Chapter 2.8, Problem 6E

Interpretation Introduction

Interpretation:

To sketch the solutions $\text{x(t)}$ for $\text{t}\ge \text{0}$, for the initial value problem ${\dot{\text{x}}\text{ = x + e}}^{\text{-x}}$, $\text{x(0) = 0}$. The rigorous bounds on the value of $\text{x}$ at $\text{t = 1}$ is to be obtained. To compute $\text{x}$ at $\text{t = 1}$, using Euler method, correct to three decimal places, and to find how small the stepsize needs to be to obtain the desired accuracy. To compute $\text{x}$ at $\text{t = 1}$, using Runge-Kutta method. To compare the results for stepsizes $\text{Δt = 1}$, $\text{Δt = 0}\text{.1}$, and $\text{Δt = 0}\text{.01}$.

Concept Introduction:

Taylor’s series expansion for ${\text{e}}^{\text{- x}}$ is given as

${\text{e}}^{\text{- x}}\text{ = }{\displaystyle \sum _{\text{n = 0}}^{\infty }{\text{(-1)}}^{\text{n}}\frac{{\text{x}}^{\text{n}}}{\text{n!}}}\text{ = 1 - x + }\frac{{\text{x}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{x}}^{\text{3}}}{\text{3!}}\text{ + }\mathrm{..................}$

Euler method is given by

${\text{x}}_{\text{n+1}}{\text{ = x}}_{\text{n}}\text{+ f}\left({\text{x}}_{\text{n}}\right)\text{Δt}$

Runge-Kutta method is given by

${\text{x}}_{\text{n+1}}{\text{ = x}}_{\text{n}}\text{+ }\frac{\text{1}}{\text{6}}\left({\text{k}}_{\text{1}}{\text{+ 2k}}_{\text{2}}{\text{+ 2k}}_{\text{3}}{\text{+ k}}_{\text{4}}\right)$, where

$\begin{array}{l}{\text{k}}_{\text{1}}=\text{ f}\left({\text{x}}_{\text{n}}\right)\text{Δt}\\ {\text{k}}_{\text{2}}=\text{ f}\left({\text{x}}_{\text{n}}+\frac{1}{2}{\text{k}}_{\text{1}}\right)\text{Δt}\\ {\text{k}}_{\text{3}}=\text{ f}\left({\text{x}}_{\text{n}}+\frac{1}{2}{\text{k}}_2\right)\text{Δt}\\ {\text{k}}_{\text{4}}=\text{ f}\left({\text{x}}_{\text{n}}+{\text{k}}_3\right)\text{Δt}\end{array}$