Problem 38 - 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 involves integration techniques, substitution, and constant solving. Evaluate and plot to verify correctness.

Step 1: Identify the differential equation

The differential equation provided is \( \frac{d y}{d x}=(y-1)^2-0.01 \) with the initial condition \( y(0) = 1 \). This is a first-order ordinary differential equation.

Step 2: Simplify the equation

Let's rewrite the differential equation as \( \frac{d y}{d x} = (y-1)^2 - 0.01 \). Our goal is to find an explicit expression for \( y(x) \).

Step 3: Separate the variables

By separating the variables, we have \( \frac{d y}{(y-1)^2 - 0.01} = d x \). This is crucial to integrate both sides.

Step 4: Perform integration

Integrate the left side with respect to \( y \) and the right side with respect to \( x \). The left side integral becomes \( \int \frac{d y}{(y-1)^2 - 0.01} \), and the right side becomes \( \int d x \).

Step 5: Solve the integral on the left

The integral \( \int \frac{d y}{(y-1)^2 - 0.01} \) can be solved by partial fraction decomposition or substitution. Let's assume a substitution such that \( u = y -1 \). Thus, the integral becomes \( \int \frac{1}{u^2 - 0.01} du \). This can be further solved using standard integral tables or transformation.

Step 6: Integration and simplification

Continuing from the substitution, we look for a standard integral form that matches \( \int \frac{1}{u^2 - a^2} du\), which gives \( \frac{1}{2a} \ln \left| \frac{u - a}{u + a} \right| + C \), where \( a^2 = 0.01 \), so \( a = 0.1 \). Integrating, we find:\[ \frac{1}{0.2} \ln \left| \frac{y-1 - 0.1}{y-1 + 0.1} \right| = x + C \].

Step 7: Apply initial condition

Substitute the initial condition \( y(0) = 1 \) into the equation to find the constant \( C \). This simplifies the equation. Since \( y = 1 \), \( u = y - 1 = 0 \). Substitute to solve for \( C \).

Step 8: Obtain the explicit solution

After evaluating \( C \) and substituting back \( u = y - 1 \), we solve for \( y(x) \) to obtain an explicit solution form. Substituting back, the final explicit solution could be evaluated and simplified to match the structure.

Step 9: Graphing the solution

Using a graphing utility, plot \( y(x) \) with respect to \( x \) and verify the solution passes through \( (0,1) \). Compare the graph to observe changes in the neighborhood of \( (0,1) \).

Key Concepts

Initial Value Problem

In the world of differential equations, an initial value problem is a classic setup. Here, we are given a differential equation along with a starting condition, called an initial condition. This initial condition helps us find a specific solution that fits a precise point in space. In our exercise, the initial value problem is expressed as:

• Equation: \( \frac{d y}{d x} = (y-1)^2-0.01 \)
• Initial Condition: \( y(0) = 1 \)

These components help us trace a unique path or trajectory for the solution, which is not possible with just an equation alone. The initial condition, \( y(0) = 1 \), ensures that the solution curve passes through the point \((0, 1)\), grounding the abstract differential equation in a real-world context.

Variable Separation

Variable separation is a powerful tool in solving differential equations. It simplifies the equation by isolating variables on different sides, making integration possible. In our exercise, the equation is:\(\frac{d y}{d x} = (y-1)^2 - 0.01\)We can separate variables by rearranging terms to get:\[\frac{d y}{(y-1)^2 - 0.01} = d x\]This transformation is crucial because it allows us to integrate each side independently, thus making progress towards finding a solution. Essentially, by splitting the variables, we prepare the equation for the next step, which involves integration. Variable separation is particularly handy when the interaction between changes in \( x \) and \( y \) is nonlinear.

Integration Techniques

Once the equation's variables have been separated, the next step is integration. The challenge is to integrate each side of the equation with respect to its variable:

• Left Side: \( \int \frac{d y}{(y-1)^2 - 0.01} \)
• Right Side: \( \int d x \)

The integration on the left is more complex and often requires specific techniques like substitution. By using a substitution, such as \( u = y - 1 \), we get a new integral form:\[ \int \frac{1}{u^2 - 0.01} du \]This can be approached through known integral forms, involving logarithmic functions. On the other hand, integrating with respect to \( x \) is straightforward, yielding a linear function of \( x \). These integration techniques convert our separated variables into a form where we can solve for \( y \) as a function of \( x \).

Solution Behavior Analysis

Analyzing the behavior of a differential equation solution helps us to understand how it evolves over time and under given conditions. After solving and integrating, we need to incorporate the initial condition \( y(0) = 1 \) and evaluate any constants that arise. This is key:- Ensure the solution curve starts at the point \((0, 1)\).- Observe how \( y \) changes as \( x \) increases.Small changes in the differential equation or initial conditions can result in vastly different solution trajectories, illustrating the sensitivity of solutions. This analysis shows the solution's behavior in response to perturbations, and how it stabilizes or transforms. Understanding this can give insights into the equation's practical applications, like motion of particles or growth models.

Graphical Solution Representation

Graphing the solution brings the abstract, mathematical process into a visual interpretation, which makes understanding the behavior of the solution easier. By using a graphing utility:- Plot the explicit solution \( y(x) \).- Confirm the graph passes through the initial condition point \((0, 1)\).Graphing also enables us to inspect the solution's behavior around this neighborhood. We can visually see how the curve behaves, whether it extends smoothly or encounters any abrupt changes. This representation not only verifies our solution but also highlights the sensitivity to initial conditions. It provides a clear, immediate sense of the paths solutions might take, enriching our comprehension beyond just numerical outcomes.