Chapter 2: Problem 4
Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ \left(2 x y^{2}+2 y\right)+\left(2 x^{2} y+2 x\right) y^{\prime}=0 $$
Short Answer
Expert verified
Question: Determine if the given equation is exact, and if so, find the solution.
Given Equation: $$
\left(2 x y^{2}+2 y\right)+\left(2 x^{2} y+2 x\right) y^{\prime}=0
$$
Answer: The given equation is exact, and its general solution is: $$
x^2y^2 + 2xy = C
$$
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In the context of exact differential equations, partial derivatives play a crucial role. They involve the differentiation of a function with respect to one variable while holding the other variables constant. For example, for the function \(M(x, y) = 2xy^2 + 2y\), finding the partial derivative with respect to \(y\) involves treating \(x\) as a constant. This gives \(\frac{\partial M}{\partial y} = 4xy + 2\). Similarly, for \(N(x, y) = 2x^2y + 2x\), the partial derivative with respect to \(x\) is \(\frac{\partial N}{\partial x} = 4xy + 2\).
Understanding these derivatives helps us to determine the exactness of a differential equation. If the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) is satisfied, the equation is exact. In simple terms, matching partial derivatives indicates that the equation can potentially represent a gradient field, meaning an exact solution exists. This determination lays the foundation for solving the equation effectively.
Partial derivatives are a stepping stone for exploring multivariable calculus, offering insight into how functions change around a certain point and how these changes relate to solving complex equations.
Understanding these derivatives helps us to determine the exactness of a differential equation. If the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) is satisfied, the equation is exact. In simple terms, matching partial derivatives indicates that the equation can potentially represent a gradient field, meaning an exact solution exists. This determination lays the foundation for solving the equation effectively.
Partial derivatives are a stepping stone for exploring multivariable calculus, offering insight into how functions change around a certain point and how these changes relate to solving complex equations.
Integration of Functions
Once we establish that a differential equation is exact, the next step involves integrating the functions. This means finding the antiderivative or the original function whose derivative would give us the function we're integrating. In our problem, we integrate \(M(x, y)\) with respect to \(x\) and \(N(x, y)\) with respect to \(y\).
Integrating \(M(x, y) = 2xy^2 + 2y\) with respect to \(x\) gives us \(\int M \, dx = x^2y^2 + 2xy + h(y)\). Similarly, integrating \(N(x, y) = 2x^2y + 2x\) with respect to \(y\) results in \(\int N \, dy = x^2y^2 + 2xy + g(x)\). These expressions involve adding arbitrary functions \(h(y)\) or \(g(x)\), representing constants that might come from integration.
Integrals provide a "smoothing out" process, where instead of focusing on the rate of change (as derivatives do), we look at cumulative effects, or the total amounts. Understanding integration helps us construct solutions for exact differential equations, where we equate integrals and adjust these constant terms to find a consistent representation of the solution.
Integrating \(M(x, y) = 2xy^2 + 2y\) with respect to \(x\) gives us \(\int M \, dx = x^2y^2 + 2xy + h(y)\). Similarly, integrating \(N(x, y) = 2x^2y + 2x\) with respect to \(y\) results in \(\int N \, dy = x^2y^2 + 2xy + g(x)\). These expressions involve adding arbitrary functions \(h(y)\) or \(g(x)\), representing constants that might come from integration.
Integrals provide a "smoothing out" process, where instead of focusing on the rate of change (as derivatives do), we look at cumulative effects, or the total amounts. Understanding integration helps us construct solutions for exact differential equations, where we equate integrals and adjust these constant terms to find a consistent representation of the solution.
Differential Equation Solutions
Solving exact differential equations relies on both partial derivatives and integration to piece together a coherent solution. After confirming the exactness of an equation, the ultimate goal is to find a solution that satisfies this specific arrangement. Generally, the solution is a relationship involving the variables, expressed as a constant value.
In our example, integration results in the expressions \(x^2y^2 + 2xy + h(y)\) and \(x^2y^2 + 2xy + g(x)\). By comparing these, we realize \(h(y)\) and \(g(x)\) must be constants for the expressions to match. Thus, our solution simplifies to \(x^2y^2 + 2xy = C\), where \(C\) is a constant that might depend on initial conditions or constraints imposed on the problem.
This final equation forms a family of curves where each curve satisfies the original differential equation. Understanding how to derive this through checking exactness with partial derivatives and resolving through integration forms the basis for solving more complex systems involving differential equations. This process underscores the interconnectedness of calculus concepts, demonstrating their practical utility in addressing real-world problems.
In our example, integration results in the expressions \(x^2y^2 + 2xy + h(y)\) and \(x^2y^2 + 2xy + g(x)\). By comparing these, we realize \(h(y)\) and \(g(x)\) must be constants for the expressions to match. Thus, our solution simplifies to \(x^2y^2 + 2xy = C\), where \(C\) is a constant that might depend on initial conditions or constraints imposed on the problem.
This final equation forms a family of curves where each curve satisfies the original differential equation. Understanding how to derive this through checking exactness with partial derivatives and resolving through integration forms the basis for solving more complex systems involving differential equations. This process underscores the interconnectedness of calculus concepts, demonstrating their practical utility in addressing real-world problems.
