
Euler's Method
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What is Euler method? |
|
Example:
The blue line shows the actual graph of a function. The green curve is an approximation using Euler's method. It is the collection of line segments as a result of Euler's Method. Each time Euler method is used another point is created and thus another line segment. |
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Note:
-Generally, the approximation gets less accurate the further you are away from the initial value.
-Better accuracy is achieved when the points in the approximation are closer together.
-Better accuracy is achieved when the points in the approximation are closer together.
What are the three things needed in order to use Euler's method?
- Initial point- You must be given a starting point (x0,y0)
- $\Delta x $- You must be given the change in x or step size. You are either given the step size directly or given information to find it. For example, if you have to go from x=2 to x=2.6 using three equal step sizes. You know the step size is .2
*Remember- the smaller your $\Delta x$the better your approximation
- The differential equation – You have to know the slope of each individual line segment so that you can find $\Delta y$
How to Use Euler's Method
Whenever you are given a problem that requires Euler's Method, create the chart below :
$$ (x,y)$$ $\Delta x \text{ or }dx$ $$ \frac{dy}{dx} $$ $$ dx (\frac{dy}{dx})=dy $$ $$(x+dx, y +dy) $$
General Steps
Step 1
The first 3 columns initial point, dx , and
$(\frac{dy}{dx})
$
are given to you in the problem. Just fill in that information
Step 2
The $\Delta x $or dx remains the same for each iteration of Euler's method so you can actually fill the entire column with the same number.
Step 4
Add the dx to the initial x value (column 1 and 2), and add dy to the initial y value (column 1 and 4) to fill in the last column
Step 5
Re-write the point in the last column in the first column of the second row and repeat steps 2 through 5 until you reach the desired x value.
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So how do we use this method to solve a specific problem? |
Example 1
Using separable equations techniqueGiven $\frac{dy}{dx} = 2(x-1)$and the point (1,0) is a point on the curve, find an equation in the form y = f(x) and use it to evaluate f(3).
$\begin{align}
& \int \frac{dx}{dy} = \int 2(x-1)
\\
& y = (x-1)^2 + c
\\
\end{align}
$
| using the point (1,0) | $0 = (1 - 1)^2 + C \text{ so C = 0} $ |
| Making the equation | $y = (x-1)^2 $ |
| Using x = 3 | $y = (3-1)^2 =4
$ so when x = 3 , y = 4 |
Example 1
Using Euler's Method
Given $\frac{dy}{dx}$= 2(x-1) and the point (1,0) is a point on the curve, find an equation in the form y = f(x) and use it to evaluate f(3).
If we are going from x =1 to x =3 using 5 steps that means the $\Delta x = \frac{3-1}{4} =\frac{2}{4} = .5 $
| $ (x,y)$ | $\Delta x \text{ or }dx$ | $ \frac{dy}{dx} $ | $ dx (\frac{dy}{dx})=dy $ | $(x+dx, y +dy) $ |
| (1,0) | .5 | 0 | ||
| .5 | ||||
| .5 | ||||
| .5 |
Steps 3 and 4
| $ (x,y)$ | $\Delta x \text{ or }dx$ | $ \frac{dy}{dx} $ | $ dx (\frac{dy}{dx})=dy $ | $(x+dx, y +dy) $ |
| (1,0) | .5 | 0 | 0 | (1+.5,0+0) |
| .5 | ||||
| .5 | ||||
| .5 |
Step 5
| $ (x,y)$ | $\Delta x \text{ or }dx$ | $ \frac{dy}{dx} $ | $ dx (\frac{dy}{dx})=dy $ | $(x+dx, y +dy) $ |
| (1,0) | .5 | 0 | 0 | (1+.5,0+0) |
| (1.5,0) | .5 | |||
| .5 | ||||
| .5 |
Repeat
| $ (x,y)$ | $\Delta x \text{ or }dx$ | $ \frac{dy}{dx} $ | $ dx (\frac{dy}{dx})=dy $ | $(x+dx, y +dy) $ |
| (1,0) | .5 | 0 | 0 | (1.5,0+0) |
| (1.5,0) | .5 | 1 | .5 | (2, .5) |
| (2, .5) | .5 | |||
| .5 |
Repeat Again
Complete the Last Row
Complete the Last Row
| $ (x,y)$ | $\Delta x \text{ or }dx$ | $$ \frac{dy}{dx} $$ | $$ dx (\frac{dy}{dx})=dy $$ | $$(x+dx, y +dy) $$ |
| (1,0) | .5 | 0 | 0 | (1.5,0) |
| (1.5,0) | .5 | 1 | .5 | (2 , .5) |
| (2, .5) | .5 | 2 | 2* .51 | (2+.5, 1+.5) (2.5,1.5) |
| (2.5,1.5) | .5 | 3 | 3 *.5 1.5 | (2.5+.5,1.5+1.5) (3,3) |
The Red Curve
- shows the actual solution from the differential equation.
y = (x -1)2 and f(3) = 4 The Line Segments below the curve show the solution from Euler's Method. Line 1 connects (1,0) and (1.5,0) Line 2 connects (1.5,0) and (2,.5) Line 3 connects (2,.5) and (2.5,1.5) Line 4 connects (2.5,1.5) and (3,3) Here f(3) = 3 Again since the curve is concave up, our solution from Euler's method is less than the actual function. |
(Mouse over/click to enlarge image)
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Also, we can see the further we get form the initial point (1,0), the less accurate our approximation is. The lines with the arrow indicate the difference.
| How to Use the TI-89 |
| Mode- Graph should be differential equation |
| Select y=, you will see:
t0 = (place initial x value here) y1' = (differential equation goes here) yi1 = ( initial y value goes here) Note: When using the differential equation mode make sure you use y1 for y and t for x. |
| Now, while in the "y=" screen press the diamond |
| Change solution method to "Euler" and Fields to "SLPFLD" |
| Now select "window and change the step to the desired step size. Adjust the rest of the new window accordingly. (Disregard the last 3 options) |
| The graph will now show the slope field and Euler Approximation. |
| To check your solution, hit F3 and enter the desired x-value , the lower right corner will display the corresponding y-value. |
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