Question

Solve \(\left(\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+\frac{1}{y}\right) d x\) \(+\left(\frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}-\frac{x}{y^{2}}\right) d y=0\)

Short answer

#Summary# The given equation is a total differential equation, as it satisfies the condition of equal partial derivatives. We found the potential function, \(F(x,y) = \sqrt{x^2+y^2} + \ln(x) - \ln(y) + \frac{x}{y} + \text{constant}\), by integrating both expressions. The general solution is in the form of a level curve, \(F(x,y) = C\), and is given by: $$\sqrt{x^2+y^2} + \ln(x) - \ln(y) + \frac{x}{y} = C.$$

Step-by-step solution

Check if the equation is a total differential equation

To check if the given equation is a total differential equation, we must check whether: $$ \frac{\partial}{\partial y}\left(\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+\frac{1}{y}\right)=\frac{\partial}{\partial x}\left(\frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}-\frac{x}{y^{2}}\right) $$ Calculate both partial derivatives.

Calculate the partial derivatives

First, calculate the partial derivative with respect to \(y\): $$ \frac{\partial}{\partial y}\left(\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+\frac{1}{y}\right) = -\frac{x^2 \cdot y}{(x^2 + y^2)^{3/2}} - \frac{1}{y^2} $$ Next, calculate the partial derivative with respect to \(x\): $$ \frac{\partial}{\partial x}\left(\frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}-\frac{x}{y^{2}}\right) = -\frac{x \cdot y^2}{(x^2 + y^2)^{3/2}} - \frac{1}{y^2} $$

Compare partial derivatives

Compare both partial derivatives: $$ -\frac{x^2 \cdot y}{(x^2 + y^2)^{3/2}} - \frac{1}{y^2} = -\frac{x \cdot y^2}{(x^2 + y^2)^{3/2}} - \frac{1}{y^2} $$ Since both expressions are equal, the equation is a total differential equation.

Find the potential function

The potential function \(F(x,y)\) can now be found by integrating both expressions: $$ F(x,y) = \int \left(\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+\frac{1}{y}\right) dx + \int \left(\frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}-\frac{x}{y^{2}}\right) dy $$ Integrating with respect to \(x\): $$ \int \left(\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+\frac{1}{y}\right) dx = \sqrt{x^2+y^2} + \ln(x) + \frac{x}{y} + \text{constant}_1(y) $$ Integrating with respect to \(y\): $$ \int \left(\frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}-\frac{x}{y^{2}}\right) dy = \sqrt{x^2+y^2} - \ln(y) + \frac{x}{y} + \text{constant}_2(x) $$ Combining the results, we get the potential function: $$ F(x,y) = \sqrt{x^2+y^2} + \ln(x) - \ln(y) + \frac{x}{y} + \text{constant} $$

Find the level curve

Since the equation is a total differential equation, we can find its solution by finding the level curves of the potential function. This means that the solution will be of the form: $$ F(x,y) = C $$ for constant \(C\). Therefore, we have: $$ \sqrt{x^2+y^2} + \ln(x) - \ln(y) + \frac{x}{y} = C $$ This represents the general solution of the given equation.

Key concepts

Partial Derivatives

Partial derivatives play a critical role in the analysis of multivariable functions, such as the total differential equation encountered in the exercise. When dealing with functions of two or more variables, such as \(f(x, y)\), the partial derivative with respect to one variable, say \(x\), measures how the function changes as \(x\) changes while keeping the other variable \(y\) constant.

In the context of our exercise, we computed the partial derivatives with respect to both variables to ensure that the equation was indeed a total differential equation. This verification was carried out because for an equation to be classified as a total differential equation, the cross partial derivatives need to be equal. That is, the partial derivative of the first function with respect to \(y\) must equal the partial derivative of the second function with respect to \(x\).

Why Are Partial Derivatives Important?

• They help verify whether a two-variable function is a potential function.
• They facilitate the check for exactness in a total differential equation.
• Understanding each partial derivative's behavior provides insights into the function's slope along the axes of variables.

Potential Function

A potential function, sometimes known as a scalar field, is a function \(F(x, y)\) whose gradients (partial derivatives) are equal to the components of a vector field. In simpler terms, it represents the original function from which the total differential equation is derived.

Once we determine that a differential equation is total, we can find a potential function that, when differentiated, reproduces the original differential equation. To solve an equation of this kind, we need to integrate the given functions with respect to their respective variables, which we did step by step in the provided solution.

Integrating to Find a Potential Function

The idea is to reverse the differentiation process. For the given equation, integration was performed separately with respect to \(x\) and \(y\), and constants of integration were included. These constants may functionally depend on the opposite variable due to integrating a partial derivative.

Level Curves

Level curves, or contour lines, are curves that represent points on a graph where a function \(f(x, y)\) takes on a constant value. These curves are crucial when analyzing multivariable functions because they can help visualize complex relationships in two dimensions.

In the last step of the solution, we sought the level curve of the potential function, which graphically corresponds to a 'slice' or a contour of the 3D surface defined by the potential function. The equation \(F(x,y) = C\), where \(C\) is a constant, describes these level curves and represents the general solution to the total differential equation.

Visualizing Solutions through Level Curves

Level curves can be an effective tool to extract insights regarding the distribution and behavior of solutions in the plane. They effectively transform a 3D surface into a series of 2D slices, making it easier to understand the interaction between variables in solutions to differential equations.