Question

Two steps of Euler's method For the following initial value problems, compute the first two approximations \(u_{1}\)and\(u_{2}\) given by Euler's method using the given time step. $$y^{\prime}(t)=t+y, y(0)=4 ; \Delta t=0.5$$

Short answer

The first two approximations using Euler's method are \(u_1 = 6\) and \(u_2 = 9.25\).

Step-by-step solution

Find the derivative at the initial point

Calculate the derivative \(y^{\prime}(0)\) using the initial condition: $$y^{\prime}(0)=0 + 4 = 4$$

Estimate the value at t=0.5 using Euler's method

Using the Euler's method formula \(y(t_1) = y(t_0) + \Delta t y^{\prime}(t_0)\), where \(t_1 = t_0 + \Delta t\): $$u_1 = 4 + 0.5 \cdot 4 = 4 + 2 = 6$$

Find the derivative at t=0.5

Calculate the derivative \(y^{\prime}(0.5)\) using the estimated value u1: $$y^{\prime}(0.5)=0.5 + 6 = 6.5$$

Estimate the value at t=1 using Euler's method

Using the Euler's method formula \(y(t_2) = y(t_1) + \Delta t y^{\prime}(t_1)\), where \(t_2 = t_1 + \Delta t\): $$u_2 = 6 + 0.5 \cdot 6.5 = 6 + 3.25 = 9.25$$ The first two approximations using Euler's method are \(u_1 = 6\) and \(u_2 = 9.25\).

Key concepts

Initial Value Problems

An initial value problem is a type of problem in mathematics where you are given a differential equation and an initial condition. The goal is to find the solution that satisfies both.
In simple terms, it's like having a road map with a starting point. Knowing where you begin is crucial because differential equations have many possible solutions, and this initial condition helps you pinpoint the exact one.
Consider it as if you have multiple paths to reach a destination; the initial value helps you choose the correct path right from the start.
This is exactly what we do in Euler's method when we start our approximations with an initial point, like in the given exercise where at time zero, the value is 4.

Differential Equations

Differential equations are mathematical equations involving derivatives, which represent rates of change.
These equations describe a variety of phenomena, from the motion of planets to the growth of populations and more.
The derivative gives us information on how one quantity changes with respect to another—typically time in many problems.
In the original exercise, the differential equation tells us how the function changes as time progresses.

• For example: In the equation, the term \(y'(t) = t + y\) tells us that the rate of change of \(y\) (slope of the tangent to the graph at any point \(t\)) is dependent on both the current time \(t\) and the current value \(y\).

Numerical Approximation

Numerical approximation involves using algorithms to find approximate solutions to mathematical problems that are otherwise difficult to solve analytically.
This is particularly useful when an exact answer is either impossible or impractical to compute by hand.
Euler's method is one such numerical technique that's widely used to approximate solutions to ordinary differential equations.
In this method, a simple formula is used iteratively to get values that approximate the solution at successive time steps. Euler's method is quite basic, yet effective for understanding the trajectory of a function over time.
In the exercise, each approximation, like \(u_1\) and \(u_2\), gives us a closer approximation to the true solution.

Time Step

The time step, often denoted as \(\Delta t\), is a crucial part of numerical methods like Euler's method.
It determines how far forward in time we calculate the solution in each step of the approximation.
A smaller time step often results in a more accurate approximation, but this comes at the cost of increased computational time and effort.

• For example, in the exercise given, the time step \(\Delta t = 0.5\) means that each approximation employs information at intervals of 0.5 units along the time axis.

Choosing a time step involves balancing the need for computational efficiency with the need for accuracy.
Too large a time step might miss crucial information leading to a misleading approximation, while too small a step can be computationally expensive even though it's highly accurate.