Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\left(x \cos y-e^{-\sin y}\right) d y+d x=0$$

To understand partial derivatives, we must first grasp what a derivative is. In simple terms, a derivative measures how a function changes as its input changes. When dealing with functions of multiple variables, we take partial derivatives. A partial derivative focuses on the change with respect to one variable, keeping the others constant.
Here's an example: For a function of two variables, say \( z = f(x,y) \), the partial derivatives are \( \frac{\text{∂}z}{\text{∂}x} \) and \( \frac{\text{∂}z}{\text{∂}y} \).
These derivatives show how \( z \) changes with \( x \) and \( y \) respectively.

Let's use the given differential equation: \( (x \text{cos} \, y - e^{-\text{sin} \, y}) dy + dx = 0 \). We rewrite this in standard form: \( M(x, y) dx + N(x, y) dy = 0 \). Here, \( M(x,y) = 1 \) and \( N(x, y) = x \text{cos} \, y - e^{-\text{sin} \, y} \).
To check for exactness, we compute the partial derivatives: \( \frac{\text{∂} M}{\text{∂} y} = 0 \) and \( \frac{\text{∂} N}{\text{∂} x} = \text{cos} \, y \). Since these are not equal, the equation isn't exact.
Understanding and calculating partial derivatives is essential in verifying whether a differential equation is exact or not.

When facing a non-exact differential equation, we can often use an integrating factor to transform it into an exact one. An integrating factor is typically a function of \( x \) or \( y \). Applying it multiplies both sides of the differential equation, potentially converting it into an exact equation.

Consider a function \( \text{μ}(y) \):\[ \text{μ}(y) = e^{\text{−∫ P(y) dy}} \]
Where \( P(y) \) represents terms derived from the partial derivatives of the original non-exact equation. In our example, the integrating factor will depend on \( y \), as shown in step 3: \( \text{μ}(y) = e^{\text{−∫ (cos \, y) dy}} \).
This approach ensures that once we apply the integrating factor, the new equation can be easily integrated, providing a clear path to the solution. Understanding the concept and application of an integrating factor helps simplify seemingly complex differential equations.

Non-exact differential equations do not meet the criteria where \( \frac{\text{∂}M}{\text{∂}y} = \frac{\text{∂}N}{\text{∂}x} \). This means we can't immediately integrate them. However, with techniques such as using an integrating factor, they can be converted to an exact form.

The given differential equation \( (x \text{cos} \, y - e^{-\text{sin} \, y}) dy + dx = 0 \) was identified as non-exact due to the mismatch between \( \frac{\text{∂} M}{\text{∂} y} \) and \( \frac{\text{∂} N}{\text{∂} x} \). Thus, we must find an integrating factor.
The process may involve trial and error or insight into the structure of the equation. Once identified and applied, the differential equation transforms into exact form, making it easier to solve.
Understanding how to work with non-exact differential equations enhances problem-solving skills and mathematical intuition. This knowledge is particularly valuable for advanced studies in differential equations and their applications.