Question

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ (y / x+6 x) d x+(\ln x-2) d y=0, \quad x>0 $$

Short answer

Question: Determine the general solution to the given exact differential equation: \((y/x + 6x)dx + (\ln x - 2)dy = 0\). Answer: The general solution to the given exact differential equation is \(y\ln{x} + 3x^2 - 2y = k\), where \(k\) is an arbitrary constant.

Step-by-step solution

Check for Exactness

First, let's rewrite the equation in a more common notation for differential equations: $$ M(x, y)dx + N(x, y)dy = 0 $$ Here, \(M(x, y) = \frac{y}{x} + 6x\) and \(N(x, y) = \ln x - 2\). Now let's calculate the partial derivatives of \(M\) and \(N\) with respect to \(x\) and \(y\): $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(\frac{y}{x} + 6x) = \frac{1}{x} $$ $$ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(\ln x - 2) = \frac{1}{x} $$ Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the differential equation is exact.

Find Potential Function

Now that we've confirmed the equation is exact, we need to find the potential function \(\Psi(x, y)\) such that: $$ \frac{\partial \Psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \Psi}{\partial y} = N(x, y) $$ First we find the antiderivative of M with respect to x. Integrating \(M(x, y)\) with respect to \(x\): $$ \Psi(x, y) = \int M(x, y) \, dx = \int \left(\frac{y}{x} + 6x \right) dx = y\ln{x} + 3x^2 + H(y) $$ Here \(H(y)\) is the integration constant, which is allowed since we integrate with respect to \(x\). Now differentiate \(\Psi\) with respect to \(y\): $$ \frac{\partial \Psi}{\partial y} = \frac{\partial}{\partial y}\left(y\ln{x} + 3x^2 + H(y)\right) = \ln x + H'(y) $$ Since \(\frac{\partial \Psi}{\partial y} = N(x, y)\), we can equate: \(N(x,y) = \ln x + H'(y)\). As \(H'(y)\) does not consist of x, it should be equal to the constant value -2. Thus, we have $$ H(y) = -2y + C $$ Now substitute H(y) back into the potential function: $$ \Psi(x, y) = y\ln{x} + 3x^2 - 2y + C $$

Determine Solution

The solution to the exact differential equation is given by the level curves of \(\Psi(x, y)\). Thus, $$ \Psi(x,y) = k $$ $$ y\ln{x} + 3x^2 - 2y = k $$ This is the general solution to the given exact differential equation, where \(k\) is an arbitrary constant.

Key concepts

Partial Derivatives

When we deal with functions of multiple variables, such as in our differential equation, the concept of partial derivatives comes into play. Partial derivatives measure how a function changes as only one of the variables is varied while others are held constant.

For instance, if we have a function described by two variables, say, \( f(x, y) \), the partial derivative with respect to \( x \) is written as \( \frac{\partial f}{\partial x} \). Here, we're focusing only on how \( f \) changes with \( x \), ignoring any changes related to \( y \) and vice versa for \( \frac{\partial f}{\partial y} \).

In the given exercise, computing the partial derivatives of \( M(x, y) \) and \( N(x, y) \) helps us determine whether the differential equation is exact. If these partials are equal, as they are in our example, this indicates the existence of a potential function that can be used to find a solution.

Potential Function

The potential function is a powerful tool in solving exact differential equations. It is a function \( \Psi(x, y) \) whose gradients, regarding \( x \) and \( y \), correspond to the components of the differential equation.

To find this function, we integrate each part of the differential equation. The integration is done with respect to one variable at a time, considering the other variable as a constant, which may introduce an arbitrary function of the other variable as a 'constant' of integration.

The potential function provides not just the solution but a deeper understanding of the behavior of the system described by the differential equation, as it encapsulates the whole system's state in a single expression.

Integration of Differential Equations

Integration is a critical operation in mathematics, particularly when we aim to solve differential equations. In the context of differential equations, integration is not merely finding an antiderivative, but a systematic approach to uncovering the potential function from the given components of the equation.

In our example, we integrated \( M(x, y) \) with respect to \( x \) to find a part of \( \Psi \) that corresponds to \( M \), then differentiated this integrated expression with respect to \( y \) to find the 'missing piece', the function \( H(y) \). This method shows the interconnectedness of differentiation and integration in finding solutions to exact differential equations.

Level Curves

Level curves, or contour lines, are a graphical representation of a 3D surface on a 2D plane, where each curve represents points at which a function has the same value, commonly referred to as 'level'. They are especially useful in fields such as geography, engineering, and physics.

In the scenario of solving exact differential equations, the level curves of the potential function \( \Psi \) represent the solution. Each curve is associated with a constant value that the potential function takes, defining a relationship between the variables \( x \) and \( y \). Essentially, the curves are a family of solutions, and any one of them, selected by a specific value of \( k \), will satisfy the original differential equation, as seen in the final form of the solution in our exercise.