Question

Estimate the following solutions using Euler's method with \(n=5\) steps over the interval \(t=[0,1] .\) If you are able to solve the initialvalue problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler's method. How accurate is Euler's method? $$ y^{\prime}=y^{2} \ln (x+1), y(0)=1 . \text { Exact solution is } y=-\frac{1}{(x+1)(\ln (x+1)-1)} $$

Short answer

Euler's method gives \( y \approx 1.91 \) at \( x = 1 \), but the exact solution is undefined.

Step-by-step solution

Setup the differential equation for Euler's Method

The given differential equation is \( y' = y^2 \ln(x+1) \) with the initial condition \( y(0) = 1 \). We are asked to use Euler's method over the interval \( t = [0, 1] \) with \( n = 5 \) steps. Our step size \( h \) can be calculated as \( h = \frac{1}{n} = \frac{1}{5} = 0.2 \).

Calculate the increment

Euler's method increments involve updating the next value \( y_{i+1} \) using the formula:\[y_{i+1} = y_i + h \cdot y'(x_i, y_i)\]For the function \( y^2 \ln(x+1) \), the increment for each step is \( 0.2 \cdot y_i^2 \ln(x_i + 1) \).

Perform Euler's Method Iteratively

Starting at \( x_0 = 0 \) and \( y_0 = 1 \):- **Step 1:** \( x_1 = 0.2 \), \( y_1 = 1 + 0.2 \cdot 1^2 \cdot \ln(0.2 + 1) = 1.0367 \).- **Step 2:** \( x_2 = 0.4 \), \( y_2 = 1.0367 + 0.2 \cdot (1.0367)^2 \cdot \ln(0.4 + 1) = 1.1043 \).- **Step 3:** \( x_3 = 0.6 \), \( y_3 = 1.1043 + 0.2 \cdot (1.1043)^2 \cdot \ln(0.6 + 1) = 1.2199 \).- **Step 4:** \( x_4 = 0.8 \), \( y_4 = 1.2199 + 0.2 \cdot (1.2199)^2 \cdot \ln(0.8 + 1) = 1.4216 \).- **Step 5:** \( x_5 = 1.0 \), \( y_5 = 1.4216 + 0.2 \cdot (1.4216)^2 \cdot \ln(1.0 + 1) = 1.9100 \).

Evaluate the Exact Solution at x = 1

Calculate the exact solution using the formula provided:\[y = -\frac{1}{(x+1)(\ln(x+1)-1)}\]At \( x = 1 \),\[ y = -\frac{1}{2 \cdot (0)} = \text{undefined (note: this suggests a singularity was reached)} \].

Compare Euler's Method with Exact

Since the exact solution is undefined at \( x = 1 \) due to a singularity, we can’t directly compare it with the Euler's approximation, but we can discuss the behavior: Euler's method predicted \( y_5 \approx 1.9100 \), indicating a divergence from the true nature of the solution as it approaches the singularity.

Key concepts

Numerical Analysis

Numerical analysis is all about finding approximate solutions for mathematical problems that can't be solved exactly.
When dealing with complex issues like differential equations, sometimes analytical solutions are not readily available.
In such cases, numerical methods like Euler's method become extremely valuable. Euler's method, for instance, helps to approximate the solutions to differential equations.
By using step-by-step calculations incrementally, numerical analysis allows us to estimate the behavior of the function over a specific interval.
It's incredibly useful in science and engineering, where exact solutions may not always exist. In the exercise, we implemented Euler's method to estimate the differential equation's solution across five small steps.
Ultimately, numerical analysis provides insight into the probable behavior of solutions, even in the face of complexity and singularities.

Differential Equations

Differential equations are equations that involve derivatives of a function.
They express how a quantity changes over time and are pivotal in fields such as physics, engineering, and economics.
The specific equation given here, \( y' = y^2 \ln(x+1) \), links the rate of change of \( y \) to its value through the logarithmic term.To solve a differential equation, we often look for a function \( y \) whose derivative matches the equation we are given.
This can be challenging, particularly when straightforward algebra isn't enough to find the answer.
That's where methods like Euler's come into play, letting us approximate the solution for complex differential equations. Understanding the nature of differential equations helps us to conceptualize how solutions behave over time.
They provide a powerful way to model real-world situations where change is a key component.

Initial Value Problem

In mathematics, an initial value problem (IVP) is a type of differential equation paired with a starting condition.
It can be symbolically represented as: \( y' = f(x, y) \), with \( y(a) = b \).
This means we want to find the function \( y(x) \) that satisfies the differential equation and the initial condition.The initial condition gives us a specific point \( (a, b) \) that the function must pass through.
This point is crucial in guiding the approximation method, essentially anchoring the solution.In this exercise, the initial value \( y(0) = 1 \) served as our starting point.
Euler's method built from this starting point, iterating step-by-step to estimate the solution.
Initial value problems are common across many scientific domains, helping to predict future states of dynamic systems.

Exact Solution Comparison

Comparing numerical solutions with exact solutions is essential to understanding their accuracy.
While Euler's method provided an approximation, certain problems like the singularity at \( x = 1 \) can prevent this comparison.In the given task, the exact solution \( y = -\frac{1}{(x+1)(\ln(x+1) - 1)} \) presented a challenge.
At \( x = 1 \), the solution is undefined, showing a divergence in behavior due to a singularity.By comparing the Euler's method results with the nature of the exact solution, we learn about the limitations and reliability of the numerical technique.
When the exact solution is undefined, such insights are crucial for evaluating the prediction and approximation quality.
Although singularities create roadblocks in comparisons, they also highlight the critical points where caution is needed in analysis.