I have to use Euler's method on a problem..I don't really know what I'm doing...

This is exactly what my book says is the sequence for Euler's Method:

y(x_{i+1}) ≈ y_{i} _{+1} = y_{i} + hf(x_{i }, y_{i}), for i = 0, 1, 2, . . .

My quesion is what is f(x_{i }, y_{i}) ???

What is there a comma doing in there???

Here is the actual question that I have to answer:

Use Euler's methos with h = 0.1 to approximate y(1). Show the first 2 steps by hand.

y' = 2xy, y(0) = 1

hectictar Mar 29, 2017

#1**+2 **

Let me see if I remember this, hectictar.....

y' = f (x, y) = 2xy

The initial condition says that y(x_{0}) = y_{0} ...so..... y(0) = 1 and (x_{0}, y_{0}) = (0, 1)

So f( x0, y0 ) = 2(x_{0})(y_{0}) = 2(0)(1) = 0

And h = .1 [ this is known as the step size ]

So....we need to find (x_{1}, y_{1})

x_{1} = x_{0} + h = 0 + .1 = .1

y_{1} = y_{0} + h f(x_{0}, y_{0}) = 1 + .1 ( 0) = 1

So......our second point (x_{1}, y_{1}) is (.1, 1) and f (x_{1}, y_{1}) = 2(.1)(1) = .2

Now.......find (x_{2}, y_{2} )

x_{2} = x_{1} + h = .1 + .1 = .2

y_{2} = y_{1} + h f(x_{1}, y_{1}) = 1 + .1 ( .2) = 1.02

So.....our third point (x_{2}, y_{2}) is (.2, 1.02) and f (x_{2}, y_{2} ) = 2(.2)(1.02) = .408

This iteration continues until we get to x_{10} = 1

I assume you have some program that you can use to do this???...it's very time consuming, by hand!!

Anyway ...... Here's a pretty good explanation of the method :

http://calculuslab.deltacollege.edu/ODE/7-C-1/7-C-1-h-c.html

CPhill Mar 29, 2017

#2**+2 **

Thank you very much CPhill! This helped a lot!

( Those are the same answers as the book for y_{1 }and y_{2} )

I actually don't have a program that can do it...my friend said that there was a website that can do this, so I will try to find that. If not I *might* can come up with something that will work in excel....

hectictar Mar 30, 2017