91Ó°ÊÓ

Partial derivatives are a key concept when dealing with functions of multiple variables, like in exact differential equations. Unlike a regular derivative, which represents the rate of change of a function with one variable, a partial derivative considers how a multivariable function changes as we vary just one of the variables while keeping others constant.
For example, in the differential equation \((2x - 1)dx + (3y + 7)dy = 0\), we have identified \(M(x, y) = 2x - 1\) and \(N(x, y) = 3y + 7\). To check if the equation is exact, we calculate the partial derivatives:

• \(\frac{\partial M}{\partial y} = 0\): Since \(M\) depends only on \(x\), changing \(y\) does not affect it, hence the derivative is zero.
• \(\frac{\partial N}{\partial x} = 0\): Similarly, \(N\) depends only on \(y\), meaning changes in \(x\) do not affect \(N\).

Both partial derivatives are equal, confirming exactness.

Integration plays a crucial role in solving exact differential equations. It helps us to find the potential function, which solves the equation as a whole. Once the exactness is confirmed, the next step involves integrating the function \(M(x, y)\) or \(N(x, y)\).
Consider \(M(x, y) = 2x - 1\): we integrate it with respect to \(x\). This gives us the function \(\psi(x, y) = \int (2x - 1) \, dx = x^2 - x + h(y)\), where \(h(y)\) is an arbitrary function of \(y\).
This method allows us to build our solution step-by-step, incorporating terms that account for all changes in the function's behavior.

The potential function \(\psi(x, y)\) is an essential piece in solving exact differential equations, as it embodies the equation in a single expression. After integrating \(M(x, y)\) with respect to \(x\), we find:

• \(\psi(x, y) = x^2 - x + h(y)\)

To complete \(\psi(x, y)\), we balance it against \(N(x, y)\), ensuring partial derivatives of \(\psi(x, y)\) align.
Differentiating \(\psi(x, y)\) with respect to \(y\) yields \(h'(y)\), which we set equal to \(N(x, y) = 3y + 7\), finding \(h(y) = \frac{3}{2}y^2 + 7y + C\) upon integration. This allows us to write a complete potential function, \(\psi(x, y) = x^2 - x + \frac{3}{2}y^2 + 7y + C\).
This function characterizes our solution fully.

The exactness condition is pivotal in determining whether a given differential equation can be solved using potential functions. It's based on a straightforward criterion: for an equation of the form \(M(x, y)dx + N(x, y)dy = 0\) to be exact, the mixed partial derivatives must be equal.
This means:

• \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)

In our example, we saw that both \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) equalled zero, confirming exactness. This equality ensures that a single potential function can truly represent the entire differential equation.
This condition simplifies the differential equation into a form where standard techniques, like integration, can efficiently find a solution.