Question

Find an integrating factor; that is a function of only one variable, and solve the given equation. $$ y \sin y d x+x(\sin y-y \cos y) d y=0 $$

Short answer

Question: Determine the general solution of the given differential equation: $$y\sin y\,dx + x(\sin y - y\cos y)\,dy = 0$$ Answer: The general solution of the given differential equation can be expressed as: $$\int x\left(\frac{\sin y}{y} - \cos y\right)\,dy = C - x\sin y$$

Step-by-step solution

Identify the given differential equation

The given differential equation is: $$ y\sin y\,dx + x(\sin y - y\cos y)\,dy = 0 $$

Rewrite the equation in the form M dx + N dy = 0

Divide both sides by x*y to separate the variables: $$ \frac{\sin y}{x}\,dx + \left(\frac{\sin y}{y} - \cos y\right)\,dy = 0 $$ Now we have the equation in the form M dx + N dy = 0, where: $$ M = \frac{\sin y}{x},\qquad N = \frac{\sin y}{y} - \cos y $$

Find an integrating factor only dependent on x

We're looking for an integrating factor μ(x) that will make the given equation an exact differential equation. Exactness condition: $$ \frac{dM}{dy} - \frac{dN}{dx} = \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} $$ Calculate the partial derivatives \(\frac{\partial M}{\partial y}\), and \(\frac{\partial N}{\partial x}\): $$ \frac{\partial M}{\partial y} = \frac{\cos y}{x}, \qquad \frac{\partial N}{\partial x} = 0 $$ Now, find the integrating factor: $$ \frac{dM}{dy} - \frac{dN}{dx} = \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = \frac{\cos y}{x} $$ From this equation, we can determine that the integrating factor μ(x) is of the form: $$ \mu(x) = x \\ $$

Multiply the equation by the integrating factor and integrate

Multiply both sides of the equation by μ(x)=x: $$ \sin y\,dx + x\left(\frac{\sin y}{y} - \cos y\right)\,dy = 0 $$ Now, integrate with respect to x: $$ \int \sin y\,dx + \int x\left(\frac{\sin y}{y} - \cos y\right)\,dy = C $$ $$ x\sin y + \int x\left(\frac{\sin y}{y} - \cos y\right)\,dy = C \\ $$

Solve for the general solution

At this point, we have found an integrating factor and integrated the equation. To find the general solution, we just need to solve for y: $$ x\sin y + \int x\left(\frac{\sin y}{y} - \cos y\right)\,dy = C $$ While it is difficult to find a closed-form solution for y, the general solution can be expressed as: $$ \int x\left(\frac{\sin y}{y} - \cos y\right)\,dy = C - x\sin y $$ This is the general solution of the given differential equation.

Key concepts

Exact Differential Equations

In the realm of calculus, an 'exact differential equation' is characterized by a particular harmony between two functions, known as M(x, y) and N(x, y), which are part of the differential equation M dx + N dy = 0. The crux of an exact differential equation is the condition that the cross partial derivatives of a potential function (often denoted as \( \Psi (x, y) \) ) must be equal. This means that for the equation to be exact, the partial derivative of M with respect to y must match the partial derivative of N with respect to x, represented mathematically as \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

If this condition does not hold naturally, we sometimes can employ an 'integrating factor'—a function dependent on x or y (but not both)—which, when multiplied with the differential equation, renders it exact. By applying the integrating factor, we're essentially manipulating the equation to fulfill the exactness condition, thereby making it solvable via integration. In this given exercise, identifying a suitable integrating factor allowed us to transform the differential equation into an exact one, enabling us to find its general solution.

Partial Derivatives

Partial derivatives are a foundational concept in the study of multivariable calculus, and they are critical when dealing with differential equations. Specific to their role in exact differential equations, partial derivatives measure how a function changes as one variable moves infinitesimally, while keeping all other variables constant. The notation \( \frac{\partial}{\partial x} \) represents the partial derivative with respect to x.

When we calculate the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) in the context of finding an integrating factor for a differential equation, we're essentially looking for hints about the underlying function that both M and N derive from. In our exercise, analyzing partial derivatives was crucial to identify that the equation was not initially exact, and that an integrating factor was needed to bridge the gap.

General Solution of Differential Equation

The 'general solution' of a differential equation represents a comprehensive set of all possible solutions to the equation. It includes a constant of integration, typically denoted as C, which accounts for the fact that there are infinitely many solutions, each corresponding to a different initial condition. To achieve the general solution, we integrate the differential equation, usually after making it exact, which is what happened in our exercise.

Once we obtain the general solution, it can be manipulated further to solve specific problems, often by applying initial conditions that will allow us to solve for the constant C. In complicated scenarios, such as in our exercise, the solution might not be expressible in terms of elementary functions, and it may be left in an implicit or integral form. Nonetheless, this expression still encompasses all the possible solutions to the differential equation.