Problem 4 - Advanced Engineering Mathematics

Question

Let \(M=\sin y-y \sin x\) and \(N=\cos x+x \cos y-y\) so that \(M_{y}=\cos y-\sin x=N_{x} .\) From \(f_{x}=\sin y-y \sin x\) we obtain \(f=x \sin y+y \cos x+h(y), h^{\prime}(y)=-y,\) and \(h(y)=-\frac{1}{2} y^{2} .\) A solution is \(x \sin y+y \cos x-\frac{1}{2} y^{2}=c\)

Step-by-Step Solution

Answer

The solution is \( x \sin y+y \cos x - \frac{1}{2}y^2 = c \).

Step 1: Understand the Functions

Given two functions, \( M = \sin y - y \sin x \) and \( N = \cos x + x \cos y - y \). The goal is to verify their relationship and use them to find the potential function \( f \).

Step 2: Check Condition for Exact Differential Equation

We are given that \( M_y = \cos y - \sin x \) and \( N_x = \cos y - \sin x \), implying that \( M_y = N_x \). This confirms the equation is exact, and a potential function \( f(x, y) \) exists such that \( f_x = M \) and \( f_y = N \).

Step 3: Integrate to Find Potential Function

Start by integrating \( f_x = M = \sin y - y \sin x \) with respect to \( x \). The result is \( f(x, y) = x \sin y + y \cos x + h(y) \), where \( h(y) \) is an arbitrary function of \( y \).

Step 4: Differentiate with Respect to y

To find \( h(y) \), differentiate \( f(x, y) = x \sin y + y \cos x + h(y) \) with respect to \( y \) to get \( f_y = x \cos y + \cos x + h'(y) \).

Step 5: Match with Given N

Given \( f_y = N = \cos x + x \cos y - y \), equate this to the result from Step 4: \( x \cos y + \cos x + h'(y) = \cos x + x \cos y - y \).

Step 6: Solve for h(y)

By comparing, \( h'(y) = -y \). Integrate to find \( h(y) = -\frac{1}{2}y^2 + C \). Since constant \( C \) is usually absorbed in the arbitrary constant for the solution set, we set \( C=0 \).

Step 7: Write the General Solution

The solution to the differential equation is \( f(x, y) = x \sin y + y \cos x - \frac{1}{2}y^2 = c \), where \( c \) is an arbitrary constant.

Key Concepts

Potential Function

When dealing with exact differential equations, a potential function is a pivotal component. This function, often denoted as \( f(x, y) \), serves as an intermediary that consolidates the components of a differential equation. For our problem, the relationship between \( M = \sin y - y \sin x \) and \( N = \cos x + x \cos y - y \) is crucial. They must satisfy the equality \( M_y = N_x \) to confirm that a potential function exists. This equality, \( \cos y - \sin x = \cos y - \sin x \), proves that our equation is indeed exact. Consequently, the potential function \( f(x, y) \) can be constructed to satisfy \( f_x = M \) and \( f_y = N \). In this scenario, we have:

• \( f_x = \sin y - y \sin x \)
• \( f_y = \cos x + x \cos y - y \)

Finding the potential function sets up the pathway to solving the entire differential equation.

Integration

Integration plays a key role in uncovering the potential function from its partial derivatives. For this problem, we integrate the expression \( f_x = M = \sin y - y \sin x \) with respect to \( x \). This process results in the function \( f(x, y) = x \sin y + y \cos x + h(y) \). Here, \( h(y) \) is an arbitrary function that accounts for any terms solely dependent on \( y \), which ensures that the integration is complete.

• Integrating \( \sin y - y \sin x \) with respect to \( x \) produces \( x \sin y + y \cos x \).
• \( h(y) \) is an added function of \( y \) that is determined later by matching the original function \( N \).

Through integration, we start constructing the potential function piece by piece, getting closer to the full solution.

Differential Equation Verification

Verifying a differential equation involves ensuring the potential function accurately reflects both original functions \( M \) and \( N \). For verification, we differentiate our integrated potential function \( f(x, y) = x \sin y + y \cos x + h(y) \) with respect to \( y \). Achieving \( f_y = x \cos y + \cos x + h'(y) \), we compare this with \( N = \cos x + x \cos y - y \). The matching process reveals:

• The terms \( x \cos y \) and \( \cos x \) match in both expressions.
• This lets us conclude \( h'(y) = -y \), thus ensuring the functions are equivalent upon differentiation.

Differentiation and verification are critical; they ensure our potential function maintains coherence with the initial differential equation provided.

General Solution

The journey through potential functions and verification culminates in the general solution of the exact differential equation. In our exercise, solving for \( h(y) \) results in \( h(y) = -\frac{1}{2}y^2 + C \), where \( C \) serves as an arbitrary constant usually included in the constant of integration. Wrapping up these computations and integrations, we reach the elegant general solution: \( f(x, y) = x \sin y + y \cos x - \frac{1}{2}y^2 = c \). This final equation represents the family of solutions determined by any variation of \( c \). The general solution efficiently encompasses all potential behaviors of the system described by the differential equation. By understanding each step leading to it, students grasp how differential problems transform into complete mathematical representations, encapsulating infinitely many solutions.