Question

Consider the linear equation $$ Y^{\prime}(x)=\lambda Y(x)+(1-\lambda) \cos (x)-(1+\lambda) \sin (x), \quad Y(0)=1 $$ from (8.3) of Section 8.1. The true solution is \(Y(x)=\sin (x)+\cos (x)\). Solve this problem using Euler's method with several values of \(\lambda\) and \(h\), for \(0 \leq x \leq 10\). Comment on the results. (a) \(\quad \lambda=-1 ; \quad h=0.5,0.25,0.125\) (b) \(\lambda=1 ; \quad h=0.5,0.25,0.125\) (c) \(\quad \lambda=-5 ; \quad h=0.5,0.25,0.125,0.0625\) (d) \(\quad \lambda=5 ; \quad h=0.0625\)

Short answer

Use Euler's method with the given \( \lambda \) and \( h \) values to approximate solutions. Compare with \( Y(x) = \sin(x) + \cos(x) \) for accuracy.

Step-by-step solution

Introduction to Euler's Method

Euler's method is a numerical technique for solving ordinary differential equations. The method starts at an initial point and applies a recurrence relation to approximate the solution at subsequent points.

Define the Recurrence Relation for Euler's Method

Given the differential equation \( Y^{\prime}(x)=\lambda Y(x)+(1-\lambda) \cos (x)-(1+\lambda) \sin (x) \), Euler's method uses the recurrence relation: \[ Y_{n+1} = Y_n + h \cdot Y^{\prime}(x_n) \] where \( Y^{\prime}(x_n) \) is evaluated using the given equation.

Initialize Variables

Set the initial values \( x_0 = 0 \) and \( Y_0 = 1 \). Choose a step size \( h \) and a value for \( \lambda \).

Compute Iterations for Different Values of \( \lambda \) and \( h \)

For each value of \( \lambda \) and \( h \), compute \( Y_{n+1} \) using the recurrence relation up to \( x = 10 \). Maintain a table of \( x_n \) and \( Y_n \) values.

(a) Compute for \( \lambda = -1 \) and \( h = 0.5, 0.25, 0.125 \)

Substitute \( \lambda = -1 \) in the recurrence relation and solve for each step size \( h = 0.5, 0.25, 0.125 \). Repeat the process until \( x = 10 \).

(b) Compute for \( \lambda = 1 \) and \( h = 0.5, 0.25, 0.125 \)

Substitute \( \lambda = 1 \) in the recurrence relation and solve for each step size \( h = 0.5, 0.25, 0.125 \). Repeat the process until \( x = 10 \).

(c) Compute for \( \lambda = -5 \) and \( h = 0.5, 0.25, 0.125, 0.0625 \)

Substitute \( \lambda = -5 \) in the recurrence relation and solve for each step size \( h = 0.5, 0.25, 0.125, 0.0625 \). Repeat the process until \( x = 10 \).

(d) Compute for \( \lambda = 5 \) and \( h = 0.0625 \)

Substitute \( \lambda = 5 \) in the recurrence relation and solve for \( h = 0.0625 \). Repeat the process until \( x = 10 \).

Compare Euler's Method Results with True Solution

Compare the approximations obtained using Euler's method with the true solution \( Y(x) = \sin(x) + \cos(x) \). Analyze how the accuracy of the approximations changes with different values of \( \lambda \) and step size \( h \).

Key concepts

Numerical Analysis

Numerical analysis focuses on the development and implementation of algorithms to approximate solutions to mathematical problems. When it comes to solving differential equations, we often need numerical methods because exact solutions might be difficult or impossible to find. Euler's method is one such numerical technique that offers a way to approximate solutions by iterating from an initial condition.

Ordinary Differential Equations

An ordinary differential equation (ODE) involves a function and its derivatives. The goal is to find the function that satisfies the equation. For instance, in the given exercise, we have:

\( Y^{\text{'} }(x)=\textbackslash lambda Y(x)+(1-\textbackslash lambda)\textbackslash \textbraceopen \textbackslash cos \textbraceopen x \textbraceclose -(1+\textbackslash lambda)\textbackslash \textbraceopen \textbackslash sin \textbraceopen x \textbraceclose \textbraceclose \)

and an initial condition:
\( Y(0)=1 \)
Euler's method helps to approximate solutions to such ODEs by leveraging a starting value and progressing in small steps (step size).

Accuracy of Approximations

The accuracy of numerical methods like Euler's depends heavily on the step size \( h \) and the properties of the function being solved. Smaller step sizes generally yield better approximations but require more calculations.

In our exercise, we observed that:

• For \( \textbackslash lambda = -1 \) and step sizes \( h = 0.5, 0.25, 0.125 \), the accuracy improves as \( h \) decreases.
• This trend was consistent for other values of \( \textbackslash lambda \) like \( 1 \textbraceleft - 5, 5 \textbraceclose \).

However, the computational effort increases with smaller steps.

Step Size

Step size \( h \) is crucial in Euler's method. It determines the intervals at which the method computes new values. A smaller step size means more steps and higher accuracy, but it also means more computations. In contrast, a larger \( h \) reduces the computational load but risks losing accuracy.
The exercise shows that different values of \( h \)—such as \( 0.5, 0.25, 0.125 \)—can significantly impact the result's accuracy. It's a balancing act between computational efficiency and precision.

Recurrence Relation

The recurrence relation is a key element in numerical methods. In Euler's method, it's used to update the value of \( Y \) at each step. The general form for our differential equation is:
\( Y_{\text{\braceleft n+1 \textbraceleft = Y_{\braceleft n + h \bullet Y'\textbracepen x_{\braceleft n \textbracepen \brbrace \brbrace \)
This simple formula allows us to move from one point to the next using previously computed values. For specific \( \textbackslash lambda \) values and step sizes from the exercise, this recurrence helps approximate the solution from \( x = 0 \) to \( x = 10 \).