Exercise 8

In this problem, we'll modify Euler's method to obtain better approximations to solutions of initial value problems. This method is called the Improved Euler's method.

In Euler's method, we walk across an interval of width $\Delta t$ using the slope obtained from the differential equation at the left endpoint of the interval. Of course, the slope of the solution will most likely change over this interval. We can improve our approximation by trying to incorporate the change in the slope over the interval.

Let's again consider the initial value problem $dy/dt = y$ and $y(0) = 1$, which we will approximate using steps of width $\Delta t = 0.2$. Our first interval is therefore $0\leq t \leq 0.2$. At $t=0$, the differential equation tells us that the slope is 1, and the approximation we obtain from Euler's method is that $y(0.2)\approx y_1= 1+ 1(0.2)= 1.2$.

This gives us some idea for how the slope has changed over the interval $0\leq t\leq 0.2$. We know the slope at $t=0$ is 1, while the slope at $t=0.2$ is 1.2, trusting in the Euler's method approximation. We will therefore refine our estimate of the initial slope to be the average of these two slopes; that is, we will estimate the slope to be $(1+1.2)/2 = 1.1$. This gives the new approximation $y(1) = y_1 = 1 + 1.1(0.2) = 1.22$.

The first few steps look like what is found in Table.

1. Continue with this method to obtain an approximation for $y(1) = e$.

2. Repeat this method with $\Delta t = 0.1$ to obtain a better approximation for $y(1)$.

3. We saw that the error in Euler's method is proportional to $\Delta t$. Using your results from parts (a) and (b), what power of $\Delta t$ appears to be proportional to the error in the Improved Euler's Method?