Question

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

Short answer

The solution to the given differential equation is: $$ -y \cos x \cos y -\frac{y^4}{4} = C $$ Here, C is an arbitrary constant.

Step-by-step solution

Find the integrating factor

We are given the following differential equation: $$ \cos x \cos y dx+(\sin x \cos y-\sin x \sin y+y) dy=0 $$ Let M and N be the coefficients of dx and dy respectively: $$ M = \cos x \cos y \\ N = \sin x \cos y - \sin x \sin y + y $$ Now, let's find the partial derivative of M with respect to y and the partial derivative of N with respect to x: $$ \frac{\partial M}{\partial y} = -\cos x \sin y \\ \frac{\partial N}{\partial x} = \cos x \cos y - \cos x \sin y $$ To find an integrating factor depending on a single variable, let's look for a function where \(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=M\mu'_y\). Here, \(\mu\) is the integrating factor and \(\mu'_y\) is its derivative. $$ -\cos x \sin y - (\cos x \cos y - \cos x \sin y) = \cos x \cos y \mu'_y $$ This simplifies to: $$ \mu'_y =-1 $$ Integrate this with respect to y to obtain the integrating factor \(\mu\): $$ \mu(y) = -y + C $$ Here, C is a constant. Since the integrating factor only depends on one variable, we can leave out the constant by taking \(\mu(y) = -y\) as our integrating factor.

Multiply the given equation by the integrating factor

Now, we multiply the given differential equation by the integrating factor \(\mu(y) = -y\): $$ -y \cos x \cos y dx -y(\sin x \cos y-\sin x \sin y+y) dy = 0 $$ Simplify the equation: $$ -y \cos x \cos y dx -y^2 \sin x \cos y + y^2 \sin x \sin y - y^3 dy = 0 $$

Check if it becomes an exact differential equation

Now, assign new M and N, and check if the new equation is exact: $$ M = -y \cos x \cos y \\ N = -y^2 \sin x \cos y + y^2 \sin x \sin y - y^3 $$ Check for exactness: $$ \frac{\partial M}{\partial y} =-\cos x\cos y+y \cos x \sin y \\ \frac{\partial N}{\partial x} =-\cos x\cos y+y \cos x \sin y $$ Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the new equation is exact.

Solve the exact differential equation

We have the exact differential equation: $$ -y \cos x \cos y dx -y^2 \sin x \cos y + y^2 \sin x \sin y - y^3 dy = 0 $$ Now, we can integrate the equation on both sides. Integrate M with respect to x and N with respect to y: $$ F(x, y) = \int M dx = -y \cos x \cos y \\ F(x, y) = \int N dy = -\frac{y^3}{3} \sin x \cos y + \frac{y^3}{3} \sin x \sin y - \frac{y^4}{4} $$ Take the intersection of both equations: $$ F(x, y) = -y \cos x \cos y - \frac{y^4}{4} + C $$ So, the function F(x,y) equals a constant. This is the solution to the given equation: $$ -y \cos x \cos y -\frac{y^4}{4} = C $$

Key concepts

Exact Differential Equation

An exact differential equation is a specific type of differential equation where the total differential of a function equals zero. The idea is that there exists some function \( F(x, y) \) such that the given differential equation can be expressed as the exact differential \( dF = 0 \). This often requires that the partial derivatives of certain components, when checked, satisfy exactness conditions.

• If a differential equation \( M(x, y)dx + N(x, y)dy = 0 \) is exact, it means there is a function \( F(x, y) \) where \( dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy \).
• In such cases, the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) must hold.

In the solution provided, we used an integrating factor to modify the original equation, so it becomes exact. This allows us to then integrate and find the function \( F \) that describes the integral curve of the differential equation. Finding such a curve means solving the equation.

Partial Derivatives

Partial derivatives play a crucial role in working with functions of several variables, especially when dealing with differential equations like we have here. A partial derivative of a function involves differentiating with respect to one variable while keeping the others constant.

• Given a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \) and with respect to \( y \) as \( \frac{\partial f}{\partial y} \).
• These derivatives are foundational for examining changes in multivariable functions and play a key role in determining exactness.
• In our exercise, the derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) were calculated to check precisely the condition of exactness for differential equations.

Understanding partial derivatives is vital for solving problems involving systems of equations and physical models represented by such calculus.

Differential Equations

Differential equations form the basis for describing various physical phenomena mathematically. They involve derivatives of functions and can be ordinary differential equations (ODEs) or partial differential equations (PDEs) depending on the situation.

• Differential equations specify how a particular quantity changes in relation to others. These equations can model a wide range of processes in engineering, physics, economics, and biology.
• The objective is often to find a function or set of functions that satisfy the given differential equation condition.
• For example, a differential equation like \( \cos x \cos y dx + (\sin x \cos y - \sin x \sin y + y)dy=0 \) expresses a relationship between changes in \( x \) and \( y \).

In this exercise, we solved such an equation by first converting it to an exact form using an integrating factor and then determining the solution function. This highlights how differential equations can be tackled using systematic mathematical approaches to uncover underlying functional relationships.