Problem 24 - Advanced Engineering Mathematics

Question

In Problems, find an implicit and an explicit solution of the given initial- value problem. \(\frac{d y}{d x}=\frac{y^{2}-1}{x^{2}-1}, \quad y(2)=2\)

Step-by-Step Solution

Answer

Implicit solution: \( \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right| \), explicit solution: \( y = \frac{x-1}{2} \).

Step 1: Identify the Separable Equation

We begin by recognizing that the given differential equation \( \frac{dy}{dx} = \frac{y^2 - 1}{x^2 - 1} \) is a separable equation. This means we can rewrite it as \( \frac{dy}{y^2 - 1} = \frac{dx}{x^2 - 1} \) to separate the variables \( y \) and \( x \).

Step 2: Integrate Both Sides

Integrate both sides of the separated equation. On the left side, we have \( \int \frac{dy}{y^2 - 1} \), which can be integrated as \( \frac{1}{2} \ln \left| \frac{y-1}{y+1} \right| \). On the right side, we have \( \int \frac{dx}{x^2 - 1} \), which is \( \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| \) plus a constant \( C \).

Step 3: Combine Results

Combine the integrated results to form the implicit solution: \( \frac{1}{2} \ln \left| \frac{y-1}{y+1} \right| = \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C \). This can be rewritten to eliminate the logarithms as: \( \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right| e^{2C} \).

Step 4: Solve for the Constant Using Initial Condition

Use the initial condition \( y(2) = 2 \). Substitute \( x = 2 \) and \( y = 2 \) into the implicit equation: \( \left| \frac{2-1}{2+1} \right| = \left| \frac{2-1}{2+1} \right| e^{2C} \), leading to \( \frac{1}{3} = \frac{1}{3} e^{2C} \). Thus, \( e^{2C} = 1 \), so \( C = 0 \).

Step 5: Derive the Explicit Solution

With \( e^{2C} = 1 \), the equation simplifies to \( \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right| \). For the explicit solution, solve for \( y \): \( \frac{y-1}{y+1} = \frac{x-1}{x+1} \) (for simplicity, assume positive solutions initially), then cross-multiply to get \( (y-1)(x+1) = (y+1)(x-1) \).

Step 6: Simplify and Extract the Explicit Solution

Expanding both sides, we get: \( yx + y = yx - y + x - 1 \). Simplifying, this yields \( 2y = x - 1 \), thus \( y = \frac{x-1}{2} \). This is the explicit solution.

Key Concepts

Separable Equations

Separable equations are a type of differential equation where you can separate the variables on opposite sides of the equation. This makes them easier to integrate and solve. When dealing with the exercise, the differential equation you're given is \( \frac{dy}{dx} = \frac{y^2 - 1}{x^2 - 1} \). This is a separable equation because you can rearrange it as \( \frac{dy}{y^2 - 1} = \frac{dx}{x^2 - 1} \), effectively separating \( y \) terms from \( x \) terms.

• By separating variables, you can integrate both sides independently.
• The goal is to find a solution that relates \( y \) and \( x \).

Once you’ve separated the terms, the next step is integration, setting the stage for finding both implicit and explicit solutions.

Initial Value Problem

An Initial Value Problem (IVP) involves finding a function that satisfies a differential equation and meets an initial condition. In our exercise, alongside the differential equation, you're provided with the condition \( y(2) = 2 \). This initial condition specifies the value of the function \( y \) when \( x = 2 \).

• IVPs are crucial in identifying a unique solution among many potential solutions to a differential equation.
• The initial condition helps determine the constant of integration.

In this problem, the initial value is used to find the constant \( C \), ensuring the solutions reflect this particular condition, leading to the specific implicit and explicit solutions found later.

Implicit Solution

An implicit solution of a differential equation is an equation where the relationship between \( y \) and \( x \) is expressed indirectly. Unlike explicit solutions, where \( y \) is isolated, implicit solutions might keep \( y \) intertwined with an equation.
For instance, after integrating the separated terms from the exercise, you achieve:\[\frac{1}{2} \ln \left| \frac{y-1}{y+1} \right| = \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C\]This form is implicit because \( y \) is not isolated. Implicit solutions can often be messy or complex, which makes finding an explicit solution preferable in practical applications. However, implicit forms are crucial in deriving explicit solutions and in situations where isolation is difficult.

Explicit Solution

An explicit solution takes an implicit relationship and solves for \( y \) in terms of \( x \) directly. In the exercise, after using the initial value condition to determine \( C = 0 \), the implicit solution simplifies. Following further algebraic manipulation, we solve:\[(y-1)(x+1) = (y+1)(x-1)\]Expanding both sides makes it possible to isolate \( y \), ultimately leading to the explicit form\[y = \frac{x-1}{2}\]This gives a clear, direct relationship making it easy to evaluate \( y \) for any given \( x \). Explicit solutions are often preferred for their straightforwardness and clarity, especially when applying solutions to practical problems or further analysis.