Problem 22 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. $$ \left(e^{x}+y\right) d x+\left(2+x+y e^{y}\right) d y=0, \quad y(0)=1 $$

Step-by-Step Solution

Answer

The solution to the initial-value problem is \( e^x + xy + 2y + e^y = 3 + e \).

Step 1: Identify the Form of the Equation

The given differential equation is \( (e^{x} + y) \, dx + (2 + x + y e^y) \, dy = 0 \). This is of the form \( P(x, y) \, dx + Q(x, y) \, dy = 0 \), with \( P(x, y) = e^x + y \) and \( Q(x, y) = 2 + x + y e^y \).

Step 2: Check for Exactness

To check if the equation is exact, compute the partial derivative \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \). We have \( \frac{\partial P}{\partial y} = 1 \) and \( \frac{\partial Q}{\partial x} = 1 \). Since these are equal, the differential equation is exact.

Step 3: Find the Potential Function

To find the potential function \( \Psi(x, y) \), integrate \( P(x, y) = e^x + y \) with respect to \( x \). This gives \( \Psi(x, y) = e^x + xy + C(y) \), where \( C(y) \) is an arbitrary function of \( y \).

Step 4: Determine \( C(y) \) from Partial Derivative

To find \( C(y) \), differentiate \( \Psi(x, y) \) with respect to \( y \) and equate to \( Q(x, y) \). This gives \( \frac{\partial \Psi}{\partial y} = x + C'(y) = 2 + x + y e^y \). Solving for \( C'(y) = 2 + y e^y \), integrate \( C'(y) \) to find \( C(y) = 2y + e^y + C_1 \).

Step 5: Write the General Solution

Substitute \( C(y) \) back into \( \Psi(x, y) \) to find \( \Psi(x, y) = e^x + xy + 2y + e^y \). Thus, the general solution is \( e^x + xy + 2y + e^y = C_2 \).

Step 6: Apply Initial Condition

Use the initial condition \( y(0) = 1 \) to find the constant \( C_2 \). Substitute \( x = 0 \) and \( y = 1 \) into \( e^x + xy + 2y + e^y = C_2 \). This gives \( e^0 + 0 \cdot 1 + 2 \cdot 1 + e^1 = C_2 \), or \( 1 + 2 + e = C_2 \). Thus, \( C_2 = 3 + e \).

Step 7: Write the Particular Solution

The particular solution that satisfies the initial value problem is \( e^x + xy + 2y + e^y = 3 + e \).

Key Concepts

Initial-Value Problems

Initial-value problems are a specific type of problem in differential equations where you are not only tasked with solving the differential equation but also finding the particular solution that passes through a given point, called the initial condition.
The initial condition is provided as a baseline to specify the exact path or solution out of many potential solutions.
In the exercise given, the initial condition was provided as \( y(0) = 1 \). This forms the basis to find the particular solution from the general solution of the differential equation.
To apply the initial condition:

• First, solve the differential equation to find the general solution.
• Then, substitute the coordinates from the initial condition into the general solution.
• This will allow you to find any unknown constants and finalize the specific solution for the given initial conditions.

By using the initial-value information, we find a more manageable, specific solution that uniquely fits our problem criteria.

Potential Function

In the context of exact differential equations, the potential function is a critical concept.
This is because an exact differential equation can be expressed as the gradient of a potential function \( \Psi(x, y) \), akin to a potential energy landscape in physics.
To find the potential function, you usually follow these steps:

• Integrate \( P(x, y) \) with respect to \( x \) to get a partial expression of \( \Psi(x, y) \).
• Introduce a function \( C(y) \) as a placeholder for parts that depend only on \( y \).
• Use the derivative of the potential function with respect to \( y \) and equate it to \( Q(x, y) \) to solve for \( C(y) \).

The resulting \( \Psi(x, y) \) will represent a surface whose curvature and changes correspond to the behavior described by the differential equation.

Partial Derivatives

Partial derivatives are key in dealing with multivariable differential equations and are used to check the exactness of a differential equation and to find potential functions.
A partial derivative represents the rate of change of a function with respect to one variable, keeping all other variables constant.
Here's how partial derivatives play a role in the process:

• Calculate \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) to determine the exactness of the differential equation.
• If those two derivatives are equal, the equation is considered \'exact\'.
• Partial derivatives are fundamental when finding \( C(y) \) to complete the potential function.

In this exact context, using partial derivatives helps convert a system of equations into solvable forms, and aids in tracing the behavior of solutions in complex multivariable terrains.