Problem 13 - Differential Equations and Linear Algebra

Question

Solve the given differential equation. $$[1+\ln (x y)] d x+x y^{-1} d y=0$$

Step-by-Step Solution

Answer

The short version of the solution is as follows: The given differential equation is not exact, so we find an integrating factor \(\mu = \frac{x-1}{x(1+\ln(xy))}\). After multiplying the equation by the integrating factor and simplifying, we obtain an exact differential equation \((x-1) dx + (x^2-x)y^{-1}dy = 0\). Solving the exact differential equation, we find the particular solution: \(y = e^{\frac{1}{2}x^2-x}\).

Step 1: Identify the type of differential equation

The differential equation given is: \[[1+\ln (xy)] dx + x y^{-1} dy = 0\] Notice that it can be written as M(x, y) dx + N(x, y) dy = 0, where M(x, y) = 1+\ln(xy) and N(x, y) = xy^{-1}. This form suggests that the given equation might be an exact differential equation.

Step 2: Check if the given equation is exact

To check if the given equation is exact, we should check if the following condition holds: \[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\] First, we need to find the partial derivatives: \(\frac{\partial M}{\partial y} = \frac{1}{xy}\), and \(\frac{\partial N}{\partial x} = \frac{1}{y}\) It appears that \(\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}\). Hence, the given equation is not exact.

Step 3: Find an integrating factor

In order to solve the non-exact differential equation, we need to find an integrating factor (IF). We can compute the integrating factor for this equation as the exponent of the integrating factor (IF): \[\mu = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}\] Computing the IF, we get: \[\mu = \frac{\frac{1}{y} - \frac{1}{xy}}{1+\ln(xy)}\] Simplifying the expression, we obtain: \[\mu = \frac{x-1}{x(1+\ln(xy))}\]

Step 4: Multiply the equation by the integrating factor

Multiply the given differential equation by the integrating factor: \[\left(\frac{x-1}{x(1+\ln(xy))}\right)[(1+\ln (xy)) dx + xy^{-1}dy] = 0\] Simplifying, we obtain the exact differential equation: \[((x-1) dx + (x^2-x)y^{-1}dy) = 0\]

Step 5: Solve the exact differential equation

Now that we have an exact differential equation, we can find its solution. To do that, we can integrate M(x, y) with respect to x and N(x, y) with respect to y. We will find two functions and combine them in a single function to solve the equation. First, integrate M(x, y) with respect to x: \[\int (x-1) dx = \frac{1}{2}x^2-x+C_1(y)\] Next, integrate N(x, y) with respect to y: \[\int (x^2-x)y^{-1} dy = (x^2-x)\ln(y) + C_2(x)\] Now, combine both functions and find the constant terms: \[C_1(y) = 0 \text{ and } C_2(x) = 0 \Rightarrow \frac{1}{2}x^2-x = (x^2-x)\ln(y)\] Finally, just isolate y to get the particular solution: \[y = e^{\frac{1}{2}x^2-x}\]

Key Concepts

Exact Differential Equation

An exact differential equation is a type of equation that can be expressed in the form \( M(x, y)dx + N(x, y)dy = 0 \). This equation is considered exact if the partial derivative of \( M \) with respect to \( y \) equals the partial derivative of \( N \) with respect to \( x \). This condition ensures that there exists a function \( \phi(x,y) \) whose differential equals the left side of the equation.When dealing with an exact differential equation:

• Identify \( M(x, y) \) and \( N(x, y) \) from the original differential expression.
• Check exactness by verifying that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

In cases where this condition holds, you can find the solution by integrating \( M \) with respect to \( x \) and \( N \) with respect to \( y \), combining results appropriately to find \( \phi(x,y) \). For example, if \( M(x, y) = x-1 \) and \( N(x, y) = (x^2-x)y^{-1} \), the exactness can be checked and solved accordingly.

Integrating Factor

When a differential equation is not exact, as identified in steps like checking the derivatives, an integrating factor can be introduced to make it exact. This transforming multiplier enables one to solve the previously non-exact equations.To find an integrating factor \( \mu \):

• Apply \( \mu = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} \) and simplify the expression as needed.
• Multiply the entire differential equation by this \( \mu \) to convert it into an exact form.

In the provided exercise, after obtaining \( \mu = \frac{x-1}{x(1+\ln(xy))} \), multiplying this to the original expression helps transform it back into an exact differential equation. This enables us to proceed using the techniques available for exact equations.

Partial Derivatives

Partial derivatives are crucial in the analysis of differential equations. They measure how a function changes as its variables are changed individually. In examining differential equations, they help determine whether an equation is exact.Here's how partial derivatives were used in the solution:

• \( \frac{\partial M}{\partial y} = \frac{1}{xy} \) and \( \frac{\partial N}{\partial x} = \frac{1}{y} \) were calculated.
• By comparing these results, it was found that the original equation was not exact since \( \frac{1}{xy} eq \frac{1}{y} \) .

Partial derivatives hence serve as a diagnostic tool for evaluating and manipulating differential equations to find solutions or necessary adjustments like finding integrating factors.

Solution Techniques

Solving differential equations involves a series of structured steps. With the provided differential equation, these steps include recognizing the non-exact form, finding an integrating factor, adjusting the equation, and then resolving the exact form of the equation.Key techniques include:

• Identifying whether the provided equation is exact or not.
• Using the integrating factor to convert the equation into an exact one.
• Integrating \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \), then combining them to find a single function that describes the solution.

This was exemplified by integrating \( M(x, y) = x-1 \) and \( N(x, y) = (x^2-x)y^{-1} \), leading to the solution \( y = e^{\frac{1}{2}x^2-x} \). These solution techniques are broadly applicable to a variety of differential equation problems, adapting as needed based on the equation type.