91Ó°ÊÓ

Differential equations are often not exact in their given form, which means their paths of integration do not align neatly. An integrating factor is a function that can be multiplied with a differential equation to make it exact, allowing for straightforward solutions. For a given non-exact differential equation, finding the right integrating factor is crucial. This factor transforms the differential equation into a form where standard solution techniques can be applied.
In the exercise, the integrating factor provided is \( \mu(x, y) = xy \), which multiplies with the original differential equation to convert it into a form where the paths of integration align, making it exact. The result of applying this integrating factor ensures that corresponding partial derivatives match, thus allowing us to solve the equation under more manageable conditions.

To determine whether a differential equation is exact, partial derivatives play a key role. Specifically, for an equation \( M(x, y) \) and \( N(x, y) \), if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact. Partial derivatives measure how a function changes as each of its input variables changes, with all other variables held constant.
In this exercise, we start by calculating the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) from the terms of the differential equation. For the initial equation, these derivatives did not equate, indicating the equation was not exact. This verifies the need for an integrating factor to make the equation solvable through exact methods.

Exactness verification is the process of checking if a differential equation is in a solvable form without additional factors or methods. This can be verified by ensuring that the order of integration does not affect the equation's solution—that is, verifying if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
After applying the integrating factor in our exercise, we again compute partial derivatives and verify that they are indeed equal. This confirmed exactness allows us to find a potential function \( \Phi(x, y) \) such that the level curves of this function represent the solutions to the differential equation. This step ensures the integration process can proceed seamlessly to arrive at the solution.

The general solution of an exact differential equation provides a comprehensive representation of all possible solutions the equation can have. Once exactness is achieved, you integrate \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \), combining them to find a function \( \Phi(x, y) \). This yields a potential function whose level curves correspond to solutions.
For the transformed exact equation from our exercise, integrating leads to the potential function: \( \Phi(x, y) = -\frac{x^2 y^2}{2} \sin x + 2y^2 \cos x + C \). Here, \( C \) represents a constant that can be any real number, embodying a family of solutions. Solving \( \Phi(x, y) = 0 \) gives us the general solution, portraying the equation's diverse possible behaviors.