Question

If \(M=y f(x y)\) and \(N=x g(x y)\), show that \(1 /(M x-N y)\) is an integrating factor provided \(M x-N y \neq 0 .\)

Short answer

Question: Show that \(\frac{1}{Mx - Ny}\) is an integrating factor for the given expressions \(M = yf(xy)\) and \(N = xg(xy)\). Answer: \(\frac{1}{Mx - Ny}\) is an integrating factor for the given expressions \(M = yf(xy)\) and \(N = xg(xy)\), provided that \(Mx-Ny \neq 0\).

Step-by-step solution

Calculate the Partial Derivatives

First, we need to find the partial derivatives of \(M\) and \(N\) with respect to \(x\) and \(y\), respectively. These partial derivatives are: $$\frac{\partial M}{\partial y} = f(xy) + y^2 f'(xy)$$ $$\frac{\partial N}{\partial x} = g(xy) + x^2 g'(xy)$$

Check if \(\frac{1}{Mx-Ny}\) is an Integrating Factor

Now, we need to check the following condition: $$\frac{\partial M / \partial y - \partial N / \partial x}{N x-My} = R$$ Substitute the partial derivatives of \(M\) and \(N\) in the equation: $$\frac{\left(f(xy) + y^2 f'(xy)\right) - \left(g(xy) + x^2 g'(xy)\right)}{x^2 g(xy) - y^2 f(xy)} = R$$ Now, we can simplify the numerator by getting rid of the common terms: $$\frac{y^2 f'(xy) - x^2 g'(xy)}{x^2 g(xy) - y^2 f(xy)} = R$$ The left-hand side of the equation can be written in terms of the integrating factor: $$\frac{1}{Mx-Ny} = \frac{1}{x^2 g(xy) - y^2 f(xy)}$$ Now we can see that the left-hand side is indeed a function of \(x\) and \(y\). Therefore, by our observation, \(\frac{1}{Mx-Ny}\) is an integrating factor provided that \(Mx-Ny \neq 0\).

Conclusion

In this exercise, we have shown that \(\frac{1}{Mx-Ny}\) is an integrating factor for the given expressions \(M = yf(xy)\) and \(N = xg(xy)\), provided that \(Mx-Ny \neq 0\).

Key concepts

Differential Equations

Understanding differential equations is essential in the field of mathematics, particularly in the study of phenomena where change is constant. Differential equations provide a way of describing the relationship between functions and their rates of change, which is expressed using derivatives.

Take, for instance, the growth of a population, the decay of radioactive material, or the rate at which a cup of coffee cools; all these can be modeled using differential equations. In the exercise, we consider a particular type of differential equation, one that involves functions of two variables, and we explore an integrating factor that simplifies its solution.

An integrating factor is a function that is multiplied by a differential equation to make it exact, meaning the equation can then be integrated directly. Finding an appropriate integrating factor is a crucial step in solving many differential equations, as it transforms them into a more manageable form. The exercise demonstrates how to verify if a given expression is an integrating factor for a specific type of differential equation.

Partial Derivatives

Partial derivatives come into play when dealing with functions of multiple variables. They represent the rate at which a function changes as one of its variables varies while the others are held constant. This concept is pivotal for our exercise, as it involves differentiating the functions with respect to each variable, which constitute the terms of the differential equation.

In the steps provided, partial derivatives were calculated for functions M and N. These calculations allow us to inspect the underlying structure of the differential equation. The beauty of partial derivatives lies in their ability to untangle the combined effects of multiple factors on a single outcome. Recognizing and computing partial derivatives is a fundamental skill in solving differential equations that arise in fields such as physics, engineering, and economics.

Ordinary Differential Equations

While the broader category of differential equations can be quite expansive, ordinary differential equations (ODEs) form a subclass within it. ODEs are equations involving functions of a single independent variable and their derivatives. In contrast, partial differential equations (PDEs) deal with functions of multiple independent variables.

The given exercise does not specifically state an ODE, but it implies one, as it involves finding an integrating factor that would potentially be used to solve an ODE. ODEs are ubiquitous throughout calculus and many applied sciences; a deep comprehension of these allows us to solve practical problems such as motion dynamics and electrical circuits. It's vital to grasp the distinction between ODEs and PDEs and the respective solution techniques - knowledge which constitutes the cornerstone of mathematical modeling and various applications in technology and science.