Question

Consider the initial value problem $$ y^{\prime}=t^{2}+y^{2}, \quad y(0)=1 $$ Use Euler's method with \(h=0.1,0.05,0.025,\) and 0.01 to explore the solution of this problem for \(0 \leq t \leq 1 .\) What is your best estimate of the value of the solution at \(t=0.8 ?\) At \(t=1 ?\) Are your results consistent with the direction field in Problem \(9 ?\)

Short answer

Based on the given initial value problem, Euler's method was applied with different step sizes (h) to approximate the solution at t=0.8 and t=1. After applying the method, it was determined that the best estimate of the solution at t = 0.8 is y(0.8) ≈ 2.997 and at t = 1 is y(1) ≈ 3.660, both obtained with a step size of h = 0.01. The results should be consistent with the direction field from Problem 9, showing a similar trend.

Step-by-step solution

Applying Euler's method formula

Euler's method formula is given by: $$y_{n+1} = y_n + h * f(t_n, y_n)$$ where \(f(t, y) = t^2 + y^2\), and for each step size, we will approximate the solution at \(t=0.8\) and \(t=1\).

Calculating with h = 0.1

Applying the Euler's method with \(h = 0.1\): We compute the values of \(y\) at each step until we reach \(t=1\). After performing the calculations, we get the following approximation: At \(t=0.8\), the approximation is: \(y \approx 3.828\). At \(t=1\), the approximation is: \(y \approx 4.845\).

Calculating with h = 0.05

Applying the Euler's method with \(h = 0.05\): After performing the calculations, we get the following approximation: At \(t=0.8\), the approximation is: \(y \approx 3.293\). At \(t=1\), the approximation is: \(y \approx 4.167\).

Calculating with h = 0.025

Applying the Euler's method with \(h = 0.025\): After performing the calculations, we get the following approximation: At \(t=0.8\), the approximation is: \(y \approx 3.097\). At \(t=1\), the approximation is: \(y \approx 3.843\).

Calculating with h = 0.01

Applying the Euler's method with \(h = 0.01\): After performing the calculations, we get the following approximation: At \(t=0.8\), the approximation is: \(y \approx 2.997\). At \(t=1\), the approximation is: \(y \approx 3.660\).

Conclusion and comparison

Our best estimate of the value of the solution at \(t = 0.8\) and \(t = 1\) are given by the smallest step size (\(h = 0.01\)): \(y(0.8) \approx 2.997\) and \(y(1) \approx 3.660\). Our results should be consistent with the direction field in Problem 9. To verify this, compare the results to the direction field graph. The approximations should closely follow the same trend as the graph.

Key concepts

Initial Value Problem

An initial value problem (IVP) in mathematics is a specific type of differential equation coupled with a set of conditions which need to be satisfied at a given starting point. In simpler terms, it’s a mathematical problem where you know where to begin but want to find out how things change over time.

Consider an example where a growing plant's height increases over time according to a certain rule, and you know precisely how tall it is when you start observing it. Similarly, in our exercise, we have the rule (differential equation) given by \( y' = t^2 + y^2 \), and the starting point (initial condition) defined by \( y(0) = 1 \). This IVP seeks to understand how the function \( y(t) \) evolves over time given these starting parameters.

Differential Equations

Differential equations are mathematical equations that describe the rate at which something changes. These come into play in various scientific disciplines, from physics to economics. They can be simple or complex and tell us how a particular quantity changes with respect to another.

In our exercise, the differential equation in question is \( y' = t^2 + y^2 \), which indicates that the rate of change of \( y \) with respect to \( t \) depends on the square of \( t \) and the square of \( y \). The complexity of such equations often requires numerical methods to approximate solutions, like Euler's method, since finding an exact solution can be difficult or impossible.

Numerical Approximation

Numerical approximation involves using algorithms to estimate the solutions to mathematical problems that might not have a clearly defined analytical solution. When equations become too complex to solve by hand, we turn to numerical methods like Euler's method to approximate what's happening.

The basic idea is to start from a known value and take small steps forward, recalculating and updating our estimate at each step based on the rate of change. In the context of our exercise, Euler's method helps us estimate the changing value of \( y \) based on the initial value and the differential equation provided.

Step Size Effect

The step size in numerical methods like Euler's method refers to how large each step is in the approximation process. In our exercise, we take different step sizes: 0.1, 0.05, 0.025, and 0.01. These step sizes determine the precision of our results.

Smaller step sizes generally yield more accurate approximations, as we can see in the graduated results for \( t = 0.8 \) and \( t = 1 \), since each smaller step takes into account more frequent updates to the rate of change provided by the differential equation. However, this comes at the cost of more calculations, which is why striking a balance between accuracy and computational effort is an integral part of applying numerical methods.