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Exact Equations Explained
Exact equations play a crucial role in the study of differential equations within calculus. Understanding them will provide you with a strong foundation for mastering more complex mathematical concepts.
What Are Exact Differential Equations?
An exact differential equation is a specific type of first-order differential equation that can be expressed in the form:\[M(x, y) \,dx + N(x, y) \,dy = 0\]In this form, functions \(M(x, y)\) and \(N(x, y)\) must satisfy a particular condition for the equation to be exact.
Exact Differential Equation Condition: For an equation \(M(x, y) \,dx + N(x, y) \,dy = 0\) to be exact, the following condition must hold:\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]
Consider the following differential equation:\[3x^2 + 2xy \,dx + (x^2 + y^2) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (3x^2 + 2xy)}{\partial y} = 2x\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^2 + y^2)}{\partial x} = 2x\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.
Always check the exactness condition before attempting to solve a differential equation.
Importance of Exact Equations in Calculus
Exact differential equations are significant because they can often be solved more easily than other types of differential equations. The solutions to exact equations can provide valuable insights into various applied mathematical problems.
Potential Function: A function \(\phi(x, y)\) such that \(d\phi(x, y) = M(x, y) \,dx + N(x, y) \,dy\).
Returning to our previous example, find a potential function \(\phi(x, y)\) such that:\[\frac{\partial \phi}{\partial x} = M(x, y) = 3x^2 + 2xy\]\[\frac{\partial \phi}{\partial y} = N(x, y) = x^2 + y^2\]We integrate \(\frac{\partial \phi}{\partial x}\) with respect to \(x\):\[\phi(x, y) = \int (3x^2 + 2xy) \,dx = x^3 + x^2y + f(y)\]Now, differentiate this with respect to \(y\) and compare with \(\frac{\partial \phi}{\partial y}\):\[\frac{\partial \phi}{\partial y} = x^2 + f'(y) = x^2 + y^2\]Thus, \(f'(y) = y^2\) and \(f(y) = \frac{y^3}{3}\). So the potential function is:\[\phi(x, y) = x^3 + x^2y + \frac{y^3}{3}\]
Exact equations are particularly useful in physics, engineering, and economics. They help solve problems relating to electric potential, temperature distribution, fluid flow, and more. Learning to identify and solve exact differential equations is a fundamental skill that opens the door to numerous real-world applications.Additionally, understanding the concept of potential functions and their relationship to exact equations can further enable you to explore gradient fields and their properties, essential concepts in vector calculus.
How to Solve Exact Differential Equations
Solving exact differential equations involves recognising the form of the equation and following a specific set of steps to find the solution.
Recognising Exact Differential Equations
To determine if a first-order differential equation is exact, you need to write it in the form:\[M(x, y) \,dx + N(x, y) \,dy = 0\]Then you check for the exactness condition.
Exactness Condition: For an equation \(M(x, y) \,dx + N(x, y) \,dy = 0\) to be exact, the following must be true:\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]
Consider the differential equation:\[2xy \,dx + (x^2 + 2y) \,dy = 0\]To check for exactness, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (2xy)}{\partial y} = 2x\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^2 + 2y)}{\partial x} = 2x\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.
Ensure you always verify the exactness condition before proceeding to solve the equation.
Steps for Solving an Exact Differential Equation
Once you have verified that a differential equation is exact, you can solve it by finding a potential function.Follow these steps:
- Integrate \(M(x, y)\) with respect to \(x\) to find part of the potential function \(\phi(x, y)\).
- Include an unknown function of \(y\), say \(g(y)\), into \(\phi(x, y)\).
- Differentiate the potential function \(\phi(x, y)\) with respect to \(y\) and equate it to \(N(x, y)\).
- Solve for \(g(y)\).
- Combine the results to form the complete potential function.
Consider the equation from the last example:\[2xy \,dx + (x^2 + 2y) \,dy = 0\]Follow these steps:1. Integrate \(M(x, y) = 2xy\) with respect to \(x\):\[\phi(x, y) = \int 2xy \,dx = x^2y + h(y)\]2. Differentiate \(\phi(x, y)\) with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = x^2 + h'(y) = x^2 + 2y\]3. Solve for \(h'(y)\):\[h'(y) = 2y\] and \[h(y) = y^2\]4. Combine the results to obtain:\[\phi(x, y) = x^2y + y^2\]5. The solution to the differential equation is given by:\[\phi(x, y) = C\] or \[x^2y + y^2 = C\]
The method of solving exact differential equations by finding potential functions is not only precise but also highly intuitive. This technique is closely related to the concept of gradients in multivariable calculus. Understanding and visualising these gradients can further enrich your comprehension and appreciation of exact differential equations.The symmetry and structure in exact differential equations provide powerful tools for solving complex problems in various fields, ranging from engineering to economics, thereby demonstrating their extensive utility and relevance.
Integrating Factors Method
If the given differential equation is not exact, an integrating factor may be used to convert it into an exact equation. This method involves multiplying the differential equation by a function, known as the integrating factor, which makes the equation exact.Finding the appropriate integrating factor can often simplify the problem significantly.
Exact Differential Equations Examples
Examples help solidify your understanding of exact differential equations. By working through different types of examples, you can grasp the nuances in solving these equations.
Basic Examples
Starting with simpler cases can make more complex problems easier to handle later on.
Consider the differential equation:\[2xy \,dx + (x^2 + y) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (2xy)}{\partial y} = 2x\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^2 + y)}{\partial x} = 2x\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.To find the solution, we need a potential function \(\phi(x, y)\):Integrate \(2xy\) with respect to \(x\):\[\phi(x, y) = \int 2xy \,dx = x^2y + h(y)\]Differentiate with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = x^2 + h'(y) = x^2 + y\]Thus,\[h'(y) = y\] and \[h(y) = \frac{y^2}{2}\]The potential function is:\[\phi(x, y) = x^2y + \frac{y^2}{2}\]The solution to the differential equation is:\[x^2y + \frac{y^2}{2} = C\]
Working through similar basic examples can help you become more comfortable with the exactness condition and potential functions.
Complex Examples
Once you master basic examples, you can tackle more complex exact differential equations.
Consider the differential equation:\[(3x^2y + y^3) \,dx + (x^3 + 3xy^2 + y) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (3x^2y + y^3)}{\partial y} = 3x^2 + 3y^2\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^3 + 3xy^2 + y)}{\partial x} = 3x^2 + 3y^2\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.Integrate \(3x^2y + y^3\) with respect to \(x\):\[\phi(x, y) = \int (3x^2y + y^3) \,dx = x^3y + xy^3 + h(y)\]Dfferentiate with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = x^3 + 3xy^2 + h'(y) = x^3 + 3xy^2 + y\]Thus,\[h'(y) = y\] and \[h(y) = \frac{y^2}{2}\]The potential function is:\[\phi(x, y) = x^3y + xy^3 + \frac{y^2}{2}\]The solution to the differential equation is:\[x^3y + xy^3 + \frac{y^2}{2} = C\]
Problems can get significantly harder, requiring careful handling of integrals and derivatives.
Examples with Step-by-Step Exact Equation Solutions
Detailed step-by-step examples help you understand the process of solving exact differential equations.
Consider the differential equation:\[(2x + 3y^2) \,dx + (6xy - 4) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (2x + 3y^2)}{\partial y} = 6y\]\[\frac{\partial N}{\partial x} = \frac{\partial (6xy - 4)}{\partial x} = 6y\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.To solve it, follow these steps:1. Integrate \(2x + 3y^2\) with respect to \(x\):\[\phi(x, y) = \int (2x + 3y^2) \,dx = x^2 + 3xy^2 + h(y)\]2. Differentiate with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = 6xy + h'(y) = 6xy - 4\]3. Solve for \(h'(y)\):\[h'(y) = -4\] and \[h(y) = -4y\]4. The potential function is:\[\phi(x, y) = x^2 + 3xy^2 - 4y\]5. The solution to the differential equation is:\[x^2 + 3xy^2 - 4y = C\]
In these examples, the potential functions diverge based on the original functions in the differential equation. By mastering these steps, you'll increase your ability to handle diverse types of differential equations. The connections between partial derivatives and the exactness condition illustrate the beauty and utility of calculus in solving real-world problems. From physics to engineering, the ability to solve exact differential equations is an essential skill.
Practical Applications of Exact Equations
Exact differential equations have numerous practical applications across various fields. By understanding how to solve them, you can unlock solutions to complex problems in engineering, physics, and economics.
Engineering Applications
In engineering, exact differential equations are used to model and solve problems involving fluid dynamics, heat transfer, and structural analysis.
Consider an engineering problem where the temperature distribution along a rod is described by the differential equation:\[k \frac{\text{d}T}{\text{d}x} + T \frac{\text{d}x}{\text{d}t} = 0\]To determine the temperature \(T\) along the rod over time, you identify it as an exact equation and employ the methods to solve it.
Always verify the exactness condition before attempting to solve an equation to avoid errors.
The understanding of heat transfer is vital in designing heat exchangers, engines, and even electronic devices. By mastering exact differential equations, you can contribute to creating more efficient and safe designs.
Physics Applications
In physics, exact differential equations are used to describe phenomena in classical mechanics, thermodynamics, and electromagnetism.
Consider the motion of a particle in a potential field described by the equation:\[F(x, y) \,dx + G(x, y) \,dy = 0\]By determining the potential function, you can find the path of the particle and understand its behaviour within the field.
In physics, potential functions simplify the understanding of forces and energy.
Exact differential equations can also be applied to quantum mechanics and relativity. The principles of exactness and potential functions extend to wave functions and spacetime curvature, showing the broad relevance of these mathematical concepts.
Economics Applications
In economics, exact differential equations can model economic growth, optimisation problems, and market equilibria.
Consider an economic model where the rate of change of an investment value is described by:\[rV \,dV + b \,dI = 0\]By solving this exact equation, you can determine how the investment changes over time and optimise economic outcomes.
Economic models often rely on differential equations to make predictions and inform policies.
Advanced economic models consider multiple variables and constraints, using exact differential equations to find equilibria and optimal solutions. This mathematical approach helps economists understand market dynamics and inform decisions at both micro and macroeconomic levels.
Exact equations - Key takeaways
- Exact Differential Equations Condition: For an equation
M(x, y) dx + N(x, y) dy = 0
to be exact, it must satisfy∂M/∂y = ∂N/∂x
. - Potential Function: A function
φ(x, y)
such thatdφ(x, y) = M(x, y) dx + N(x, y) dy
. - Solving Exact Equations: Verify the exactness condition, integrate
M(x, y)
with respect tox
, differentiate with respect toy
, and solve for the resulting unknown function. - Integrating Factors: If an equation is not exact, multiply by an integrating factor to make it exact.
- Applications: Exact differential equations are used in physics (e.g., potential fields), engineering (e.g., heat distribution), and economics (e.g., investment growth models).
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