Problem 30 - Differential Equations and Linear Algebra

Question

Determine the values of the constants \(r\) and \(s\) such that \(I(x, y)=x^{r} y^{s}\) is an integrating factor for the given differential equation. $$\left(y^{-1}-x^{-1}\right) d x+\left(x y^{-2}-2 y^{-1}\right) d y=0$$

Step-by-Step Solution

Answer

The values of the constants are \(r = -1\) and \(s = 1\), making the integrating factor \(I(x, y) = x^{-1}y^1 = \frac{y}{x}\).

Step 1: Write down the given differential equation and function

Our given differential equation is: \[\left(y^{-1}-x^{-1}\right) dx+\left(x y^{-2}-2 y^{-1}\right) dy=0\] And we want to find the values of r and s such that the function \[I(x, y) = x^r y^s\] is an integrating factor for the given differential equation.

Step 2: Write the given differential equation in the form M(x, y) dx + N(x, y) dy = 0

We can rewrite the given differential equation as follows: \[M(x, y) dx + N(x, y) dy = 0\] Where \[M(x, y) = y^{-1} - x^{-1}\] and \[N(x, y) = x y^{-2} - 2 y^{-1}\]

Step 3: Multiply the differential equation by the integrating factor and check exactness

Let's multiply the given differential equation by the integrating factor, I(x, y) = x^r y^s: \[\left(x^r y^s(y^{-1}-x^{-1})\right) dx + \left(x^r y^s(x y^{-2}-2 y^{-1})\right) dy = 0\] Now we need to check if the new differential equation is exact. For that, we need to check if the following condition holds: \[\frac{\partial(M\cdot I(x, y))}{\partial y} = \frac{\partial(N\cdot I(x, y))}{\partial x}\]

Step 4: Calculate partial derivatives and find r and s

Let's calculate the partial derivatives: \[\frac{\partial(M\cdot I(x, y))}{\partial y} = \frac{\partial\left(x^r y^s(y^{-1}-x^{-1})\right)}{\partial y}\] \[\frac{\partial(N\cdot I(x, y))}{\partial x} = \frac{\partial\left(x^r y^s(x y^{-2}-2 y^{-1})\right)}{\partial x}\] Solve these partial derivatives and find r and s such that the exactness condition holds. By solving the partial derivatives, we get: \[\frac{\partial(M\cdot I(x, y))}{\partial y} = -rsx^{r-1}y^{s-2} + sx^{r-1}y^{2s-1}\] \[\frac{\partial(N\cdot I(x, y))}{\partial x} = -rsx^{r-1}y^{s-1} + sx^{r}y^{2s-2}\] Now, we need to find r and s such that: \[-rsx^{r-1}y^{s-2} + sx^{r-1}y^{2s-1} = -rsx^{r-1}y^{s-1} + sx^{r}y^{2s-2}\]

Step 5: Solve the equation to find the values of r and s

By comparing the coefficients of powers of x and y, we can derive: \[\begin{cases} r - 1 = r \\ s - 2 = 2s - 1 \\ \end{cases}\] Solve these equations to get the values of r and s. By solving the system of equations, we get: \[r = -1\] \[s = 1\] Therefore, the integrating factor I(x, y) = x^r y^s is: \[I(x, y) = x^{-1}y^1 = \frac{y}{x}\]

Key Concepts

Differential Equation

A differential equation is a mathematical expression that involves functions and their derivatives. The core idea is to understand how a function changes. For example, in the given exercise, we have the equation: \( \left(y^{-1}-x^{-1}\right) dx + \left(x y^{-2}-2 y^{-1}\right) dy = 0 \). This expression is composed of derivatives \( dx \) and \( dy \), representing small changes in \( x \) and \( y \).
An equation like this tells us about the relationship between the functions \( x \) and \( y \) and their rates of change. Solving a differential equation like this one often involves finding a function or functions that satisfy the equation in a particular domain.

There are many types of differential equations, and each type requires specific methods for finding a solution. The given equation falls under first-order differential equations, which involves the first derivative of the function.

Exactness Condition

The exactness condition helps us determine whether a differential equation can be solved by the method of exact equations. An equation is exact if it can be written in the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \) and satisfies the exactness condition, which is:

• \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \)

This condition ensures that there is a function \( F(x, y) \) such that its total differential, \( dF = M \, dx + N \, dy \), will satisfy the equation. In the context of the exercise, multiplying both terms of the original differential equation by an integrating factor \( I(x, y) = x^r y^s \) aims to convert it into an exact equation. To validate this transformation, we check the exactness condition with the new differential equation: \( \frac{\partial(M \cdot I(x, y))}{\partial y} = \frac{\partial(N \cdot I(x, y))}{\partial x} \).
Solving this confirms whether the selected integrating factor truly makes the equation exact.

Partial Derivatives

Partial derivatives play a crucial role in solving differential equations, particularly when analyzing the exactness of an equation. They represent the derivative of a function with respect to one of several variables, holding the others constant.

In our example,

• \( \frac{\partial(M \cdot I(x, y))}{\partial y} \) represents the derivative of the product \( M \cdot I(x, y) \) concerning \( y \).
• \( \frac{\partial(N \cdot I(x, y))}{\partial x} \) represents the derivative of the product \( N \cdot I(x, y) \) concerning \( x \).

These derivatives are essential for checking the exactness condition, where their equality ensures that the differential equation can be integrated. It involves computing these derivatives accurately to understand how each term contributes to the overall behavior of the equation.

System of Equations

A system of equations consists of multiple equations that are solved together since they share common variables. In the process of finding the integrating factor, we often end up with a system of equations derived from matching coefficients and conditions.

For instance, to find the values of \( r \) and \( s \) such that the differential equation multiplied by the integrating factor becomes exact, we derive systems like:

• \( r - 1 = r \)
• \( s - 2 = 2s - 1 \)

Solving these equations involves finding values of \( r \) and \( s \) that satisfy all the conditions simultaneously.
In our case:

• The solution \( r = -1 \) and \( s = 1 \) helped us construct the integrating factor \( I(x, y) = \frac{y}{x} \).

This solution demonstrates how algebraic manipulation and logical reasoning are applied to solve such systems, leading to a deeper understanding of how to simplify and solve differential equations.