Question

Show that if \(\left(N_{x}-M_{y}\right) / M=Q,\) where \(Q\) is a function of \(y\) only, then the differential equation $$ M+N y^{\prime}=0 $$ has an integrating factor of the form $$ \mu(y)=\exp \int Q(y) d y $$

Short answer

Answer: The integrating factor for a first order inexact differential equation is given by \(\mu(y) = \exp \int Q(y) dy\).

Step-by-step solution

Express the differential equation in the exact form

To do this, rearrange the given differential equation as \(M dx + N dy = 0\). This form allows us to identify the exact coefficients (M and N). Thus, we have: $$ M dx + N dy = 0 $$

Find the partial derivatives of the coefficients

For the given condition \((N_x - M_y) / M = Q\), we need to find the partial derivative \(M_y\) and \(N_x\). Taking the partial derivative of M with respect to y, we get \(M_y\). Similarly, taking the partial derivative of N with respect to x, we get \(N_x\).

Use the given condition to derive an equation for \(\mu(y)\)

Using \((N_x - M_y) / M = Q\), we can write the equation: $$ \frac{\mu'}{\mu} = Q $$ Where \(\mu'\) represents the derivative of the integrating factor \(\mu\) with respect to y.

Integrate the equation for \(\mu(y)\) to find the integrating factor

To arrive at the answer, we need to integrate the equation. We will integrate both sides of the equation with respect to y: $$ \int\frac{\mu'}{\mu} dy = \int Q(y) dy $$ Now, integrating the left-hand side of the equation, we get: $$ \ln|\mu(y)| = \int Q(y) dy + C $$ To find \(\mu(y)\), we can take the exponential on both sides: $$ \mu(y) = \exp \int Q(y) dy \cdot \exp(C) $$ Since the goal is to find an integrating factor that satisfies the equation \(\mu(y) = \exp \int Q(y) dy\), we can ignore the exponential of constant factor \(\exp(C)\). Therefore, the integrating factor is: $$ \mu(y) = \exp \int Q(y) dy $$ This is the desired solution of the problem.

Key concepts

Exact Differential Equations

An exact differential equation is a special type of first-order differential equation that can be solved by integration directly. It has the form
\[ M(x, y)dx + N(x, y)dy = 0 \]
where \( M \) and \( N \) are functions of \( x \) and \( y \). For this equation to be exact, the partial derivatives of \( M \) with respect to \( y \), denoted \( M_y \), and \( N \) with respect to \( x \), denoted \( N_x \), must be equal. In other words, \( M_y = N_x \). This signifies that there exists a potential function \( F(x,y) \) such that \( dF = Mdx + Ndy \), allowing us to integrate directly to find the solution. The problem we're addressing involves finding an integrating factor when the equation is not initially exact but can be made exact through multiplication by the appropriate integrating factor.

Partial Derivatives

Partial derivatives are used to measure how a function changes as one of its variables changes while keeping the other variables constant. If we have a function \( f(x, y) \), then the partial derivative with respect to \( x \), denoted \( f_x \) or \( \frac{\partial f}{\partial x} \), is determined by differentiating \( f \) with only \( x \) varying. Similarly, the partial derivative with respect to \( y \), designated as \( f_y \) or \( \frac{\partial f}{\partial y} \), is found by differentiating with only \( y \) changing. In our exercise, the given condition involved partial derivatives and was essential to determining the form of the integrating factor for the differential equation.

Integrating Factors

An integrating factor is a function, usually denoted by \( \mu \), that when multiplied by a non-exact differential equation, makes it exact. In other words, it transforms a non-exact equation into an exact one. Finding the integrating factor is often the key to solving a first-order linear non-exact differential equation. The usual steps to find an integrating factor involve identifying a function that depends on only one of the variables (usually \( x \) or \( y \)) such that when it is multiplied by the original equation, it satisfies the exactness condition. The equation \( \frac{\mu'}{\mu} = Q \) comes from manipulating the differential equation into a form where \( \mu \) can be isolated and then integrated to find the solution. In the context of our exercise, once the integrating factor \( \mu(y) \) is determined, it is used to make the initial differential equation exact, paving the way for finding a solution.

First-Order Ordinary Differential Equations

First-order ordinary differential equations (ODEs) are equations involving a function and its first derivative. They are classified as 'ordinary' because they involve derivatives with respect to only one independent variable. These types of equations often describe various phenomena in physics, engineering, and other sciences. The general form of a first-order ODE is \[ y' + p(x)y = g(x) \], where \( p \) and \( g \) are given functions, and \( y' \) represents the derivative of \( y \) with respect to \( x \). Solving a first-order ODE typically involves finding a function \( y(x) \) that satisfies the equation. There are several methods to solve such equations, with integrating factors being one powerful technique, especially when the first-order ODE is not exact to begin with, as illustrated in our provided exercise.