§7.3 Euler's method

Introduction

Euler's Method

The error in Euler's method

Summary

Euler's method approximates solutions of $\frac{dy}{dt} = f(t,y)$ by repeatedly applying

For the initial value problem $\frac{dy}{dt} = \frac{1}{2}(y+1)$ with $y(0) = 0$, the tangent line at $t = 0$ is

Applying Euler's method with $y(0) = 6$ to $\frac{dy}{dt} = 6y-y^2$ produces

The differential equation $\frac{dy}{dt} = 6y-y^2$ has equilibrium solutions

For $\frac{dy}{dt} = f(t) = 2t-1$ with $y(0) = 0$ and $\Delta t = 0.2$, the Euler value $y_5$ coincides with

Alice's coffee ($k = 0.5$) and Bob's coffee ($k = 0.1$) both start at $100^\circ$F in a $70^\circ$ room. Their initial cooling rates are

When the room temperature oscillates as $T_r = 70+10\sin t$, Euler approximations for Alice ($k=0.5$) and Bob ($k=0.1$) show that

For $\frac{dy}{dt} = 6y-y^2$ with $y(0) = 1$ and $\Delta t = 0.2$, compute the first approximation $y_1$.

Continuing with $y_1 = 2$ at $t = 0.2$, compute the second Euler approximation $y_2$ at $t = 0.4$, still with $\Delta t = 0.2$.

For $\frac{dy}{dt} = 2t-1$ with $y(0) = 0$ and $\Delta t = 0.2$, compute the first Euler approximation $y_1$.

The exact solution of $\frac{dy}{dt} = 2t-1$, $y(0) = 0$ is $y = t^2-t$, while Euler with $\Delta t = 0.2$ ends at $y_5 = -0.2$. Compute the error $y(1)-y_5$.

For $\frac{dy}{dt} = y$, $y(0)=1$, Euler gave $E_{0.2} = 2.4883$ and $E_{0.1} = 2.5937$. Assuming $y(1)-E_{\Delta t} = K\Delta t$, solve the two equations and report the improved $y(1)$, rounded to three decimals.

For $\frac{dy}{dt} = 6y-y^2$ with $y(0) = 1$, the solution increases toward the stable equilibrium $y = 6$ and never crosses it.

The error of Euler's method at a fixed target time is proportional to $\Delta t$, so halving the step size roughly halves the error.

For the improved Euler's method, experiments such as $I_{0.2} = 2.7027$ and $I_{0.1} = 2.7141$ approximating $e$ suggest the error is proportional to $\Delta t^2$.

For $\frac{dy}{dt} = 2t-1$, changing the initial value from $y(0) = 0$ to $y(0) = 1$ simply adds $1$ to every Euler value, reflecting that all solutions differ by a constant.