Question

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ (x \ln y+x y) d x+(y \ln x+x y) d y=0 ; \quad x>0, \quad y>0 $$

Short answer

Given Differential Equation: $$(x \ln y+x y) d x+(y \ln x+x y) d y=0 ; \quad x>0, \quad y>0$$ Answer: Yes, the given differential equation is exact. The solution of the given exact differential equation is $$x^2 \ln y + \frac{1}{2}x^2y = C$$, where C is the integration constant.

Step-by-step solution

Identify M(x, y) and N(x, y)

First, let's identify the functions M(x, y) and N(x, y) from the given equation: Here, $$ M(x, y) = x \ln y + xy $$ and $$ N(x, y) = y \ln x + xy $$

Calculate the partial derivatives of M and N

Now, find the partial derivatives of M and N with respect to x and y: $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x \ln y + xy) = x\frac{1}{y} + x $$ and $$ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(y \ln x + xy) = y\frac{1}{x} + y $$

Check if the equation is exact

Compare the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) to determine if the equation is exact: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \Rightarrow x\frac{1}{y} + x = y\frac{1}{x} + y $$ The condition holds true; therefore, the given differential equation is exact.

Find the potential function

Now, integrate M(x, y) with respect to x and N(x, y) with respect to y to find the potential function (\(\psi(x, y)\)): $$ \psi(x, y) = \int M(x, y) dx = \int (x \ln y + xy) dx = x^2 \ln y + \frac{1}{2}x^2y + h(y) $$ $$ \psi(x, y) = \int N(x, y) dy = \int (y \ln x + xy) dy = x^2 \ln x + \frac{1}{2}x^2y + k(x) $$ Comparing the two expressions for \(\psi(x, y)\), we get: $$ \psi(x, y) = x^2 \ln y + \frac{1}{2}x^2y + C $$ Here, C is the integration constant.

Find the solution

As we have found the potential function, the solution of the given exact differential equation will be: $$ \psi(x, y) = C \Rightarrow x^2 \ln y + \frac{1}{2}x^2y = C $$ This is the solution of the given exact differential equation.

Key concepts

Partial Derivatives

Partial derivatives are a tool in calculus used to examine how a function changes as its variables change. They are essential when dealing with functions of multiple variables. In this context, for a function with variables \(x\) and \(y\), the partial derivative with respect to \(x\) is denoted as \(\partial M / \partial x\), and reflects how the function changes as \(x\) varies while keeping \(y\) constant.

Similarly, \(\partial M / \partial y\) measures changes concerning \(y\). In our exercise, these derivatives help determine if the given differential equation is exact.

• Equation Exactness: To check if an equation is exact, you need to compare the partial derivatives of two functions \(M(x, y)\) and \(N(x, y)\). If \(\partial M / \partial y = \partial N / \partial x\), the equation is exact. This equality signifies that a potential function exists.
• Calculation Example: For the functions in our problem, \(\partial M / \partial y\) and \(\partial N / \partial x\) were computed, matched, and confirmed the equation as exact.

Potential Function

The concept of a potential function is tied closely to exact differential equations. When an equation is exact, a potential function, often denoted \(\psi(x, y)\), can be identified.

This function is central because it consolidates the expressions \(M\) and \(N\) into one mathematical representation.

• Finding \(\psi(x, y)\): To find the potential function, integrate \(M(x, y)\) regarding \(x\) and \(N(x, y)\) with respect to \(y\). Each integration often introduces arbitrary functions of the other variable which are eliminated during the matching step.
• Example Integration: In our solution, integrating \(M(x, y)\) gives \(x^2 \ln y + \frac{1}{2}x^2y + h(y)\), while \(N(x, y)\) provides \(x^2 \ln x + \frac{1}{2}x^2y + k(x)\). The overlapping parts confirm the potential function as \(x^2 \ln y + \frac{1}{2}x^2y + C\).

Through this process, the potential function represents the general solution to the differential equation.

Integration Constants

Integration constants play a vital role in integrating functions, especially when discovering potential functions. These constants account for the indefinite aspect of integration and help in accurately defining the family of potential functions.

In our exercise, constants emerge during integration because indefinite integrals are involved, and these constants represent possible translations of the curve along the y-axis without altering the shape.

• Role in Potential Functions: When both \(M(x, y)\) and \(N(x, y)\) are integrated separately, integration constants, are added to each resulting expression. These constants are crucial for arriving at a complete potential function \(\psi(x, y)\).
• Unified Constant: After matching expressions, leftover components like \(h(y)\) or \(k(x)\) reduce when checked against each other, and a single constant \(C\) is retained, representing the general integrative constant in the final function.

Thus, this process ensures that the family of solutions is accurately represented through the function.

Solution of Differential Equations

The process of solving differential equations, specifically exact differential equations, revolves around verifying exactness and finding a potential function. An exact differential equation always allows its solution to be expressed in terms of a potential function.

Key steps involve verifying equation exactness, which means the partial derivatives \(\partial M / \partial y\) and \(\partial N / \partial x\) are equal. This ensures we can proceed to find an actual solution.

• General Solution: Once the potential function \(\psi(x, y)\) is determined, the equation \(\psi(x, y) = C\) provides a complete solution to the differential equation. This solution implies a relationship between \(x\) and \(y\) that satisfies the given equation for any constant \(C\).
• Interpreting the Solution: In the given problem, the solution \(x^2 \ln y + \frac{1}{2}x^2y = C\) encompasses all possible solutions, representing an implicit relationship within the context of differential equations.

Solving such equations systematically helps establish a deep connection between calculus concepts and practical problem-solving.