# Euler's Method

• Sep 30th 2007, 03:29 PM
xfyz
Euler's Method
Use Euler's method with step size .2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = 1-xy, y(0)=0.

Can someone help me with this thanks
• Oct 1st 2007, 09:54 AM
TKHunny
This is plug-n-chug.

Use the given point and calculated slope to create a linear approximation of the next point.

Point: (0,0) Slope: 1 - (0)(0) = 1 Line: y = x Next Point: y = 0.2

Point: (0.2,0.2) Slope: 0.96 Line: y-0.2 = 0.96*(x-0.2) New Point: y = 0.392

And so on. In this case, you should see that it wanders off a bit. I get an absolute error of around 0.06 by the time you're done.

Note: You don't REALLY have to write the equation of the line in each step. That was just for your benefit. Use a little algebra to create more convenient formulations, maybe $y_{next}\;=\;y_{prev}\;+\;y'_{prev}*(x_{next}\;-\;x_{prev})$.
• Oct 1st 2007, 03:06 PM
xfyz
Quote:

Originally Posted by TKHunny
This is plug-n-chug.

Use the given point and calculated slope to create a linear approximation of the next point.

Point: (0,0) Slope: 1 - (0)(0) = 1 Line: y = x Next Point: y = 0.2

Point: (0.2,0.2) Slope: 0.96 Line: y-0.2 = 0.96*(x-0.2) New Point: y = 0.392

And so on. In this case, you should see that it wanders off a bit. I get an absolute error of around 0.06 by the time you're done.

Note: You don't REALLY have to write the equation of the line in each step. That was just for your benefit. Use a little algebra to create more convenient formulations, maybe $y_{next}\;=\;y_{prev}\;+\;y'_{prev}*(x_{next}\;-\;x_{prev})$.

sorry you just confused me with those. where are you getting y = x and the value y = .392 for the 2nd part..?