Problem 37 - 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 \((0,1)\). $$ \frac{d y}{d x}=(y-1)^{2}+0.01, \quad y(0)=1 $$

Step-by-Step Solution

Answer

The explicit solution is \( y(x) = 1 + \sqrt{0.01} \tan(\sqrt{0.01}x) \).

Step 1: Understand the Differential Equation

We are given the first-order ordinary differential equation \( \frac{dy}{dx} = (y - 1)^2 + 0.01 \) with the initial condition \( y(0) = 1 \). Our task is to find an explicit solution that satisfies this initial condition.

Step 2: Analyze the Behavior at the Initial Condition

At \( x = 0 \) and \( y = 1 \), the right-hand side of the differential equation becomes \( (1 - 1)^2 + 0.01 = 0.01 \). This indicates that near \( (0, 1) \), the change in \( y \) with respect to \( x \) is positive and small.

Step 3: Solve the Differential Equation

To solve \( \frac{dy}{dx} = (y-1)^2 + 0.01 \), separate variables: \( \frac{1}{(y-1)^2 + 0.01} dy = dx \). Integrating both sides with respect to their respective variables will yield an expression involving \( y \) and \( x \).

Step 4: Integration Process

Recognize that the integral \( \int \frac{1}{(y-1)^2 + 0.01} \, dy \) is in the form of a standard integral similar to the arctangent. Use the substitution \( u = y - 1 \) and integrate: \( \int \frac{1}{u^2 + 0.01} \, du = \frac{1}{\sqrt{0.01}} \tan^{-1}\left( \frac{u}{\sqrt{0.01}} \right) + C \).

Step 5: Solve for y(x) Explicitly

Solving \( \tan^{-1}\left( \frac{y-1}{\sqrt{0.01}} \right) = \sqrt{0.01}x + C \) for \( y \) gives \( y - 1 = \sqrt{0.01} \tan(\sqrt{0.01}x + C) \). Therefore, \( y = 1 + \sqrt{0.01} \tan(\sqrt{0.01}x + C) \).

Step 6: Determine the Constant C

Apply the initial condition \( y(0) = 1 \) to find \( C \). Plug in \( x = 0 \) and \( y = 1 \) into the equation: \( 1 = 1 + \sqrt{0.01} \tan(C) \). Thus, \( C = 0 \).

Step 7: Final Explicit Solution

With \( C = 0 \), the solution becomes \( y(x) = 1 + \sqrt{0.01} \tan(\sqrt{0.01}x) \).

Step 8: Graphing the Solution Around (0,1)

Use a graphing utility to plot the function \( y(x) = 1 + \sqrt{0.01} \tan(\sqrt{0.01}x) \) near \((0,1)\). Note how the solution curve behaves in this neighborhood, marking the small effect of the constant term on the overall dynamics.

Key Concepts

Initial-Value Problem

An initial-value problem involves a differential equation and a specific initial condition that the solution must satisfy. In this context, we identify our differential equation as \( \frac{dy}{dx} = (y - 1)^2 + 0.01 \) with the initial condition \( y(0) = 1 \). This initial condition allows us to determine a particular solution among the family of solutions generated by a differential equation.
It's crucial because it pinpoints the exact trajectory of the solution curve. Without this, the problem would have multiple solutions, each differing by a constant. Setting \( y(0) = 1 \) ensures that the initial point on our solution curve is clearly defined, anchoring the entire solution plot.

Ordinary Differential Equation

An ordinary differential equation (ODE) contains one or more functions of one independent variable and its derivatives. In our problem, we encounter a first-order ODE, indicating that the highest derivative involved is the first derivative \( \frac{dy}{dx} \).
Oftentimes, first-order ODEs describe a wide variety of processes such as exponential growth, cooling, and population dynamics. Solving an ODE typically involves finding an explicit expression that describes how the dependent variable changes with respect to the independent variable under given conditions.

Separation of Variables

"Separation of variables" is a technique for solving differential equations. Here, we rearrange the equation so that each variable and its derivative are on separate sides of the equation. In our case, we've reorganized \( \frac{dy}{dx} = (y - 1)^2 + 0.01 \) into \( \frac{1}{(y-1)^2 + 0.01} dy = dx \), allowing us to integrate both sides separately.
This method is particularly useful for equations where the rate of change can be expressed as a product of functions, one involving only the dependent variable and the other involving only the independent variable. Successful use of this method results in an integrable equation that leads to an explicit solution.

Integration

Integration is a mathematical process used to find a function given its derivative. It is the reverse operation of differentiation. In our differential equation, after separating variables, we faced the integral \( \int \frac{1}{(y-1)^2 + 0.01} \, dy \).
Recognizing as a form akin to the arctangent function, we substitute and integrate corresponding to this format. The integral led us to an expression involving the arctangent function, which played a key role in targeting the explicit solution of \( y(x) \). Understanding different types of integrals and their corresponding antiderivatives is fundamental in solving ODEs.

Arctangent Function

The arctangent function, denoted as \( \tan^{-1} \), is the inverse of the tangent function. It emerges in solving integrals that involve terms in the form of \( \frac{1}{a^2 + t^2} \). In this problem, it appears when integrating the separated equation \( \int \frac{1}{(y-1)^2 + 0.01} \, dy \).
The substitution \( u = y - 1 \) transforms the integral into a familiar arctangent form, yielding \( \frac{1}{\sqrt{0.01}} \tan^{-1}\left( \frac{u}{\sqrt{0.01}} \right) + C \). The application of the arctangent function is a pivotal step that brings us closer to finding \( y \) in terms of \( x \) comprehensively.