4
$\begingroup$

I have to do this take home test for school before break ends and it is on Slope Fields and Euler's Method. I'm having trouble understanding and/or working with Euler's Method.

These are the two problems I am stuck on:

2.$\text{}$ Use Euler's Method with $\Delta x=0.1$, $\dfrac{dy}{dx}=2x-y$ and $y=0$ when $x=1$ to find the value of $y$ when $x=1.3$.

A) 0.6      B) 0.2      C) 0.4      D) 0.8

3.$\text{}$ Find the first three approximations (by hand) $y_1$, $y_2$, $y_3$ using Euler's Method for the initial value problem

$\frac{dy}{dx}=1+y,\quad y(0)=1.$

(original image)

Thanks for any help or tips in advance!

  • 0
    Hint: can you watch a lecture on the method and then show us your work so we don't outright provide answers? For example, http://www.youtube.com/watch?v=nYXclUjpFaw. There are others like that too. For the second question, can you follow the [Wiki Example](http://en.wikipedia.org/wiki/Euler_method#Example)? Regards2012-12-28

1 Answers 1

4

Euler's method in essence is just following the directional field of a differential equation by taking steps through it.

Here we have a simple case of $\frac{\textrm dy}{\textrm dx}=f(x,y)$.

Recall the intuitive idea of the derivative of a function as a function which gives the slope of the tangent to that function. Then by simple linear approximation, starting from a point $(x_0,y_0)$, we get the next point by $x_1=x+\Delta x$, $y_1=y_0+\Delta y\approx y_0+\frac{\textrm dy}{\textrm dx}(x_0,y_0)\Delta x$, where $\Delta x$ is a given in the question. Then we note that $\frac{\textrm d y}{\textrm d x}(x,y)=f(x,y)$, because this is a given.

So to conclude, we start with a point $(x_0,y_0)$, and for every point we get the next point by $x_{n+1}=x_n+\Delta x$ and $y_{n+1}=y_n+f(x_n,y_n)\Delta x$. Simply do this for the given $f(x,y)$ and $\Delta x$ and you should be fine.

I know the point of this exercise is to learn Euler's method, but you should be able to solve this equation without too much difficulty, it is a first order linear differential equation.

I hope this was clear enough, else just post a comment.

  • 0
    No problem, but that was a bit fast to have read it all I think. Have a good read of it and decide if you're still as happy with it then :). Like I said just post a comment otherwise.2012-12-28