Shared12 - Euler's Method Assignment / Euler's Method Notes.sagewsOpen in CoCalc
This material was developed by Aaron Tresham at the University of Hawaii at Hilo and is

### Prerequisites:

• Intro to Sage
• Differential Equations

# Euler's Method

Euler's Method is an algorithm used to construct approximate solutions to a differential equation of the form $\displaystyle\frac{dy}{dx}=f(x,y)$ starting at an initial point $(x_0,y_0)$.

Since the differential equation $\displaystyle\frac{dy}{dx}=f(x,y)$ tells us the slope of the tangent line at any point on the xy-plane, we can find the slope at $(x_0,y_0)$ and move along the tangent line some distance to a point $(x_1,y_1)$. Since the solution curve is close to its tangent line (as long as we're not too far from the point of tangency), the point $(x_1,y_1)$ is almost on the solution curve.

Now we repeat the process. Find the tangent line at $(x_1,y_1)$ using the differential equation, follow it for a short distance, and find a new point $(x_2,y_2)$. This point is also close to the solution curve.

Repeat the process as many times as you like.

The process is the same each time, so we can develop an iterated formula and automate the process.

Let's determine how to get from $(x_n,y_n)$ to $(x_{n+1},y_{n+1})$.

First, we need to find the tangent line at $(x_n,y_n)$.

In general, the tangent line to a function $y(x)$ at the point $x=a$ has equation $TL(x)=y(a)+y\,’(a)(x-a)$.

In this case, the derivative is given by the differential equation, $a=x_n$, and $y(a)=y_n$, so we have $TL(x)=y_n+f(x_n,y_n)(x-x_n)$.

Therefore, $y_{n+1}=TL(x_{n+1})=y_n+f(x_n,y_n)(x_{n+1}-x_n)$

We will move along the tangent line the same horizontal distance each step of the process. In other words, $x_{n+1}-x_n$ is a constant, called the “step size.” We will call this $h$.

In summary, we have the following:

Euler's Formula

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

$y_{n+1}=y_n+f(x_n,y_n)\cdot h$

## Example 1

Use Euler's Method to approximate the solution curve to the differential equation $\displaystyle\frac{dy}{dx}=x\cdot y$ that passes through the point $(0,1)$. Plot the approximation for $0\le x \le 2$.

We'll start with a small example by hand, and then we'll let the computer do the work.

We will use just 5 steps. That means the step size is $h=\frac{2-0}{5} = \frac{2}{5}$.

We'll start with $x_0=0$ and $y_0=1$, and then we will calculate new x- and y-coordinates with the formulas $x_{n+1}=x_n+h$ $y_{n+1}=y_n+x_n\cdot y_n\cdot h$

x y
$x_0=0$ $y_0=1$
$x_1=0+\frac{2}{5}=\frac{2}{5}$ $y_1=1+0\cdot 1\cdot \frac{2}{5}=1$
$x_2=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}$ $y_2=1+\frac{2}{5}\cdot 1\cdot \frac{2}{5}=\frac{29}{25}$
$x_3=\frac{4}{5}+\frac{2}{5}=\frac{6}{5}$ $y_3=\frac{29}{25}+\frac{4}{5}\cdot \frac{29}{25}\cdot \frac{2}{5}=\frac{957}{625}$
$x_4=\frac{6}{5}+\frac{2}{5}=\frac{8}{5}$ $y_4=\frac{957}{625}+\frac{6}{5}\cdot \frac{957}{625}\cdot \frac{2}{5}=\frac{35409}{15625}$
$x_5=\frac{8}{5}+\frac{2}{5}=2$ $y_5=\frac{35409}{15625}+\frac{8}{5}\cdot \frac{35409}{15625}\cdot \frac{2}{5}=\frac{1451769}{390625}$

Now let's plot these six points.

point([(0,1),(2/5,1),(4/5,29/25),(6/5,957/625),(8/5,35409/15625),(2,1451769/390625)])


The six points above are approximately on the solution curve. If we connect the points with straight lines, we will have an approximate solution curve.

Of course, just 5 steps is not enough to get a good approximation, so we'll use the computer with many more steps.

%var y
f(x,y)=x*y            #this is the function given by the differential equation
x0=0                  #initial value of x given in the problem
y0=1                  #initial value of y given in the problem
x_end=2               #the x-value you want to stop at
n=50                  #number of points to calculate
h=(x_end-x0)/n        #this calculates the step size for you
xlist=[x0];ylist=[y0] #we will use lists to keep track of all the x's and y's
p=point((x0,y0))      #we'll start keeping track of the points to graph
for i in range(n):    #here we apply Euler's Formula
xlist+=[xlist[i]+h]
ylist+=[ylist[i]+RR(f(xlist[i],ylist[i])*h)]  #Note: RR converts to a floating-point number to avoid Sage taking too much time with exact values.
p=p+point((xlist[i+1],ylist[i+1]))+line([(xlist[i],ylist[i]),(xlist[i+1],ylist[i+1])])
show(p)


Here is a plot of our approximation (blue) along with the actual solution (red).

p+plot(e^(1/2*x^2),xmin=x0,xmax=x_end,color='red')


We can make the approximation better by increasing $n$ (this decreases the step size).

If we want to plot the approximation past $x=2$, then we can change x_end. Of course, the approximation is going to get worse when we are farther away from our starting point.

The interactive box below allows us to change $n$ and x_end. Experiment with different values.

## Example 2

Consider the initial value problem $\displaystyle\frac{dy}{dx}=y+x,\quad y(0)=0$.

Use Euler's Method to approximate $y(2)$.

I will copy and paste the formulas from above, skipping the plot:

%var y
f(x,y)=y+x  #dy/dx = y+x
x0=0        #initial x-value = 0
y0=0        #initial y-value = 0
x_end=2     #we want to stop at x = 2
n=50        #we'll try 50 for now
h=(x_end-x0)/n
xlist=[x0];ylist=[y0]
for i in range(n):
xlist+=[xlist[i]+h]
ylist+=[ylist[i]+RR(f(xlist[i],ylist[i])*h)]
N(ylist[n]) #notice that ylist[n] is the last point calculated, in this case y(2)

4.10668334627831

We have found that $y(2)\approx4.1067$.

Let's try a higher value of n and see what happens.

%var y
f(x,y)=y+x
x0=0
y0=0
x_end=2
n=100        #we'll try n=100 this time
h=(x_end-x0)/n
xlist=[x0];ylist=[y0]
for i in range(n):
xlist+=[xlist[i]+h]
ylist+=[ylist[i]+RR(f(xlist[i],ylist[i])*h)]
N(ylist[n])

4.24464611825234

Now we have $y(2)\approx4.2446$.

Let's find the actual value. First, solve the differential equation.

y=function('y',x)
desolve(derivative(y,x)==y+x,y,[0,0])

-x + e^x - 1

Now plug in $x=2$.

F(x)=-x + e^x - 1 #I'll call the solution F(x)
F(2)
N(F(2))

e^2 - 3 4.38905609893065

So $y(2)=4.38905609893065$.

Notice that increasing $n$ has gotten us closer to the actual answer. Let's increase $n$ one more time and see if we can get at least the first decimal place correct.

%var y
f(x,y)=y+x
x0=0
y0=0
x_end=2
n=500        #we'll try n=500 this time
h=(x_end-x0)/n
xlist=[x0];ylist=[y0]
for i in range(n):
xlist+=[xlist[i]+h]
ylist+=[ylist[i]+RR(f(xlist[i],ylist[i])*h)]
N(ylist[n])

4.35963717586897

Here is a summary of our results:

n Approximation
$50$ $4.10668334627831$
$100$ $4.24464611825234$
$500$ $4.35963717586897$

The actual value is $e^2-3\approx 4.38905609893065$.