91Ó°ÊÓ

The partial derivative is a fundamental concept in calculus, especially when dealing with functions of multiple variables. It's akin to taking a regular derivative, but with a twist: you keep all other variables constant except the one you are differentiating with respect to. This is crucial in multivariable calculus and in exact differential equations.

Consider a function like \( M(x, y) = 2x - y \sin(xy) + ky^4 \). When we take the partial derivative of \( M \) with respect to \( y \), we essentially regard \( x \) as a constant:

• The term \( 2x \) treated as a constant becomes zero when differentiated with respect to \( y \).
• For \( -y \sin(xy) \), apply the product and chain rules resulting in \(-\sin(xy) - xy \cos(xy) \).
• The term \( ky^4 \) with a standard power rule derivation, yields \( 4ky^3 \).

This process allows us to analyze changes specific to \( y \), irrespective of \( x \), pivotal in assessing the relationships between the variables in an equation.

A differential equation links a function with its derivatives, and it can describe complex phenomena such as sound, heat, or motion. In our exercise, we work with a specific type called an exact differential equation.

Exact differential equations take the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \). For such an equation to be termed exact, a balance condition must be satisfied. This involves comparing partial derivatives: the partial derivative of \( M(x, y) \) with respect to \( y \) and \( N(x, y) \) with respect to \( x \).

• If these two conditions are equal, the differential equation is exact, meaning there exists a function \( F(x, y) \) such that \( dF = 0 \).
• This elegant feature, when met, implies a deep relationship between the components and allows us to solve the differential using integration methods.

Understanding these principles supports the structured approach in finding values of parameters like \( k \), as demonstrated in this exercise.

Solving differential equations requires applying mathematical principles strategically. In this exercise, determining \( k \) to make the equation exact involves deriving mathematical solutions via differentiation.

First, we determine the partial derivatives of \( M \) and \( N \), which are expressions of how each component varies relative to another variable. This step includes:

• For \( M(x, y) \), compute \( \frac{\partial M}{\partial y} = -\sin(xy) - xy \cos(xy) + 4ky^3 \).
• For \( N(x, y) \), which was \( - (20xy^3 + x \sin(xy)) \), find \( \frac{\partial N}{\partial x} = -(20y^3 + \sin(xy) + xy \cos(xy)) \).

By equating these two derivatives, you can isolate the parameter \( k \). Solving the equation
\[ -\sin(xy) - xy \cos(xy) + 4ky^3 = -(20y^3 + \sin(xy) + xy \cos(xy)) \]
leads you to \( 4ky^3 = 20y^3 \), from which \( k \) is discerned by simplifying to \( k = 5 \).
This solution showcases a neat union of calculus, algebra, and logical reasoning to resolve variables elegantly.