Question

Find all functions \(M\) such that the equation is exact. (a) \(M(x, y) d x+\left(x^{2}-y^{2}\right) d y=0\) (b) \(M(x, y) d x+2 x y \sin x \cos y d y=0\) (c) \(M(x, y) d x+\left(e^{x}-e^{y} \sin x\right) d y=0\)

Short answer

Question: Given a set of differential equations, find the function M(x, y) that makes each equation exact. (a) \(M(x, y) dx + \left(x^{2} - y^{2}\right) dy = 0\) (b) \(M(x, y) dx + 2xy \sin x \cos y dy = 0\) (c) \(M(x, y) dx + \left(e^{x} - e^{y} \sin x\right) dy = 0\)

Step-by-step solution

Find the partial derivatives of M(x, y) and N(x, y)

First, let us express the equation as \(M(x, y) - N(x, y) = 0\), where \(N(x, y) = y^{2} - x^{2}\). Next, we compute \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\), remembering that for the equation to be exact, these partial derivatives must be equal.

Set the partial derivatives equal to each other and solve for M(x, y)

Once we have the partial derivatives, we set them equal \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) Now we need to find M(x,y) with integrating and applying some constraints. Repeat the same process for (b) and (c) (b) \(M(x, y) dx + 2xy \sin x \cos y dy = 0\) (c) \(M(x, y) dx + \left(e^{x} - e^{y} \sin x\right) dy = 0\) In each case, find the M(x, y) that makes the equation exact by applying the exact equation criterion and solving for M(x, y).

Key concepts

Partial Derivatives

Understanding partial derivatives is crucial when working with multivariable functions like the one described in our exercise involving the function M(x, y). Partial derivatives measure how a function changes as only one variable changes while all others remain fixed.
For a function of two variables, M(x, y), the partial derivative with respect to x, denoted as \( \frac{\partial M}{\partial x} \), tells us how M changes in response to changes in x when y is held constant. Similarly, the partial derivative with respect to y, \( \frac{\partial M}{\partial y} \), measures the change in M with respect to changes in y when x is fixed. These concepts are fundamental in identifying exact differential equations, as they help in comparing how the mixed-rate of changes for the respective functions must align.
In the given exercise, the identification and computation of the partial derivatives of M with respect to y, \( \frac{\partial M}{\partial y} \), plays a pivotal role. By determining this derivative and comparing it to the derivative of the other function, N, with respect to x, students can explore whether the equation takes on the property of being exact.

Exact Equation Criterion

The exact equation criterion is at the heart of solving problems like the one presented in the exercise. An exact differential equation is one where there exists a function \( \Psi(x, y) \) such that \( d\Psi = M(x, y) dx + N(x, y) dy \). In simple terms, this means the changes in \( \Psi \) can be fully characterized by the functions M and N of x and y, respectively.
For an equation to be considered exact, there is a necessary condition that must be met: the cross partial derivatives must be equal. This means \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If this condition holds, we can then proceed to find the potential function \( \Psi(x, y) \) by integrating M and N with respect to their variables.
In our exercise, after deducing the form of N(x, y), we apply the exact equation criterion by equating the respective partial derivatives. This comparison supports us in detecting inconsistencies and helps guide us on whether an integrating factor might be required to make the initial non-exact equation into an exact one.

Integrating Factor

Sometimes equations might not immediately appear exact, necessitating the use of an integrating factor. An integrating factor is a function, \( \mu(x, y) \), which when multiplied by the non-exact equation, transforms it into an exact equation. This powerful tool allows us to leverage the properties of exact differential equations even when the equation at hand initially doesn’t meet the exact equation criterion.
To find a suitable integrating factor, one may need to engage in a bit of detective work, often involving insights into the symmetry or particular characteristics of M and N. Some situations necessitate integrating factors that depend only on x or only on y, while others call for more complex functions. Once identified and applied, you can then integrate the now-exact equation to find the general solution.
For the exercises listed, although the need for an integrating factor isn't explicitly mentioned, it's implied in cases where the exact equation criterion is not met naturally. In such instances, integrating factors play a crucial role in problem-solving, illuminating the path to uncovering the function M(x, y) that makes the differential equation exact.