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Chapter 2: Problem 36

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 \((0,1)\). $$ \frac{d y}{d x}=(y-1)^{2}, \quad y(0)=1.01 $$

Short Answer

Expert verified
The explicit solution is \(y = 1 - \frac{1}{x - 100}\).

Step by step solution

01

Understand the Differential Equation

The given differential equation is \( \frac{d y}{d x}=(y-1)^{2} \) with the initial condition \( y(0) = 1.01 \). This is a first-order ordinary differential equation (ODE) with a separable variable type.
02

Separate Variables

To solve the differential equation, separate the variables by dividing both sides by \((y-1)^2\): \[ \frac{d y}{(y-1)^{2}} = d x \].
03

Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{d y}{(y-1)^{2}} = \int d x \]. The left side integrates to \(-\frac{1}{y-1}\) and the right side integrates to \(x + C\), where \(C\) is the constant of integration.
04

Solve for y

Equating the integrated result gives us: \(-\frac{1}{y-1} = x + C\).Rearrange this equation to find \(y\): \[ y = 1 - \frac{1}{x + C} \].
05

Apply the Initial Condition

Substitute the initial condition \(y(0) = 1.01\) into the equation: \[ 1.01 = 1 - \frac{1}{0 + C} \].Solving this equation, we find \( C = -100 \).
06

Write the Explicit Solution

Substituting \(C = -100\) back into the equation for \(y\), we get the explicit solution: \[ y = 1 - \frac{1}{x - 100} \].
07

Graph the Solution Near (0,1)

Use a graphing utility to plot the curve \( y = 1 - \frac{1}{x - 100} \). Observe that near the point \((0, 1)\), the graph approaches the horizontal asymptote at \( y = 1 \), indicating a small deviation from the initial condition sensitivity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial-Value Problem
An initial-value problem in differential equations consists of finding a function that satisfies a differential equation and meets a specific condition called the initial condition. The initial condition is typically given as the value of the unknown function at a particular point. For our problem, the initial condition is \( y(0) = 1.01 \), which means that when \( x = 0 \), the value of \( y \) must be 1.01.
  • The initial condition helps us pin down the exact solution among the infinite number of possible solutions to the differential equation.
  • It determines the constant of integration that arises when solving the differential equation.
  • By using the initial condition, we ensure that the solution curve passes through the specific point \((0, 1.01)\).
Separable Variables
The method of separating variables is a common technique used for solving first-order ordinary differential equations. It involves rearranging the equation so that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the other side.
In our example, the differential equation \( \frac{d y}{d x} = (y-1)^2 \) can be rearranged by dividing both sides by \((y-1)^2\) to achieve separation: \( \frac{d y}{(y-1)^2} = d x \).
  • This separation allows for direct integration of each side with respect to its respective variable.
  • Once variables are separated, we can integrate each side of the equation, leading to the implicit form of the solution.
  • This method is particularly useful because it provides a pathway to integrate and find a general solution, which can later be turned into a specific solution by applying the initial condition.
Integration
Integration is the key step in solving differential equations once the variables have been separated. By integrating, we move from a differential equation, which involves rates of change, to a regular equation, which describes quantities.
For our separated equation \(\frac{d y}{(y-1)^2} = d x\), the integration process involves:
  • Integrating \(\int \frac{d y}{(y-1)^2}\) results in \(-\frac{1}{y-1}\), using the power rule and substitution.
  • Integrating \(\int d x\) results in \(x + C\), where \(C\) is the constant of integration often needed for initial conditions.
The integration provides us with the relationship \(-\frac{1}{y-1} = x + C\). Solving this typed of equation helps transition us towards finding an explicit solution.
First-Order ODE
A first-order ordinary differential equation (ODE) involves derivatives of a function with respect to one independent variable and is of the first order. This means the highest derivative present is the first derivative. In our example, \(\frac{d y}{d x} = (y-1)^2\), the equation involves the first derivative of \(y\) with respect to \(x\).
  • First-order ODEs are prevalent in mathematical modeling of real-world scenarios where the rate of change of a quantity is of interest.
  • They form the foundational basis for understanding more complex differential equations.
  • Simpler techniques such as separation of variables are generally applicable to solve first-order ODEs, as demonstrated in our problem.
Understanding first-order ODEs and methods to solve them, like separation of variables, is essential for analyzing systems where the rate of change is straightforward.

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