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Exact differential equations

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Intros
Lessons
  2. Partial Differentiation Review
  3. What are Exact Differential Equations and how do we solve them?
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Examples
Lessons
  1. Partial Differentiation
    Find the first order partial derivative of every variable for the following functions:
    1. f(x,y)=3x2−3y+2 f(x,y)=3x^2-3y+2
    2. N(y,z)=2y2z N(y,z)=2y^2 z
  2. Determining what is an Exact Differential Equation
    Solve the following differential equation

    2xy2−y2+(2x2y−2xy)dydx=02xy^2-y^2+(2x^2 y-2xy) \frac{dy}{dx}=0

    Using the equation: Ψ(x,y)=x2y2−xy2=2\Psi(x,y)=x^2 y^2-xy^2=2
    1. Solving Exact Equations
      Solve the following differential equation

      3exy−12y2+(3ex−xy)dydx=03e^x y- \frac{1}{2} y^2+(3e^x-xy) \frac{dy}{dx}=0

      With initial conditions y(0)=2y(0)=2
        1. Solve the following differential equation

          3x2y+2y2=3x+1+y′(2−3y−4xy−x3)3x^2 y+2y^2=3x+1+y' (2-3y-4xy-x^3)

          With initial conditions y=1,x=1y=1, x=1
        2. Verify that the solution you found is in fact the solution to the above differential equation (may be useful to do on a test).
      Topic Notes
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      Introduction to Exact Differential Equations

      Exact differential equations are a crucial concept in calculus and differential equations. These equations have a specific form that allows for a straightforward solution method. Our introduction video serves as an excellent starting point for understanding this topic, providing clear explanations and visual aids to help grasp the fundamental principles. While exact differential equations may initially seem complex, they are manageable with proper guidance and practice. The video breaks down the concept into digestible parts, making it easier for students to follow along and comprehend the material. By mastering exact differential equations, you'll gain valuable problem-solving skills applicable to various fields, including physics, engineering, and economics. Don't be intimidated by the terminology; with patience and dedication, you'll soon find yourself confidently tackling these equations. Remember, the key to success lies in understanding the underlying concepts rather than memorizing formulas.

      Understanding Partial Derivatives

      Partial derivatives are a fundamental concept in multivariable calculus, playing a crucial role in solving exact differential equations and understanding complex mathematical relationships. Unlike regular derivatives, which deal with functions of a single variable, partial derivatives are used when working with functions of multiple variables.

      The concept of partial derivatives can be best understood by considering a function f(x, y) of two variables. When we take the partial derivative of this function with respect to x, we treat y as a constant, and vice versa. This approach allows us to examine how the function changes with respect to one variable while holding the others constant.

      The notation for partial derivatives is distinct from that of regular derivatives. We use the symbol (called "del" or "partial") instead of the usual d. For example, the partial derivative of f with respect to x is written as f/x, while the partial derivative with respect to y is written as f/y.

      Let's consider an example to illustrate the difference between partial derivatives and regular derivatives. Suppose we have the function f(x, y) = x² + xy + y². To find the partial derivative with respect to x, we treat y as a constant:

      f/x = 2x + y

      Similarly, to find the partial derivative with respect to y, we treat x as a constant:

      f/y = x + 2y

      In contrast, if we were dealing with a function of a single variable, such as g(x) = x², the regular derivative would be g'(x) = 2x.

      Partial derivatives are particularly important in solving exact differential equations. These equations involve expressions that can be integrated to form a function of multiple variables. By using partial derivatives, we can verify whether a given differential equation is exact and, if so, find its solution.

      For instance, consider the differential equation (2x + y) dx + (x + 2y) dy = 0. To determine if this is an exact differential equation, we need to check if the partial derivative of the coefficient of dx with respect to y equals the partial derivative of the coefficient of dy with respect to x:

      (2x + y)/y = 1

      (x + 2y)/x = 1

      Since these partial derivatives are equal, we can confirm that this is an exact differential equation, and we can proceed to solve it using methods specific to exact equations.

      The importance of partial derivatives extends beyond solving exact differential equations. They are essential in various fields of mathematics, physics, and engineering. For example, in thermodynamics, partial derivatives are used to describe how different variables like pressure, volume, and temperature relate to each other while holding one variable constant.

      In optimization problems, partial derivatives help us find local maxima and minima of multivariable functions. The gradient, which is a vector of partial derivatives, points in the direction of steepest increase of a function and is crucial in optimization algorithms.

      Understanding and working with partial derivatives requires practice and a solid grasp of multivariable calculus. As you encounter more complex problems, you'll find that partial derivatives are an indispensable tool for analyzing and solving a wide range of mathematical and real-world problems.

      Defining Exact Differential Equations

      Exact differential equations are a specific type of first-order differential equations that can be solved using a particular method. These equations are characterized by their form and the relationship between their partial derivatives. The general form of an exact differential equation is M(x,y) + N(x,y)dy/dx = 0, where M and N are functions of both x and y.

      To understand exact differential equations, it's crucial to recognize their distinctive properties. The key feature of an exact differential equation is that it represents the total differential of a function F(x,y). This means that the left-hand side of the equation can be expressed as dF/dx + (dF/dy)(dy/dx) = 0.

      The conditions for a differential equation to be exact are fundamental to identifying and solving these equations. The primary condition is known as the integrability condition or Clairaut's Theorem. This theorem states that for a differential equation to be exact, the partial derivative of M with respect to y must equal the partial derivative of N with respect to x. Mathematically, this is expressed as M/y = N/x.

      Clairaut's Theorem is essential in determining whether a given differential equation is exact. It provides a quick and reliable method to check the exactness of an equation without having to attempt a solution. If this condition is satisfied, the equation is exact, and a specific solution method can be applied.

      Let's consider some examples to illustrate the concept of exact differential equations. An example of an exact differential equation is (2x + y)dx + (x - 3y)dy = 0. To verify its exactness, we can check if M/y = N/x. In this case, M/y = 1 and N/x = 1, satisfying Clairaut's Theorem. Therefore, this equation is exact.

      On the other hand, an example of a non-exact differential equation is (x^2 + y)dx + xdy = 0. Checking the condition, we find that M/y = 1, but N/x = 1 + x, which are not equal. Thus, this equation is not exact.

      The significance of exact differential equations lies in their solvability. When a differential equation is exact, it can be solved by finding the function F(x,y) whose total differential is given by the equation. This process involves integrating the equation with respect to one variable while treating the other as constant, and then determining the constant of integration as a function of the other variable.

      It's important to note that not all differential equations are exact. However, some non-exact equations can be made exact through the use of an integrating factor. This factor, when multiplied throughout the equation, transforms it into an exact differential equation, making it solvable using the methods for exact equations.

      In practical applications, exact differential equations appear in various fields of science and engineering. They are particularly useful in thermodynamics, where state functions like internal energy and enthalpy are often expressed as exact differentials. In physics, exact differentials are used to describe conservative force fields and potential energy functions.

      Understanding exact differential equations and the conditions for their existence, particularly Clairaut's Theorem, is crucial for solving a wide range of mathematical and physical problems. These equations provide a powerful tool for analyzing systems where the relationship between variables can be expressed as a total differential, offering insights into the underlying structure of many natural phenomena.

      Steps to Solve Exact Differential Equations

      Welcome to our tutorial on solving exact differential equations! This process can seem daunting at first, but we'll break it down into six manageable steps. By following these steps, you'll be able to tackle exact differential equations with confidence. Let's dive in!

      Step 1: Rewrite the equation in standard form
      The first step in solving exact differential equations is to rewrite the equation in standard form. This means expressing it as M(x,y)dx + N(x,y)dy = 0. This format is crucial for the subsequent steps and makes it easier to identify the M and N functions.

      Step 2: Verify exactness
      Once you have the equation in standard form, you need to verify if it's indeed exact. To do this, check if the partial derivative of M with respect to y equals the partial derivative of N with respect to x. In mathematical terms, we're checking if M/y = N/x. If this condition is met, the equation is exact, and we can proceed. If not, we'll need to use a different method.

      Step 3: Integrate M with respect to x
      Assuming the equation is exact, we now start the integration process. Begin by integrating M with respect to x, treating y as a constant. This gives us a function ψ(x,y), but it's not complete yet. We'll add a function of y, which we'll call h(y), to account for the constant of integration. So, at this point, we have ψ(x,y) = M(x,y)dx + h(y).

      Step 4: Differentiate ψ(x,y) with respect to y
      Now that we have our partial solution, we need to find h(y). To do this, we differentiate ψ(x,y) with respect to y. This should give us N(x,y) plus the derivative of h(y). In other words, ψ/y = N(x,y).

      Step 5: Solve for h(y)
      Using the result from step 4, we can now solve for h(y). Subtract N(x,y) from both sides of the equation ψ/y = N(x,y). This leaves us with h'(y) on one side. Integrate both sides with respect to y to find h(y). This completes our function ψ(x,y).

      Step 6: Find the general solution to the exact differential equation
      The final step is to set our completed ψ(x,y) equal to a constant, typically denoted as C. This gives us the general solution to the exact differential equation. The equation ψ(x,y) = C represents a family of curves that satisfy the original differential equation.

      By following these six steps, you can solve any exact differential equation. Remember, practice makes perfect! The more problems you solve, the more comfortable you'll become with this process. Don't be discouraged if it takes time to master exact differential equations can be tricky, but with patience and persistence, you'll get there.

      Key points to remember in the solving steps include carefully rewriting the equation, double-checking your exactness verification, being meticulous in your integration, and always considering the constant of integration. The process of finding the general solution ties everything together, giving you a satisfying conclusion to your problem-solving journey.

      As you work through more problems, you'll start to recognize patterns and develop intuition for these types of equations. Keep at it, and soon you'll be solving exact differential equations with ease!

      Worked Example: Solving an Exact Differential Equation

      Let's walk through a detailed, step-by-step example of solving an exact differential equation. We'll use the equation (x²y + 1)dy/dx = 1 - y²x from the transcript. This example will demonstrate the systematic process required to solve such equations.

      Step 1: Identify the equation
      Our equation is (x²y + 1)dy/dx = 1 - y²x

      Step 2: Rearrange the equation
      Move all terms to one side: (x²y + 1)dy/dx + y²x - 1 = 0

      Step 3: Identify M(x,y) and N(x,y)
      M(x,y) = y²x - 1 (coefficient of dx)
      N(x,y) = x²y + 1 (coefficient of dy)

      Step 4: Check if the equation is exact
      For an exact differential equation, M/y must equal N/x
      M/y = 2yx
      N/x = 2xy
      Since M/y = N/x, the equation is exact.

      Step 5: Find the function f(x,y)
      We can integrate M with respect to x or N with respect to y. Let's choose to integrate M:
      f(x,y) = (y²x - 1) dx = (1/2)y²x² - x + g(y)

      Step 6: Find g(y)
      Differentiate f(x,y) with respect to y: f/y = y²x + g/y
      This should equal N(x,y): x²y + 1 = y²x + g/y
      Solve for g/y: g/y = x²y + 1 - y²x

      Step 7: Integrate to find g(y)
      g(y) = (x²y + 1 - y²x) dy = (1/2)x²y² + y + C

      Step 8: Write the complete solution
      f(x,y) = (1/2)y²x² - x + (1/2)x²y² + y + C = 0

      Step 9: Simplify the solution
      (1/2)y²x² + (1/2)x²y² - x + y + C = 0
      y²x² - x + y + C = 0

      This step-by-step process demonstrates the systematic approach required to solve exact differential equations. Each step builds on the previous one, and careful attention to detail is crucial. Let's break down the importance of each step:

      1. Identifying the equation ensures we're working with the correct problem.

      2. Rearranging the equation prepares it for the exact differential equation format.

      3. Identifying M(x,y) and N(x,y) is crucial for checking exactness and finding the solution.

      4. Checking if the equation is exact determines whether we can use this method.

      5. Finding f(x,y) by integrating M or N is the core of the solution process.

      6. Finding g(y) ensures we capture all terms in the solution.

      7. Integrating to find g(y) completes the solution function.

      8. Writing the complete solution brings all parts together.

      Common Challenges and Tips for Solving Exact Differential Equations

      Solving exact differential equations can be a challenging task for many students. This section explores common difficulties encountered and provides valuable tips and strategies to overcome these challenges. By understanding these hurdles and implementing effective techniques, students can improve their problem-solving skills in this area of mathematics.

      One of the primary challenges students face is recognizing exact differential equations. Many struggle to identify when an equation is exact, often confusing it with other types of differential equations. To overcome this, it's crucial to carefully examine the equation's structure and check if it satisfies the necessary conditions for exactness. A helpful tip is to look for symmetry in the partial derivatives of M and N with respect to y and x, respectively.

      Another common difficulty is handling complex integrations that arise during the solution process. Students often find themselves stuck when faced with intricate integrals, especially those involving trigonometric or exponential functions. To address this challenge, it's essential to practice various integration techniques regularly and build a strong foundation in calculus. Utilizing integration tables and computer algebra systems can also be beneficial when dealing with particularly complex integrals.

      Verifying solutions is another area where students frequently encounter problems. Many overlook this crucial step or struggle to confirm the accuracy of their results. To overcome this challenge, it's important to develop a systematic approach to solution verification. This includes substituting the solution back into the original equation, checking boundary conditions, and ensuring the solution satisfies all given constraints.

      Students often struggle with the concept of integrating factors, which are sometimes necessary when dealing with nearly exact differential equations. To tackle this issue, it's helpful to practice identifying when an integrating factor is needed and become familiar with common integrating factor patterns. Developing a step-by-step approach to finding and applying integrating factors can significantly improve problem-solving efficiency.

      Time management during exams is another challenge many students face when solving exact differential equations. These problems can be time-consuming, leading to incomplete solutions or rushed work. To address this, it's crucial to practice solving a variety of problems under timed conditions. Developing efficient problem-solving strategies and recognizing patterns in different types of exact differential equations can help save valuable time during exams.

      To overcome these challenges, students should focus on building a strong foundation in calculus and differential equations. Regular practice with a diverse range of problems is essential for developing problem-solving skills and recognizing patterns. Utilizing online resources, textbooks, and practice exams can provide additional support and exposure to various problem types.

      In conclusion, while solving exact differential equations presents several challenges, implementing these tips and strategies can significantly improve students' abilities to tackle these problems effectively. By recognizing common pitfalls, developing systematic approaches, and practicing regularly, students can enhance their problem-solving skills and confidence in this important area of mathematics.

      Conclusion

      In summary, exact differential equations are a crucial concept in calculus, characterized by their ability to be directly integrated. Key points include recognizing the conditions for exactness, understanding the role of partial derivatives, and mastering the solution process. The theory behind exact differential equations is fundamental, as it provides insights into the nature of these equations and their applications in various fields. Students are strongly encouraged to practice solving examples to reinforce their understanding and develop problem-solving skills. The introduction video serves as an invaluable resource, offering a visual and conceptual foundation for grasping this topic. By combining theoretical knowledge with practical application and utilizing available resources like the video, students can gain a comprehensive understanding of exact differential equations. This knowledge will prove essential in tackling more advanced mathematical concepts and real-world problems in engineering and physics.

      Exact Differential Equations

      In this section, we will review the concept of exact differential equations and the necessary partial differentiation techniques required to solve them.

      Step 1: Introduction to Exact Differential Equations

      Exact differential equations are a specific type of differential equations that can be solved using the concept of exactness. These equations are not particularly difficult to solve in practice, but they do involve a significant amount of theoretical understanding. The theory behind exact differential equations helps us understand why certain methods work and provides a foundation for solving these equations effectively.

      Step 2: Importance of Theory

      Understanding the theory behind exact differential equations is crucial. While you may not need to know all the theoretical details to solve the equations, having a solid grasp of the underlying principles can make the process easier and more intuitive. The theory helps you understand why certain steps are taken and how the solutions are derived.

      Step 3: Review of Partial Derivatives

      Before diving into exact differential equations, it is essential to review partial derivatives. Partial derivatives are a fundamental concept in calculus and are used extensively in solving exact differential equations. We will go through a few examples to ensure that the concept of partial derivatives is clear and well-understood.

      Step 4: Understanding Partial Derivatives

      Partial derivatives involve taking the derivative of a function with respect to one variable while keeping the other variables constant. This is different from taking the total derivative, where all variables are considered. It is important to distinguish between partial derivatives and total derivatives, as they are used in different contexts.

      Step 5: Notation for Partial Derivatives

      In the context of exact differential equations, we will use specific notation for partial derivatives. The partial derivative of a function f f with respect to x x is denoted as fx f_x or ∂f∂x \frac{\partial f}{\partial x} . Similarly, the partial derivative of f f with respect to y y is denoted as fy f_y or ∂f∂y \frac{\partial f}{\partial y} . This notation will be used frequently in solving exact differential equations.

      Step 6: Practice with Partial Derivatives

      To ensure that you are comfortable with partial derivatives, we will go through a few practice problems. These problems will involve taking partial derivatives of various functions with respect to different variables. The goal is to reinforce your understanding and make sure you are confident in applying the concept of partial derivatives.

      Step 7: Solving Exact Differential Equations

      Once you have a solid understanding of partial derivatives, we can move on to solving exact differential equations. The process involves checking if a given differential equation is exact and then finding the solution using the concept of exactness. We will go through the steps in detail and provide examples to illustrate the process.

      Step 8: Checking for Exactness

      To determine if a differential equation is exact, we need to check if the mixed partial derivatives of the functions involved are equal. Specifically, for a differential equation of the form M(x,y)dx+N(x,y)dy=0 M(x, y)dx + N(x, y)dy = 0 , we check if ∂M∂y=∂N∂x \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} . If this condition is satisfied, the equation is exact.

      Step 9: Finding the Solution

      If the differential equation is exact, we can find the solution by integrating the functions M M and N N . The solution involves finding a potential function ψ(x,y) \psi(x, y) such that ∂ψ∂x=M \frac{\partial \psi}{\partial x} = M and ∂ψ∂y=N \frac{\partial \psi}{\partial y} = N . We will go through the integration process and show how to derive the potential function.

      Step 10: Examples and Practice

      Finally, we will work through several examples of exact differential equations. These examples will help reinforce the concepts and provide practice in applying the steps to solve exact differential equations. By the end of this section, you should have a solid understanding of exact differential equations and be able to solve them confidently.

      FAQs

      Here are some frequently asked questions about exact differential equations:

      1. What are Exact Differential Equations?

      Exact differential equations are a type of first-order differential equation that can be directly integrated to find a solution. They have the form M(x,y)dx + N(x,y)dy = 0, where the left-hand side represents the total differential of some function f(x,y).

      2. How do you know if something is an exact differential?

      To determine if a differential equation is exact, check if the partial derivative of M with respect to y equals the partial derivative of N with respect to x. Mathematically, this is expressed as M/y = N/x. If this condition is satisfied, the equation is exact.

      3. How do you find the exact differential of a function?

      To find the exact differential of a function f(x,y), calculate the partial derivatives with respect to x and y, then express it as df = (f/x)dx + (f/y)dy. This form represents the total change in the function for small changes in x and y.

      4. What is the difference between exact and inexact differentials?

      Exact differentials are the total differential of some function and can be integrated directly to find that function. Inexact differentials cannot be integrated directly to find a function. The key difference lies in whether the integrability condition (M/y = N/x) is satisfied.

      5. Is work an exact differential?

      Work is not always an exact differential. In conservative force fields, work is an exact differential and can be expressed as the change in potential energy. However, in non-conservative force fields, work is path-dependent and is not an exact differential.

      Prerequisite Topics for Exact Differential Equations

      Understanding exact differential equations is a crucial step in mastering differential equations. However, to fully grasp this concept, it's essential to have a solid foundation in certain prerequisite topics. Two key areas that significantly contribute to your understanding of exact differential equations are the integrating factor technique and modeling with differential equations.

      The integrating factor technique is a fundamental method used to solve first-order linear differential equations. This technique is particularly relevant to exact differential equations because it provides a systematic approach to transforming non-exact differential equations into exact ones. By mastering the integrating factor technique, you'll develop the skills needed to identify when a differential equation is not exact and how to make it exact. This knowledge is invaluable when dealing with more complex exact differential equations.

      Moreover, the integrating factor technique introduces you to the concept of multiplying both sides of a differential equation by a carefully chosen function. This process is similar to the method used in solving exact differential equations, where we often need to manipulate the equation to ensure it meets the criteria for exactness. Understanding this similarity will help you draw parallels between different solution methods and enhance your problem-solving skills.

      Equally important is the ability to model real-world scenarios using differential equations. This skill, often referred to as modeling with differential equations, provides context and practical applications for exact differential equations. By exploring various applications, you'll gain insight into why exact differential equations are significant and how they can be used to solve real-world problems.

      Understanding the modeling process also helps you interpret the solutions of exact differential equations in practical terms. This is crucial because many exact differential equations arise from physical, biological, or economic systems. Being able to translate between the mathematical representation and the real-world scenario it describes is a valuable skill that enhances your overall comprehension of the subject.

      Furthermore, modeling with differential equations often involves dealing with initial conditions or boundary conditions. This experience is directly applicable to exact differential equations, where such conditions are frequently used to determine specific solutions from the general solution. Familiarity with these concepts will make it easier for you to solve and interpret exact differential equations in various contexts.

      In conclusion, by thoroughly understanding the integrating factor technique and modeling with differential equations, you'll be well-prepared to tackle exact differential equations. These prerequisite topics provide the necessary mathematical tools and contextual understanding to approach exact differential equations with confidence and insight. As you progress in your studies, you'll find that this strong foundation will continue to support your learning and problem-solving abilities in more advanced topics within differential equations.

      Recall Partial Derivatives:

      Notation:

      dFdx=Fx\frac{dF}{dx}=F_x or
      dFdx=Fy\frac{dF}{dx}=F_y

      Exact Differential Equations:
      Sometimes a differential equation is not solvable via the separable equations technique.

      Let's suppose the differential equation we want to solve is of the form:

      M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y) \frac{dy}{dx}=0


      Now let's imagine we have another function of xx and yy, denoted as Ψ(x,y)\Psi (x,y) "psi".

      It is demonstrated throughout the video that: ddxΨ(x,y)=∂Ψ∂x+∂Ψ∂ydydx\frac{d}{dx} \Psi(x,y)=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}\frac{dy}{dx}

      So if we can find a Ψ(x,y)\Psi(x,y) such that Ψx=M(x,y)\Psi_x=M(x,y) and Ψy=N(x,y)\Psi_y=N(x,y), then

      ddxΨ(x,y)=∂Ψ∂x+∂Ψ∂ydydx=\frac{d}{dx} \Psi(x,y)=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}\frac{dy}{dx}= Ψx+Ψydydx=\Psi_x+\Psi_y \frac{dy}{dx}=M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y) \frac{dy}{dx}=0

      Given that ∂Ψ∂x≠dΨdx\frac{\partial \Psi}{\partial x} \neq \frac{d \Psi}{dx}


      So our solution to solving the differential equation is just

      Ψ(x,y)=c\Psi(x, y)=c


      Finding this Ψ(x,y)\Psi(x, y):

      Ok, so provided that Ψ(x,y)\Psi(x, y) is continuous and it's first derivative is continuous as well, we should have the following equality hold

      Ψxy=Ψyx\Psi_{xy}=\Psi_{yx}


      (Remember this from Calculus 3, if not just trust me!)

      So,

      Ψxy=(Ψx)y=(M)y=My\Psi_{xy}= (\Psi_x)_y=(M)_y=M_y
      Ψyx=(Ψy)x=(N)x=Nx\Psi_{yx}= (\Psi_y)_x=(N)_x=N_x
      ⟹My=Nx\Longrightarrow M_y=N_x

      So a differential equation will be exact if and only if My=NxM_y=N_x

      Then we can solve for our Ψ(x,y)\Psi(x, y) by integrating MM with respect to xx, or NN with respect to yy.

      And once we have Ψ(x,y)\Psi(x, y) we have solved the differential equation just equate:
      Ψ(x,y)=c\Psi(x, y)=c and that will be the answer
      Basic Concepts
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