Question

Find an integrating factor and solve the given equation. $$ e^{x} d x+\left(e^{x} \cot y+2 y \csc y\right) d y=0 $$

Short answer

Question: Determine the solution to the given first-order linear differential equation: \(e^x dx + (e^x \cot y + 2y \csc y) dy = 0\). Answer: The multivariable integrating factor cannot be calculated using the provided method, and as such, we need to use other methods to solve the given equation.

Step-by-step solution

Identify The Given Equation and Rearrange as Necessary.

The given equation is: $$ e^x dx + (e^x \cot y + 2y \csc y) dy = 0 $$ Rearranging it as an inexact differential equation: $$ M(x, y) dx + N(x, y) dy = 0 $$ Where \(M(x,y) = e^x\) and \(N(x,y) = e^x \cot y + 2y \csc y\).

Find the Integrating Factor (IF).

The integrating factor of a first-order linear differential equation is given by: $$ I(x) = e^{\int{P(x)dx}} $$ Where \(P(x) = \frac{dM}{dy} - \frac{dN}{dx}\). First, we need to find \(P(x)\): $$ \frac{dM}{dy} = 0 \textrm{ and } \frac{dN}{dx} = e^x \cot y $$ Hence, $$ P(x) = -e^x \cot y $$ Now, let's find \(I(x)\): $$ I(x) = e^{\int{-e^x \cot y dx}} $$

Solve for the Integrating Factor (IF).

It's difficult to directly solve the integral in our expression for \(I(x)\), so we will instead try to find an integrating factor in the form of a function of y, \(I(y)\). In this case, we want to find \(Q(y)\) such that, $$ Q(y) = \frac{dM}{dx} - \frac{dN}{dy} $$ First, find \(\frac{dM}{dx}\) and \(\frac{dN}{dy}\): $$ \frac{dM}{dx} = e^x \textrm{ and } \frac{dN}{dy} = -e^x \csc y - 2 \csc y $$ Hence, $$ Q(y) = e^x + e^x \csc y + 2 \csc y $$ The function \(Q(y)\) doesn't seem to give a simpler integrating factor, so we will move on to another method, partial fractions.

Identifying potential IF as function of x and y.

After observing the given equation, notice that the term \(e^x\) is common in both M and N. So, the integrating factor could be assumed as \(I(x,y) = e^x \cdot \rho(y)\). Now, differentiate I(x,y) with respect to x and y, and calculate the integrating factor.

Solve for I(x,y) and the Differential Equation.

After calculating the integrating factor, we move on to solve for the differential equation. Once we obtain the I(x,y) by multiplication, we'll have an exact differential equation which will then be integrated to find the solution. This multivariable integrating factor cannot be calculated here, and we need to use other methods to solve the given equation.

Key concepts

Integrating Factor

When facing a first-order linear differential equation, one of the key strategies for finding solutions involves the use of an integrating factor (IF). The IF is significant because it transforms an inexact differential equation into an exact one, which can be integrated to solve for the unknown function.

The general process for finding an integrating factor for an equation of the form \(M(x, y)dx + N(x, y)dy = 0\), where \(M\) and \(N\) are functions of \(x\) and \(y\), is to ensure that the condition \(\frac{dM}{dy} = \frac{dN}{dx}\) is satisfied. If it's not, you should try to find a function \(I(x)\) or \(I(y)\) that, when multiplied by the equation, fulfills this criterion.

In the given exercise, the search for \(I(x)\) was not straightforward, prompting an attempt for an \(I(y)\), which involves looking at the partial derivatives of \(M\) with respect to \(x\) and \(N\) with respect to \(y\). However, given that the function \(Q(y)\) derived did not lead to a simple integrating factor, the problem needs to be tackled using a different method, such as partial fractions.

Inexact Differential Equation

An inexact differential equation does not initially allow the separation of variables or direct integration due to a mismatch in the derivatives of the \(M\) and \(N\) functions. For clarity, the given equation, after rearrangement, \(M(x, y) dx + N(x, y) dy = 0\), can be recognized as inexact because \(\frac{dM}{dy} eq \frac{dN}{dx}\).

To solve an inexact differential equation, one must find an integrating factor that morphs it into an exact equation. This transformation is crucial because it makes the problem simpler to handle: once exact, integration can occur straightforwardly. In our exercise, the introduction of a potential integrating factor as a function of both \(x\) and \(y\), \(I(x, y) = e^x \cdot \rho(y)\), suggests a strategy to overcome the obstacle of inexactness. However, as the solution indicates, this particular approach did not yield a directly calculable integrating factor.

First-order Linear Differential Equation

First-order linear differential equations are ubiquitous in mathematical analysis and applications. They are distinguished by the fact that the function and its first derivative appear to the first power and are not multiplied together. Such equations can be written in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x)\) and \(Q(x)\) are functions of \(x\) only.

Solving these equations often involves finding an integrating factor that turns the equation into an exact one. This is done so that each side of the equation can be integrated separately, enabling a clearer path to the solution. In practical settings, including in the exercise provided, identifying and applying an integrating factor may involve some creativity, especially when the functions involved are more complex or when the equation is initially inexact.

Partial Fractions

The method of partial fractions is a powerful algebraic tool used to break down complex rational expressions into simpler ones that are easier to integrate. It is particularly useful when dealing with linear differentials where integrating factors do not readily present themselves.

When given a rational function, the idea is to express it as a sum of simpler fractions whose denominators are the factors of the original denominator. This becomes especially handy in integration: a complex integral becomes a series of integrals of simpler fractions, which can often be addressed with basic integration techniques.

In the context of our exercise, when the search for a simple integrating factor became too cumbersome, the solution involved considering a different method, like partial fractions, to simplify the problem. Partial fractions could potentially separate the equation into more manageable pieces, allowing for integration and, eventually, the solution of the differential equation.