Problem 25 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. $$ \begin{aligned} &\left(y^{2} \cos x-3 x^{2} y-2 x\right) d x+\left(2 y \sin x-x^{3}+\ln y\right) d y=0 \\ &y(0)=e \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( y^2 \sin x - x^3y - x^2 + y \ln y - y = 0 \).

Step 1: Verify the Differential Form

Start by defining the differential form as \( M(x,y) \, dx + N(x,y) \, dy = 0 \), where \( M = y^2 \cos x - 3x^2 y - 2x \) and \( N = 2y \sin x - x^3 + \ln y \). Check the conditions for an exact differential equation by computing the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).

Step 2: Check for Exactness

Calculate \( \frac{\partial M}{\partial y} = 2y \cos x - 3x^2 \) and \( \frac{\partial N}{\partial x} = 2y \cos x - 3x^2 \). Because \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the differential equation is exact. This means there exists a potential function \( \Psi(x, y) \) such that \( \frac{\partial \Psi}{\partial x} = M \) and \( \frac{\partial \Psi}{\partial y} = N \).

Step 3: Integrate with Respect to x

Integrate \( M(x, y) = y^2 \cos x - 3x^2 y - 2x \) with respect to \( x \) to find part of \( \Psi(x, y) \). The integration gives \( \Psi(x, y) = y^2 \sin x - x^3y - x^2 + g(y) \), where \( g(y) \) is an arbitrary function of \( y \).

Step 4: Differentiate with Respect to y

Differentiate the expression for \( \Psi(x, y) = y^2 \sin x - x^3y - x^2 + g(y) \) with respect to \( y \) to find \( \frac{\partial \Psi}{\partial y} = 2y \sin x - x^3 + g'(y) \).

Step 5: Equalize to N and Solve for \( g'(y) \)

Set \( \frac{\partial \Psi}{\partial y} = N(x, y) = 2y \sin x - x^3 + \ln y \), leading to the equality \( 2y \sin x - x^3 + g'(y) = 2y \sin x - x^3 + \ln y \). Solving for \( g'(y) \) gives \( g'(y) = \ln y \).

Step 6: Find g(y) by Integrating g'(y)

Integrate \( g'(y) = \ln y \) with respect to \( y \) to find \( g(y) = y \ln y - y + C \), where \( C \) is a constant.

Step 7: Write the Complete Potential Function \( \Psi(x, y) \)

Substitute \( g(y) \) back into \( \Psi(x, y) \) to get \( \Psi(x, y) = y^2 \sin x - x^3y - x^2 + y \ln y - y + C \).

Step 8: Solve for C using Initial Condition

Use the initial condition \( y(0) = e \) to solve for \( C \). Substitute \( x=0 \) and \( y=e \) into \( \Psi(x, y) = C \) to find \( C = -e(1-\ln e) = 0 \).

Step 9: Write the Final Equation

Thus, the solution to the initial-value problem is \( y^2 \sin x - x^3y - x^2 + y \ln y - y = 0 \).

Key Concepts

Understanding the Initial-Value Problem

An initial-value problem in differential equations involves not just finding a solution but also making sure that this solution satisfies specific conditions at a given point. In our exercise, we have the initial condition \( y(0) = e \). This means when \( x = 0 \), the value of \( y \) should be \( e \).

To solve such a problem, it's essential to harness both the differential equation itself and the initial condition. This ensures the solution we derive is applicable and uniquely determined by the starting information provided. Essentially, initial-value problems tie down the arbitrary constants that can arise during integration, refining the solution to a specific curve through the given initial point. This approach is crucial for exactness in real-world scenarios.

What is a Potential Function?

In the realm of exact differential equations, the potential function plays a crucial role. If you picture an exact differential equation as a gradient, then the potential function is akin to the surface from which this gradient arises.

For equations that can be expressed in the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \), these are exact if there exists a potential function \( \Psi(x, y) \) such that its partial derivatives align with the functions \( M \) and \( N \). Therefore:

• \( \frac{\partial \Psi}{\partial x} = M(x, y) \)
• \( \frac{\partial \Psi}{\partial y} = N(x, y) \)

By identifying such a function, the problem transforms into finding \( \Psi(x, y) = C \) for a constant \( C \). This means that the solution of the differential equation translates into identifying the level curves of the potential function, which simplifies what might otherwise be a complex integration task.

The Role of Integration in Solving Differential Equations

Integration is a central tool in solving differential equations, especially when working with potential functions. During the solution process for an exact differential equation, we first integrate with respect to one of the variables, such as \( x \).

For example, in our problem, the partial derivative \( M(x, y) \) is integrated with respect to \( x \) to unveil part of the potential function \( \Psi(x, y) \). This practice takes advantage of the fact that integrating with respect to one variable treats other variables as constants, simplifying the task.

• The result will express part of the potential function, with an arbitrary function of the other variable, \( g(y) \), commonly included.
• The latter represents integration constants manifested as functions due to multivariable nature.

After finding this partial potential function, we continue by comparing corresponding derivatives to find \( g(y) \). This step initially simplifies and orderly constructs complex mathematical relationships.

The Importance of Partial Derivatives

Partial derivatives allow us to examine the behavior of multivariable functions by focusing separately on each variable's impact. In exact differential equations, these derivatives are used to check whether a differential equation is indeed exact. Simply put, they are required to verify the relationship between the terms in the equation.

For such verification, partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) are taken. If they are equivalent, then the equation is exact. This means it stems from a potential function, confirming integrability.

• Helps identify simplifications that might not be apparent at first glance.
• Enables the construction of potential functions, crucial for finding solutions to exact equations effectively.

Partial derivatives simplify our understanding of how changes in one variable influence the entire system encapsulated by the function. This analysis provides essential insights into the exactness, allowing integration and other derivative-related computations to solve the differential equation effectively.