Question

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ \frac{x d x}{\left(x^{2}+y^{2}\right)^{3 / 2}}+\frac{y d y}{\left(x^{2}+y^{2}\right)^{3 / 2}}=0 $$

Short answer

The solution to the given exact differential equation is: $$ \frac{1}{(x^2+y^2)\sqrt{(x^2+y^2)}} + C = 0 $$

Step-by-step solution

Determine M and N in the given equation

Identify the terms M(x, y) and N(x, y) in the given equation: $$ \frac{x dx}{(x^2 + y^2)^{3/2}} + \frac{y dy}{(x^2 + y^2)^{3/2}} = 0 $$ So, we have: $$ M(x, y) = \frac{x}{(x^2 + y^2)^{3/2}} \quad \text{and} \quad N(x, y) = \frac{y}{(x^2 + y^2)^{3/2}} $$

Test for exactness

To determine if the given equation is exact, check if: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Find the partial derivatives with respect to y and x: $$ \frac{\partial M}{\partial y} = \frac{-3x^2y}{(x^2+y^2)^{5/2}} \\ \frac{\partial N}{\partial x} = \frac{-3x^3}{(x^2+y^2)^{5/2}} $$ Simplify and observe that: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ This means the differential equation is exact.

Finding the potential function

Integrate M(x, y) with respect to x and N(x, y) with respect to y: $$ \int M dx = \int \frac{x}{(x^2 + y^2)^{3/2}} dx \\ \int N dy = \int \frac{y}{(x^2 + y^2)^{3/2}} dy $$ The integrals are similar. Using substitution, let: $$ u = x^2 + y^2 \Rightarrow du = 2x dx $$ For M: $$ \int \frac{1}{2u^{3/2}} du = -\frac{1}{u\sqrt{u}} = -\frac{1}{(x^2+y^2)\sqrt{(x^2+y^2)}} $$ And for N, let: $$ v = x^2 + y^2 \Rightarrow dv = 2y dy $$ So, $$ \int \frac{1}{2v^{3/2}} dv = -\frac{1}{v\sqrt{v}} = -\frac{1}{(x^2+y^2)\sqrt{(x^2+y^2)}} $$ Therefore, the potential function F(x, y) is: $$ F(x, y) = -\frac{1}{(x^2+y^2)\sqrt{(x^2+y^2)}} + C $$

Determine the solution

The solution to the exact differential equation is: $$ F(x, y) = C \\ -\frac{1}{(x^2+y^2)\sqrt{(x^2+y^2)}} = C $$ Hence, the solution to the differential equation is: $$ \frac{1}{(x^2+y^2)\sqrt{(x^2+y^2)}} + C = 0 $$

Key concepts

Partial Derivatives

Partial derivatives are fundamental to calculus involving functions of several variables. In an exact differential equation, understanding partial derivatives helps determine the behavior of those functions. Given a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), measures how the function changes as \( x \) changes, while keeping \( y \) constant. Similarly, the partial derivative with respect to \( y \) is denoted as \( \frac{\partial f}{\partial y} \).
When dealing with exact equations, we check the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) to determine exactness. This indicates that the mixed partial derivatives are equal, which is crucial for the equation to be exact, leading us to a potential function. The partial derivatives link the components \( M(x, y) \) and \( N(x, y) \), blending these variables in a balanced manner.

Potential Function

The potential function, often a vital goal when working with exact differential equations, represents a function \( F(x, y) \) such that its gradient (or differential) matches the given components. When the differential of this potential function is split into \( dF = Mdx + Ndy \), each component of the equation can be traced back to a derivative of \( F \).

• For \( M(x, y) \), we have \( \frac{\partial F}{\partial x} = M \).
• For \( N(x, y) \), \( \frac{\partial F}{\partial y} = N \).

Finding this potential function involves integrating \( M \) with respect to \( x \) and \( N \) with respect to \( y \). It serves as a central piece to solve the differential equation, converting it into a simpler algebraic form.

Integration

Integration is the tool we use to find the potential function in exact differential equations. Here, we need to perform integration over the components \( M \) and \( N \).
For example, consider our previous function \( M(x, y) = \frac{x}{(x^2 + y^2)^{3/2}} \). To integrate this term with respect to \( x \), the variable part affected by the integration is \( x \), while treating \( y \) as a constant.
The process includes:

• Setting a substitution variable \( u = x^2 + y^2 \), simplifying the expression.
• Finding the new differential, such as \( du = 2x dx \).
• Integrating the rewritten expression, like \( \int \frac{1}{2u^{3/2}} du \), resulting in simpler expressions.

Repeating similar steps for \( N \) ensures we develop the function further, combined to form the entire \( F(x, y) \).

Exactness Test

The exactness test is a crucial step to ascertain whether a differential equation is indeed exact. This determines if it's possible to derive a potential function. The test involves calculating partial derivatives of the terms \( M \) and \( N \), and checking the equality:
If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), then the equation is exact. This condition tells us that there exists some potential function \( F(x, y) \) in which these partial matches occur naturally. If they are not equal, modifications or other techniques might be necessary to render the equation exact.