Problem 4 - Differential Equations and Linear Algebra

Question

Determine whether the given differential equation is exact. $$2 x e^{y} d x+\left(3 y^{2}+x^{2} e^{y}\right) d y=0$$

Step-by-Step Solution

Answer

The given differential equation is exact, as the partial derivatives match: \(\frac{\partial M}{\partial y} = 2xe^y\) and \(\frac{\partial N}{\partial x} = 2xe^y\).

Step 1: Compute the partial derivative of M with respect to y

We are given M(x, y) = 2xe^y, so we compute the partial derivative with respect to y: \(\frac{\partial M}{\partial y} = 2x\frac{d}{dy}(e^y) = 2xe^y\)

Step 2: Compute the partial derivative of N with respect to x

We are given N(x, y) = 3y^2 + x^2e^y, so we compute the partial derivative with respect to x: \(\frac{\partial N}{\partial x} = 3\cdot0 + 2xe^y = 2xe^y\)

Step 3: Compare the partial derivatives

Now we compare the partial derivatives we computed in steps 1 and 2: \(\frac{\partial M}{\partial y} = 2xe^y \\\) \(\frac{\partial N}{\partial x} = 2xe^y\) As both partial derivatives match, we can conclude that the given differential equation is exact.

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial when dealing with multivariable functions, as they are indicative of a function's rate of change with respect to one variable while keeping other variables constant. In the context of exact differential equations, checking partial derivatives plays a significant role.

For example, when examining the function M(x, y) = 2xe^y, its partial derivative with respect to y, denoted as \( \frac{\partial M}{\partial y} \), is found by treating x as a constant and differentiating M solely with respect to y. This process is akin to ordinary differentiation, just applied to one variable at a time. Notably, the existence and equality of specific partial derivatives ensure that an equation is exact, serving as a test for exactness.

Differential Equation

A differential equation relates an unknown function to its derivatives. Understanding these equations is vital as many physical laws and relationships can be modeled using them. They span from simple separable equations to more complex ones, like the one in question:

\[2 x e^{y} dx + (3 y^{2} + x^{2} e^{y}) dy = 0\]
This particular equation is an example of an exact differential equation, where the solution involves finding a potential function whose partial derivatives correspond to the components of the equation. If a differential equation is accompanied by conditions known as boundary or initial conditions, it becomes a boundary-value or initial-value problem, respectively.

Exponential Function

The exponential function, often denoted as e to the power of a variable (like \(e^y\)), is notable for its constant rate of growth and its frequent occurrence in natural phenomena, such as population growth and radioactive decay.

In our differential equation, \(e^y\) indicates the equation's non-linearity and its variable-dependent growth rate. Exponential functions are special because their derivative is proportional to the function itself, making them invaluable in solving many types of differential equations. Moreover, the exponential function can also serve as an integrating factor, a tool we use to make some non-exact equations exact.

Integrating Factor

An integrating factor is a function used to transform a non-exact differential equation into an exact one, making it solvable through integration. The integrating factor is usually denoted as \(\mu(x, y)\) and is chosen such that it depends only on x or only on y to maintain simplicity in the computations.

When multiplied by the original differential equation, the integrating factor creates perfect differentials on both sides, which can then be integrated to find the solution. However, in the exercise we're considering, since both partial derivatives of \(M\) with respect to y and \(N\) with respect to x are equal, the equation is already exact, nullifying the need for an integrating factor in this instance.