Question

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ (2 x+3)+(2 y-2) y^{\prime}=0 $$

Short answer

Question: Determine if the given equation is exact and find its general solution if it is exact: $$(2x + 3) + (2y - 2)y' = 0.$$ Answer: The given equation is exact, and its general solution is $$x^2 + 3x + y^2 - 2y + C = 0.$$

Step-by-step solution

Identify the given equation's coefficients

Identify the given equation's coefficients M(x, y) and N(x, y) as: $$ M(x, y) = 2x + 3, \quad N(x, y) = 2y - 2. $$

Check for exactness

In order to check for exactness, calculate the partial derivatives of M with respect to y and N with respect to x: $$ \frac{\partial M}{\partial y} = 0, \quad \frac{\partial N}{\partial x} = 0. $$ Since these partial derivatives are equal, the given equation is exact.

Find the potential function

To find the potential function, integrate M with respect to x and N with respect to y. Then, we will compare these integrations to find the potential function. Integrate M with respect to x: $$ \int M(x, y) dx = \int (2x + 3) dx = x^2 + 3x + g(y), $$ where g(y) is a function of y. Integrate N with respect to y: $$ \int N(x, y) dy = \int (2y - 2) dy = y^2 - 2y + f(x), $$ where f(x) is a function of x. Now, compare the two integrations and find the potential function \(\psi(x, y)\): $$ \psi(x, y) = x^2 + 3x + y^2 - 2y + C. $$

Find the solution

Since the given equation is exact and the potential function \(\psi(x, y)\) has been found, we can find the general solution of the given equation by setting the potential function equal to a constant: $$ x^2 + 3x + y^2 - 2y + C = 0. $$ This last equation represents the general solution of the given exact differential equation.

Key concepts

Exactness Condition

An exact differential equation is a type of differential equation where the problem can be reduced to finding a single function whose derivatives match the components of the equation. To determine if an equation is exact, the 'Exactness Condition' must be satisfied.

In the context of this problem, you have a differential equation of the form \[ M(x, y) + N(x, y) y' = 0. \] For this equation to be exact, the partial derivatives of these functions must be equal. Specifically,

• \( M(x, y) = 2x + 3 \)
• \( N(x, y) = 2y - 2 \)

We take the partial derivative of \( M(x, y) \) with respect to \( y \) and \( N(x, y) \) with respect to \( x \).

The differential equation is exact if: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \] In this problem, both derivatives are zero, which confirms the exactness. Understanding this condition is crucial as it transforms what might seem a complex differential equation into a solvable problem using straightforward calculus techniques.

Potential Function

Once we ascertain the exactness of the differential equation, the next step is finding what’s known as the potential function, often denoted as \( \psi(x, y) \). The potential function is fundamentally the function whose differential yields the original equation's terms.

To find \( \psi(x, y) \), we perform integrations similar to solving a function's antiderivative. We start by integrating \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \):

• Integrate \( M = 2x + 3 \) with respect to \( x \):
  \[ \int (2x + 3) \ dx = x^2 + 3x + g(y). \]
• Integrate \( N = 2y - 2 \) with respect to \( y \):
  \[ \int (2y - 2) \ dy = y^2 - 2y + f(x). \]

Here, \( g(y) \) and \( f(x) \) are functions emerging from each respective integration, accounting for potential dependency on other variables.

Comparing these results, we identify the overlapping parts to express \( \psi(x, y) \) fully. Practically, it evolves into:
\[ \psi(x, y) = x^2 + 3x + y^2 - 2y + C. \] A constant \( C \) is added as we are looking for a family of functions, demonstrating the variety of potential solutions.

Integration in Differential Equations

Integration is a vital tool in solving exact differential equations, enabling us to backtrack from derivatives to the original function, revealing potential solutions.

In the case of exact differential equations, integration helps in constructing the potential function \( \psi(x, y) \). When integrating:

• The task is to calculate \( \int M(x, y) \ dx \) and \( \int N(x, y) \ dy \),
• Identify the common components from these integrations.

These integrations reveal the form of the potential function. The approach opposes straightforward solutions in differential equations, aiming instead to integrate strategically to circle back to the initial state of a system.

The final expression, often with a constant term \( C \), represents a family of solutions, revealing how integration elucidates the complete behavior implied by the original differential equation.