Problem 34 - Advanced Engineering Mathematics

Question

Show that an implicit solution of $$ 2 x \sin ^{2} y d x-\left(x^{2}+10\right) \cos y d y=0 $$ is given by \(\ln \left(x^{2}+10\right) \csc y=c\). Find the constant solutions, if any, that were lost in the solution of the differential equation.

Step-by-Step Solution

Answer

The implicit solution is verified as given. Constant solutions occur when \(y = (\frac{2n+1}{2})\pi\), \(n \in \mathbb{Z}\).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(2x \sin^2 y \, dx - (x^2 + 10) \cos y \, dy = 0\). This equation fits the form of a first-order differential equation where separation of variables or integrating factors might be needed.

Step 2: Check Consistency with the Proposed Solution

The proposed implicit solution is \(\ln(x^2 + 10) \csc y = c\). Differentiate this expression implicitly to verify that it satisfies the original equation.

Step 3: Differentiate Implicitly

Taking the derivative of \(\ln(x^2 + 10) \csc y = c\):\[ \frac{d}{dx}[\ln(x^2 + 10)] \csc y + \ln(x^2 + 10) \frac{d}{dx}[\csc y] = 0. \] Compute the individual derivatives for further simplification.

Step 4: Calculate Derivatives Involved

Using the chain rule and derivative formulas, we have:\( \csc y \cdot \frac{2x}{x^2 + 10} - \ln(x^2 + 10) \csc y \cot y \frac{dy}{dx} = 0. \)This rearranges to match the original equation by multiplying all terms with \(d\):\[ 2x \sin^2 y \, dx - (x^2 + 10) \cos y \, dy = 0. \]

Step 5: Verify the Implicit Solution by Substitution

Substituting the implicit solution back into the differential equation confirms that the differentiated expression equals zero:\[ 2x \sin^2 y - (x^2 + 10) \cos y \frac{dy}{dx} = 0, \]proves the solution satisfies the original differential equation.

Step 6: Determine Constant Solutions

Constant solutions occur when \(dy/dx = 0\). From the original differential equation, setting \(- (x^2 + 10) \cos y = 0\) gives solutions where \( \cos y = 0\) or \( x = 0\). Thus, expressions \(y = (2n+1)\frac{\pi}{2}\), where \(n\) is an integer, provide constant solutions.

Key Concepts

Implicit Solution

An implicit solution in differential equations refers to a solution that links two functions without explicitly defining either one. In the context of the given differential equation, the implicit solution provided is \( \ln(x^2 + 10) \csc y = c \), where \( c \) is a constant. This equation highlights a relationship between \( x \) and \( y \) but does not isolate either variable. Implicit solutions are valuable because:

• They can often simplify complex relationships between variables when explicit solutions are difficult to derive.
• They allow for a broader understanding of how solutions behave over a range of values.

Differentiation techniques, such as the chain rule, can be used to verify that an implicit solution satisfies a differential equation, as seen in this problem.

First-Order Differential Equation

A first-order differential equation involves derivatives of the first degree, meaning it only includes the first derivative of a function. The given equation, \( 2x \sin^2 y \, dx - (x^2 + 10) \cos y \, dy = 0 \), involves only the first derivative \( dy/dx \).Understanding first-order differential equations is critical because:

• They are foundational for more complex differential equations encountered in various fields such as physics and engineering.
• They often model real-world processes like growth rates and the spread of diseases.

To solve them, methods like separation of variables or integrating factors are often employed, depending on the equation's specific structure.

Constant Solutions

Constant solutions in differential equations are solutions where the function does not change with respect to the variable; for example, \( dy/dx = 0 \). In this problem, when \( dy/dx = 0 \), it leads to the equation \( - (x^2 + 10) \cos y = 0 \). From this, one can determine constant solutions as points where either:

• \( \cos y = 0 \), which means \( y = (2n+1)\frac{\pi}{2} \), where \( n \) is an integer.
• \( x = 0 \)

Constant solutions are useful:

• For identifying equilibrium states in physical systems, where variables remain unchanged over time.
• For simplifying complex systems by examining conditions that do not vary.

Separation of Variables

Separation of variables is a method used to solve first-order differential equations. This technique involves rearranging the equation to isolate one variable and its differential on one side and the other variable and its differential on the other. The given differential equation gives us a path for separation: each term correlates identifiably to either \( x \) or \( y \). Steps for separation often include:

• Identifying and grouping terms in the equation that can be associated with one variable.
• Integrating both sides separately to find a solution.
• Using initial conditions or additional information to solve for integration constants.

This method is especially powerful for equations that can naturally divide into manageable segments, enabling straightforward integration and solution finding. It is a fundamental method because it's applicable to a broad range of simple to moderately complex differential equations.