Approximating Solutions Using Euler's Method (College Board AP® Calculus BC): Study Guide

If an exam question asks you how to improve an Euler’s method approximation, the answer will almost always involve decreasing the step size $∆x$.

It is possible to perform Euler's method with any step size and any number of steps. In this case you could go on to calculate $y_3$, $y_4$, $y_5$, etc. On the exam, however, the questions almost always involve only two steps.

You might find it useful to do your workings in a table

Let $y=f\left(x\right)$ be the solution to the differential equation $\frac{dy}{dx}=x-3y+1$ with initial condition $f\left(0\right)=2$. What is the approximation for $f\left(1\right)$ obtained by using Euler's method with a step size of $\frac{1}{2}$ starting at $x=0$?

Answer:

You need to go from $x=0$ to $x=1$ with a step size of $\frac{1}{2}$

• so there will be $\frac{1-0}{{\displaystyle \frac{1}{2}}}=2$ steps

Use the formula $y_{n+1}=y_n+∆x·f^{\text{'}}\left(x_n,y_n\right)$

• where here $∆x=\frac{1}{2}$

• and $\frac{dy}{dx}=x-3y+1$

Start with initial value $\left(x_0,y_0\right)=\left(0,2\right)$

$\left(x_0,y_0\right)=\left(0,2\right)$

Then $x_1=x_0+\mathrm{\Delta }x$

$x_1=0+\frac{1}{2}=\frac{1}{2}$

And $y_1=y_0+∆x·f^{\text{'}}\left(x_0,y_0\right)$

$y_1=2+\frac{1}{2}·\left(\left(0\right)-3\left(2\right)+1\right)=2+\frac{1}{2}\left(-5\right)=-\frac{1}{2}$

$\Rightarrow \left(x_1,y_1\right)=\left(\frac{1}{2},-\frac{1}{2}\right)$

Continue from $\left(x_1,y_1\right)=\left(\frac{1}{2},-\frac{1}{2}\right)$

$x_2=x_1+\mathrm{\Delta }x$

$x_2=\frac{1}{2}+\frac{1}{2}=1$

And $y_2=y_1+∆x·f^{\text{'}}\left(x_1,y_1\right)$

$y_2=-\frac{1}{2}+\frac{1}{2}·\left(\left(\frac{1}{2}\right)-3\left(-\frac{1}{2}\right)+1\right)=-\frac{1}{2}+\frac{1}{2}\left(3\right)=1$

$\Rightarrow \left(x_2,y_2\right)=\left(1,1\right)$

You could also do these workings in a table

So by the Euler method, $f\left(1\right)\approx y_2=1$

$f\left(1\right)\approx 1$

$f$ is a twice-differentiable function for all values of $x$, with $f\left(2\right)=1$. The table above gives values of the derivative of $f$, $f^{\text{'}}$, for selected values of $x$.

Use Euler's method, with two steps of equal size starting at $x=2$, to approximate $f\left(3\right)$. Show the computations that lead to your answer.

Answer:

You need to go from $x=2$ to $x=3$ with two steps of equal size

• so the step size will be $\mathrm{\Delta }x=\frac{3-2}{{\displaystyle 2}}=\frac{1}{2}$

Use the formula $y_{n+1}=y_n+∆x·f^{\text{'}}\left(x_n,y_n\right)$

• where here $\frac{dy}{dx}=f\text{'}\left(x\right)$

• i.e. where the value of $\frac{dy}{dx}$ depends on $x$ only

Start with initial value $\left(x_0,y_0\right)=\left(2,1\right)$

$\left(x_0,y_0\right)=\left(2,1\right)$

Then $x_1=x_0+\mathrm{\Delta }x$

$x_1=2+\frac{1}{2}=\frac{5}{2}$

And $y_1=y_0+∆x·f^{\text{'}}\left(x_0\right)$

$y_1=1+\frac{1}{2}·f^{\text{'}}\left(2\right)=1+\frac{1}{2}\left(5\right)=\frac{7}{2}$

$\Rightarrow \left(x_1,y_1\right)=\left(\frac{5}{2},\frac{7}{2}\right)$

Continue from $\left(x_1,y_1\right)=\left(\frac{5}{2},\frac{7}{2}\right)$

$x_2=x_1+\mathrm{\Delta }x$

$x_2=\frac{5}{2}+\frac{1}{2}=3$

And $y_2=y_1+∆x·f^{\text{'}}\left(x_1\right)$

$y_2=\frac{7}{2}+\frac{1}{2}·f^{\text{'}}\left(2.5\right)=\frac{7}{2}+\frac{1}{2}\left(2\right)=\frac{9}{2}$

$\Rightarrow \left(x_2,y_2\right)=\left(3,\frac{9}{2}\right)$

You could also do these workings in a table

So by the Euler method, $f\left(3\right)\approx y_2=\frac{9}{2}$

$f\left(3\right)\approx \frac{9}{2}$

Be careful with letters. In the exam (and in your calculator’s recursion app) the variables may not be x and y.

Although your graphing calculator can carry out Euler's method approximations, these questions are usually on the non-calculator part of the exam. So make sure you can do the method 'by hand' as well!

Consider the differential equation $\frac{dy}{dx}+y=x+1$ with the initial condition $y\left(0\right)=0.5$.

(a) Using your graphing calculator, apply Euler’s method with five equal steps to approximate the solution to the differential equation at $x=1$.

(b) Explain how the accuracy of the approximation in part (a) could be improved.

Answer:

(a)

First rearrange into $\frac{dy}{dx}=f^{\text{'}}\left(x, y\right)$ form

$\frac{dy}{dx}=x-y+1$

Next calculate the step size $∆x$; you need to go from $x=0$ to $x=1$ in five steps so

$∆x=\frac{1-0}{5}=0.2$

Put that information into the recursion equations $y_{n+1}=y_n+∆x·f^{\text{'}}\left(x_n, y_n\right)$ and $x_{n+1}=x_n+∆x$

$y_{n+1}=y_n+0.2·\left(x_n-y_n+1\right)$

$x_{n+1}=x_n+0.2$

Put those equations into the recursion app on your calculator, and specify the required number of steps ($n=5$)

Your calculator will output the answers in a table

Note that the last row in the table gives the result you're looking for (i.e. the Euler approximation for $y$ when $x=1$)

Round to 3 decimal places

$y\left(1\right)\approx 1.164\left(3\mathrm{d}.\mathrm{p}.\right)$

(b)

Decrease the step size by using a greater number of steps to get from 0 to 1