Question

$$ (\sqrt{x+y}+\sqrt{x-y}) d x+(\sqrt{x-y}-\sqrt{x+y}) d y=0 $$

Short answer

The given differential equation is exact, and its general solution is \(\frac{2}{3}(x+y)^{\frac{3}{2}} + \frac{2}{3}(x-y)^{\frac{3}{2}} = C\).

Step-by-step solution

1. Identify M and N Functions

Identify the functions M and N in the given equation: \[ M(x,y) = (\sqrt{x+y} + \sqrt{x-y}) \\ N(x,y) = (\sqrt{x-y} - \sqrt{x+y}) \]

2. Check if the Given Equation is Exact

Check if the given equation is exact by verifying if the following condition is true: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] Calculate the partial derivatives of M and N: \[ \frac{\partial M}{\partial y} = \frac{1}{2\sqrt{x+y}} + \frac{-1}{2\sqrt{x-y}} \\ \frac{\partial N}{\partial x} = \frac{1}{2\sqrt{x-y}} + \frac{-1}{2\sqrt{x+y}} \] Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the given equation is an exact differential equation.

3. Find the Potential Function

Integrate M with respect to x and N with respect to y: \[ \int M dx = \int (\sqrt{x+y} + \sqrt{x-y}) dx \] We integrate the 1st term with respect to x and the 2nd term with substitution: Let \(u = x-y\), then \(du = dx\): \[ \int M dx = \int \sqrt{x+y} dx + \int \sqrt{u} du \] \[ = \frac{2}{3}(x+y)^{\frac{3}{2}} + \frac{2}{3}(x-y)^{\frac{3}{2}} + f(y) \] Now integrate N with respect to y, and match it with the potential function which we found from the integration of M: \[ \int N dy = \int (\sqrt{x-y} - \sqrt{x+y}) dy \] With the substitutions \(v = x-y\) and \(w = x+y\): \[ = \int \sqrt{v} dv - \int \sqrt{w} dw \] \[ = \frac{2}{3}(x-y)^{\frac{3}{2}} - \frac{2}{3}(x+y)^{\frac{3}{2}} + g(x) \] Comparing the two results, we have: \[ \Psi(x, y) = \frac{2}{3}(x+y)^{\frac{3}{2}} + \frac{2}{3}(x-y)^{\frac{3}{2}} \]

4. Write the General Solution

Write the general solution of the exact differential equation: \[ \Psi(x, y) = C \] Substitute the potential function: \[ \frac{2}{3}(x+y)^{\frac{3}{2}} + \frac{2}{3}(x-y)^{\frac{3}{2}} = C \] This is the solution to the given exact differential equation.

Key concepts

Partial Derivatives

A partial derivative is a derivative taken of a function with respect to one variable while keeping all other variables constant. When dealing with exact differential equations, partial derivatives play a key role.
They help determine if the given equation is exact by ensuring the equality of the partial derivative of one function with respect to a variable and the partial derivative of another function with respect to the other variable.

• For example, if we have functions \( M(x,y) \) and \( N(x,y) \), the equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
• This condition ensures that there exists a potential function that can be derived from \( M \) and \( N \).

In the original problem, we calculated \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) to verify the exactness of the equation, leading us to successfully confirm it using partial derivatives.

Potential Function

In exact differential equations, a potential function \( \Psi(x, y) \) serves as the function whose level curves represent solutions to the differential equation.
Finding this potential function is a central step in solving these equations.

• To find \( \Psi(x, y) \), one typically integrates the function \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \).
• The goal is for these integrations to yield the same potential function, possibly up to an unknown function of the other variable.
• In our problem, the integrations indeed produced consistent expressions, confirming \( \Psi(x, y) = \frac{2}{3}(x+y)^{\frac{3}{2}} + \frac{2}{3}(x-y)^{\frac{3}{2}} \).

By identifying this function, we were then able to state the general solution to the differential equation in terms of this potential function.

Integrating Factors

When a differential equation is not exact, integrating factors can be used to make it exact. This technique involves finding a function by which the entire differential equation can be multiplied to achieve exactness.
Despite not requiring an integrating factor for this particular problem, let's see how the concept generally applies:

• First, determine if the equation is not exact using the condition \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \).
• Next, find a suitable integrating factor, \( \mu(x, y) \), which can depend on either \( x \) or \( y \), or both, such that multiplying the differential equation by \( \mu \) makes it exact.
• Afterward, one can proceed to find the potential function of the modified, now exact, differential equation, following the usual process.

This step is vital because not all differential equations are inherently exact, and integrating factors give a potent tool to address such challenges.