Problem 34 - Advanced Engineering Mathematics

Question

We note that \(\left(M_{y}-N_{x}\right) / N=-\cot x,\) so an integrating factor is \(e^{-} \int \cot x d x=\csc x .\) Let \(M=\cos x \csc x=\cot x\) and \(N=(1+2 / y) \sin x \csc x=1+2 / y,\) so that \(M_{y}=0=N_{x} .\) From \(f_{x}=\cot x\) we obtain \(f=\ln (\sin x)+h(y)\) \(h^{\prime}(y)=1+2 / y,\) and \(h(y)=y+\ln y^{2} .\) A solution of the differential equation is \(\ln (\sin x)+y+\ln y^{2}=c\)

Step-by-Step Solution

Answer

The solution is \( \ln(\sin x) + y + \ln(y^2) = c \).

Step 1: Identify Integrating Factor

Given that \( \frac{M_y - N_x}{N} = -\cot x \), the integrating factor is given by \( e^{-\int \cot x \, dx} = \csc x \). This integrating factor will be used to make the differential equation exact.

Step 2: Multiply Equation by Integrating Factor

The original differential expression is \( M \, dx + N \, dy = 0 \) with \( M = \cos x \, \csc x = \cot x \) and \( N = (1 + \frac{2}{y}) \sin x \, \csc x = 1 + \frac{2}{y} \). Multiply both terms by the integrating factor, \( \csc x \), to make the equation exact.

Step 3: Calculate Partial Derivatives

Calculate \( M_y = 0 \) and \( N_x = 0 \). Since the equation is exact (i.e., \( M_y = N_x \)), the integration process can proceed to find the potential function \( f(x, y) \).

Step 4: Integrate in Terms of x

Integrate \( f_x = \cot x \) with respect to \( x \). This results in \( f = \ln(\sin x) + h(y) \), where \( h(y) \) is an arbitrary function of \( y \).

Step 5: Determine h(y) by Integrating in Terms of y

Knowing \( h'(y) = 1 + \frac{2}{y} \), integrate with respect to \( y \) to find \( h(y) = y + \ln(y^2) \).

Step 6: Formulate the Solution

Substitute the expression for \( h(y) \) into \( f \), resulting in \( f(x, y) = \ln(\sin x) + y + \ln(y^2) \). The implicit solution to the differential equation is \( \ln(\sin x) + y + \ln(y^2) = c \), where \( c \) is a constant.

Key Concepts

Integrating Factor

An integrating factor is a function used to transform a non-exact differential equation into an exact one. This simplification allows us to solve the differential equation more easily. In our example, we are given that \( \left( M_{y} - N_{x} \right) / N = -\cot x \). From this, we find the integrating factor through the formula \( e^{ -\int \cot x \, dx} \). This integration results in the integrating factor \( \csc x \).

Once we multiply the original differential expression by this integrating factor, the equation becomes exact. This means the partial derivatives of the resulting expressions with respect to different variables are equal, setting the stage for solving the equation.

Partial Derivatives

Partial derivatives allow us to understand how a multi-variable function changes with respect to one variable while keeping the other constant. In the context of differential equations, we use them to verify the exactness of the modified equation.

For our problem, we have \( M = \cos x \, \csc x = \cot x \) and \( N = (1 + \frac{2}{y}) \sin x \, \csc x = 1 + \frac{2}{y} \). We compute \( M_y = 0 \) and \( N_x = 0 \). The equality of these partial derivatives (i.e., \( M_y = N_x \)) confirms that the equation is exact. This is a critical step enabling us to proceed with integration.

Implicit Solution

An implicit solution represents a relationship among variables without explicitly solving one variable in terms of others. In exact differential equations, the implicit solution is often preferred due to its comprehensiveness.

Our task involves integrating \( f_x = \cot x \) which gives us \( f = \ln(\sin x) + h(y) \), where \( h(y) \) is a function of \( y \) alone. Further solving for \( h(y) \), we find \( h'(y) = 1 + \frac{2}{y} \) and integrate this to find \( h(y) = y + \ln(y^2) \).

Substituting these back gives us the implicit solution: \( \ln(\sin x) + y + \ln(y^2) = c \), where \( c \) is a constant. This captures the relationship between \( x \) and \( y \) neatly.

Differential Equation Solution

Solving a differential equation means finding a function or set of functions that satisfy the equation. For exact differential equations, leveraging an integrating factor simplifies finding solutions.

Using the integrating factor \( \csc x \), our differential equation becomes exact. This means the partial derivatives of our expressions are equal, letting us find a potential function \( f(x, y) \). By solving the equations for \( f_x \) and determining \( h(y) \), we arrive at the solution

• \( f(x, y) = \ln(\sin x) + y + \ln(y^2) \)
• Resulting implicit equation: \( \ln(\sin x) + y + \ln(y^2) = c \)

Here, \( c \) is the constant representing the family of solutions fitting this differential equation's criteria.