearranging it to the form y' = f(y/x), performing the substitution, solving the resulting separable equation, and then transforming the solution back in terms of y. This process showcases the power of substitution techniques in simplifying differential equations. Let's begin by rearranging the equation to determine if it indeed fits the homogeneous form.

Step-by-step Solution

Rearrange the Equation: To determine if the differential equation x²y' - y³ + 2xy = 0 is homogeneous, the first step is to rearrange it to isolate y'. This will allow us to express y' as a function of x and y and then examine if it can be written in the form y' = f(y/x), which is the hallmark of a homogeneous equation. Adding y³ - 2xy to both sides and then dividing by x² gives us: y' = (y³ - 2xy) / x². This algebraic manipulation is a crucial step, as it brings the equation into a form where we can clearly see the relationship between y' and the other terms. The next step is to further simplify the right-hand side to see if it can be expressed as a function of y/x only. This involves dividing each term in the numerator by an appropriate power of x to achieve the desired form. Recognizing and performing this initial rearrangement is essential for identifying homogeneous equations and applying the correct solution technique. It sets the stage for the subsequent substitution that will simplify the equation and allow us to find a solution. The ability to manipulate equations algebraically is a fundamental skill in solving differential equations, and this step exemplifies its importance.
Check for Homogeneity: Now that we have the equation in the form y' = (y³ - 2xy) / x², we need to verify if it is indeed homogeneous. This involves checking if the right-hand side can be expressed as a function of y/x only. To do this, we can divide every term in the numerator and denominator by appropriate powers of x. In this case, we can divide each term in the numerator by x²: y' = (y³/x² - 2xy/x²). Further simplifying, we get y' = ((y/x)³x - 2(y/x)x) / x² = x(y³/x³ - 2y/x)/x² = (y³/x³ - 2y/x) / x. We made an error in simplification. Let's correct it. Divide each term in the numerator by x³, we get y' = (y³/x³ - 2xy/x³) = (y/x)³ - 2(y/x)/x. It still not work. Let's back to the first step y' = (y³ - 2xy) / x² and divide the numerator and denominator by x², we obtain: y' = (y³/x² - 2y/x) / 1. Now, let's try another way, rewrite the equation as y' = y³/x² - 2y/x. This equation is not in the form f(y/x) directly due to the x² term in the denominator of the first term. To make the function be in the form f(y/x), divide the first term by x and multiply by x, so the right side should be x(y/x)³ - 2y/x which is not a function of y/x. It seems there is an error, we need to review all the steps again. Go back to the equation y' = (y³ - 2xy) / x² and factor out y from the numerator: y' = y(y² - 2x)/x². This doesn't lead to a function of y/x directly. Let's reconsider our definition of homogeneous equations. A function f(x, y) is homogeneous of degree n if f(tx, ty) = tⁿf(x, y). Let's examine our equation F(x, y) = y³ - 2xy which has a degree 3 and 2 respectively. It is not a homogenous function. So, this equation is not homogenous. The key insight here is that the function on the right-hand side is not homogeneous of degree zero, meaning it cannot be expressed as a function of y/x alone. This realization is crucial, as it indicates that the standard technique for solving homogeneous equations is not applicable in this case. The next step involves reconsidering the type of equation we are dealing with and exploring alternative solution methods. This could involve looking for other transformations, such as substitutions, or employing numerical methods to approximate the solution. The importance of accurately identifying the type of differential equation cannot be overstated, as it dictates the appropriate solution strategy. In this instance, the initial assumption of homogeneity has been shown to be incorrect, prompting a shift in our approach. This iterative process of analysis and refinement is a hallmark of mathematical problem-solving. Now, let's analyze again the equation and try another method.
Recognizing as Non-Homogeneous and Trying a Substitution: Given that our analysis has revealed that the differential equation x²y' - y³ + 2xy = 0 is not homogeneous, we must pivot and explore alternative solution strategies. This adaptability is a crucial skill in tackling differential equations, as not all equations neatly fit into predefined categories. The failure of the homogeneity test prompts us to consider other methods, such as looking for a suitable substitution or attempting to transform the equation into a more recognizable form. One promising avenue is to investigate if a substitution can simplify the equation. This involves identifying a combination of x and y that, when replaced with a new variable, reduces the complexity of the equation. The choice of substitution often requires careful observation and pattern recognition. We might look for terms that appear repeatedly or for combinations of variables that suggest a particular transformation. Another approach is to rearrange the equation into a different form that might reveal a hidden structure. This could involve dividing by a common factor, adding or subtracting terms, or applying other algebraic manipulations. The goal is to make the equation more amenable to solution techniques. If a suitable substitution is found, the original equation can be transformed into a new equation in terms of the substituted variable and x. This new equation might be separable, linear, or of another type that we know how to solve. The process of selecting and applying a substitution is often iterative, involving trial and error. It requires a deep understanding of the various techniques for solving differential equations and the ability to recognize patterns that suggest a particular approach. Let's now delve into the equation and explore potential substitutions that might simplify it.
Attempting Substitution z = y/x and Further Analysis: Let's explore a common substitution technique for equations that resemble homogeneous equations: substituting z = y/x, which implies y = zx. Then, differentiating with respect to x, we get y' = z'x + z. Substituting these into the original equation x²y' - y³ + 2xy = 0, we have: x²(z'x + z) - (zx)³ + 2x(zx) = 0. Simplifying, we get x³z' + x²z - z³x³ + 2zx² = 0. Now, divide the entire equation by x³ to obtain: z' + z/x - z³ + 2z/x = 0. Combining terms, we have z' + 3z/x - z³ = 0. This is a separable equation in terms of z and x. Rearranging, we get z' = z³ - 3z/x. So, dz/dx = (xz³ - 3z)/x, which gives dz/(z³ - 3z/x) = dx/x or dz/(z(z² - 3/x)) = dx/x. The variable x in denominator prevent separating. Thus, substitution z = y/x doesn't work. This equation is separable but contains x in the denominator. This form is complex and does not readily lend itself to straightforward integration. The presence of x within the denominator of the term involving z² hinders the separation of variables, making the integration process significantly more challenging. This outcome highlights the fact that even seemingly promising substitutions may not always lead to a simple solution. The complexity introduced by the x term necessitates a reassessment of our strategy. We might consider alternative substitutions or other solution techniques that might be more effective in tackling this equation. The experience gained from this attempt underscores the importance of persistence and adaptability in mathematical problem-solving. Not every approach will be successful, but each attempt provides valuable insights into the nature of the equation and the challenges it presents. The key is to learn from these experiences and refine our strategy accordingly. Let's take a moment to reflect on the equation and explore other potential avenues for solution.
Rearranging and Considering a Bernoulli Equation Approach: Given the challenges encountered with the previous substitution, it's time to re-evaluate our approach to the differential equation x²y' - y³ + 2xy = 0. We've ruled out homogeneity and the standard substitution for homogeneous equations didn't lead to a simple solution. A crucial aspect of solving differential equations is recognizing patterns and forms that suggest particular solution methods. Let's rearrange the equation slightly to see if we can identify any familiar structures. Divide the entire equation by x²: y' - y³/x² + 2y/x = 0. Rearranging terms, we get y' + (2/x)y = y³/x². Now, this form bears a striking resemblance to a Bernoulli differential equation. Bernoulli equations have the general form y' + P(x)y = Q(x)yⁿ, where P(x) and Q(x) are functions of x, and n is a real number not equal to 0 or 1. In our rearranged equation, we can identify P(x) = 2/x, Q(x) = 1/x², and n = 3. Recognizing this structure is a significant step forward, as it allows us to apply the well-established technique for solving Bernoulli equations. This technique involves a substitution that transforms the nonlinear Bernoulli equation into a linear equation, which can then be solved using standard methods. The ability to recognize these patterns and apply the appropriate solution methods is a key skill in the study of differential equations. It demonstrates the importance of having a broad toolkit of solution techniques and the ability to apply them flexibly based on the characteristics of the equation at hand. Now that we've identified the equation as Bernoulli, let's proceed with the substitution and solution process for this type of equation.
Solving as a Bernoulli Equation: We've identified the equation y' + (2/x)y = y³/x² as a Bernoulli equation. To solve it, we use the standard substitution technique for Bernoulli equations. The general form of a Bernoulli equation is y' + P(x)y = Q(x)yⁿ, and the appropriate substitution is v = y^(1-n). In our case, n = 3, so we let v = y^(1-3) = y⁻². Differentiating with respect to x, we get v' = -2y⁻³y'. Now, we need to express the original equation in terms of v and v'. First, divide the Bernoulli equation by y³: y'y⁻³ + (2/x)y⁻² = 1/x². Substitute v = y⁻² and v' = -2y⁻³y', which implies y⁻³y' = -v'/2: -v'/2 + (2/x)v = 1/x². Multiply by -2 to eliminate the fraction: v' - (4/x)v = -2/x². This is now a linear first-order differential equation in terms of v and x. To solve it, we find the integrating factor, which is e^(∫(-4/x)dx) = e^(-4ln|x|) = x⁻⁴. Multiply the entire equation by the integrating factor x⁻⁴: x⁻⁴v' - 4x⁻⁵v = -2x⁻⁶. The left side is the derivative of x⁻⁴v, so we have (x⁻⁴v)' = -2x⁻⁶. Now, integrate both sides with respect to x: ∫(x⁻⁴v)' dx = ∫(-2x⁻⁶) dx. This gives us x⁻⁴v = (2/5)x⁻⁵ + C, where C is the constant of integration. This step transforms the equation into a readily integrable form, showcasing the power of the integrating factor method in solving linear differential equations. The constant of integration, C, arises from the indefinite integral and represents a family of solutions. To find the general solution, we now solve for v and then substitute back in terms of y. This systematic approach, characteristic of solving Bernoulli equations, highlights the importance of recognizing patterns and applying appropriate techniques. The next step is to isolate v and then revert to the original variable y.
Final Solution: We've arrived at the point where we have the equation x⁻⁴v = (2/5)x⁻⁵ + C. To solve for v, we multiply both sides by x⁴: v = (2/5)x⁻¹ + Cx⁴. Now, we substitute back v = y⁻²: y⁻² = (2/5)x⁻¹ + Cx⁴. Inverting both sides, we get y² = 1/((2/5)x⁻¹ + Cx⁴). Thus, y² = 1/((2 + 5Cx⁵)/(5x)), so y² = 5x/(2 + 5Cx⁵). Finally, taking the square root, we get y = ±sqrt(5x/(2 + 5Cx⁵)). This is the general solution to the differential equation x²y' - y³ + 2xy = 0. The ± sign indicates that there are two families of solutions, one positive and one negative. The solution involves a constant of integration, C, which can be determined if initial conditions are given. This final solution encapsulates the entire process of solving the Bernoulli equation, from the initial identification of the equation type to the final expression for y in terms of x. The solution is a function that satisfies the original differential equation, and it describes the relationship between the dependent variable y and the independent variable x. The complexity of the solution reflects the nonlinear nature of the original equation, but the Bernoulli method provides a systematic way to find this solution. Understanding and interpreting the solution is crucial for applying it to real-world problems, such as modeling physical phenomena or making predictions. The solution not only gives the functional relationship but also provides insights into the qualitative behavior of the system being modeled.

In this comprehensive exploration, we have navigated the intricate landscape of differential equations, tackling three distinct problems: 4xy' + y = x⁴y³, xy² + y³ = (cos x)/x, and x²y' - y³ + 2xy = 0. Each equation presented unique challenges and required the application of different techniques, highlighting the breadth and depth of the field of differential equations. For the Bernoulli equation 4xy' + y = x⁴y³, we successfully employed the Bernoulli method, which involved a clever substitution to transform the nonlinear equation into a linear one. This process showcased the power of mathematical transformations in simplifying complex problems and making them amenable to standard solution techniques. The solution we obtained provides a complete description of the behavior of the system governed by this equation. The second equation, xy² + y³ = (cos x)/x, proved to be more elusive. Our initial attempts to solve it as an exact equation and through separation of variables were unsuccessful. This experience underscored the importance of critical assessment and adaptability in problem-solving, as not all equations fit neatly into predefined categories. The challenges we faced with this equation served as a valuable reminder of the limitations of analytical methods and the need to explore alternative approaches, such as numerical methods, in certain cases. Finally, we tackled the equation x²y' - y³ + 2xy = 0, which we initially suspected might be homogeneous. However, our analysis revealed that it was not homogeneous, prompting us to explore other methods. Recognizing its Bernoulli nature, we successfully applied the Bernoulli method to obtain a solution. This journey through three diverse differential equations has provided a rich learning experience, highlighting the importance of pattern recognition, strategic problem-solving, and a deep understanding of various solution techniques. The ability to identify the type of equation, select the appropriate method, and execute the solution process is crucial for success in the study and application of differential equations. These skills are essential for anyone working in mathematics, physics, engineering, or any field where differential equations play a central role. The lessons learned from this exploration will undoubtedly serve as a valuable foundation for tackling more complex problems in the future.