Problem 35 - Advanced Engineering Mathematics

Question

Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. In Problems, find an explicit solution of the given initial- value problem. Use a graphing utility to plot the graph of each solution. Compare each solution curve in a neighborhood of \((0,1)\). \frac{d y}{d x}=(y-1)^{2}, \quad y(0)=1

Step-by-Step Solution

Answer

The solution is \(y(x) = 1\).

Step 1: Identify the Problem Type

This is an initial-value problem (IVP) for an ordinary differential equation (ODE). We are given \(\frac{dy}{dx} = (y-1)^2\) with the initial condition \(y(0) = 1\).

Step 2: Recognize the Nature of the Equation

The differential equation \(\frac{dy}{dx} = (y-1)^2\) is a first-order, non-linear ordinary differential equation.

Step 3: Solve the Differential Equation

To solve \(\frac{dy}{dx} = (y-1)^2\), separate the variables:\[ \frac{1}{(y-1)^2} dy = dx. \]Integrate both sides:\[ \int \frac{1}{(y-1)^2} dy = \int dx. \]The left side integrates to \(-\frac{1}{y-1}\) (using substitution \(u = y-1\)), and the right side integrates to \(x + C\):\[ -\frac{1}{y-1} = x + C. \]

Step 4: Apply the Initial Condition

To find the constant \(C\), use the initial condition \(y(0) = 1\):\[ -\frac{1}{0} = 0 + C. \] However, the left-hand side has a division by zero; thus, typically, this implies there is a vertical asymptote at \(x = 0\) for the solution passing through \(y(0)=1\). Moreover, since at \(x=0\), \(y=1\), the expression is initially zero, this indicates \(y=1\), for all small values of \(x\).

Step 5: Analyze the Solution's Behavior

Because \(y=1\) satisfies the IVP and no explicit valid solution was derived due to the indeterminate form, this is the solution curve. This indicates that the solution is constant at the neighborhood of \(x = 0\), hence the solution is simply \(y(x) = 1\) close to \(x=0\).

Step 6: Graphical Verification

We graph \(y(x) = 1\) to verify: in the interval near \(x=0\), the graph of the solution is constant, and \(y=1\) corresponds to the given initial condition, showing no slope change near the origin.

Key Concepts

Initial-Value Problem

An initial-value problem (IVP) is a type of problem involving differential equations where you look for a solution that satisfies both the equation itself and an initial condition. In simple terms, it's about finding a function that not only solves the differential equation but also passes through a specific point. For example, in the given problem, the differential equation is \( \frac{dy}{dx} = (y-1)^2 \), and the solution must satisfy the condition \( y(0) = 1 \). This initial condition anchors the solution curve at a specific point on the graph. It acts like a "starting point" for the solution of the differential equation, ensuring the solution is not just any function, but one that starts at a given location on the x-y plane.
Solving such problems involves two main tasks:

• Verify the integration of the differential equation.
• Apply the initial condition to find any constants that appear asymptotically.
  
  

With these in mind, you can achieve the full solution of the differential equation tailored to the specified condition.

Non-Linear Differential Equation

Non-linear differential equations are those where the unknown function, or its derivatives, appear raised to a power or multiplied together. Unlike linear differential equations, they can exhibit very complex behavior and are often more challenging to solve analytically. In our case, \( \frac{dy}{dx} = (y-1)^2 \) is non-linear because of the \((y-1)^2\) term, which introduces a square of the function \(y\).
This squaring affects how the equation behaves and often requires different solving techniques compared to linear equations. Specifically, non-linearity can cause solutions to vary dramatically with small changes in initial conditions or parameters in the equation. This sensitivity is a hallmark of non-linear systems and makes their study particularly interesting, as small variations can lead to vastly different behavior in the solutions.

Non-linear differential equations may not always have straightforward solutions or may even lead to indeterminate forms, as seen in our problem. Thus, they require careful analysis and often numerical or graphing verification to fully understand their complete solution.

First-Order Differential Equation

A first-order differential equation involves the first derivative of the unknown function but no higher derivatives. In simpler terms, it means the highest derivative in the equation is \( \frac{dy}{dx} \). For the given equation, \( \frac{dy}{dx} = (y-1)^2 \), we see it only contains \( \frac{dy}{dx} \), so it is indeed first-order.
Solving first-order equations often involves techniques such as:

• Separation of variables.
• Integration of both sides once the equation is set up.

These steps are straightforward, and they help find the solution that satisfies both the equation itself and any initial conditions. First-order differential equations are the simplest class of differential equations in terms of order, and mastering their solution methods provides a foundation for solving more complex higher-order equations.

Solution Behavior Analysis

Analyzing the behavior of a solution involves understanding how a solution behaves in different parts of its domain, especially around critical points such as the initial condition. For the equation \( \frac{dy}{dx} = (y-1)^2 \) with the initial condition \( y(0)=1 \), applying the initial condition leads to an indeterminate form, suggesting something unique: a constant solution.
The concept is simple yet intriguing; because of the initial condition \( y=1 \) at \( x=0 \), the equation effectively becomes zero. This implies that for small values of \( x \) near zero, the rate of change (slope) is zero, and therefore \( y(x) = 1 \) remains constant near \( x=0 \). This constant behavior indicates a lack of variation in the solution in this region, providing an important insight into the stability of solutions near initial conditions.

Graphically verifying these results by plotting \( y(x) = 1 \) shows that the solution is a horizontal line at \( y=1 \). This demonstrates how the initial condition influences the overall behavior of the solution curve, a critical part of solution behavior analysis in differential equations.