# Euler Method - IVP

• May 15th 2010, 12:24 AM
olski1
Euler Method - IVP
Just to clarify that I am on the right track, for the following problem:

$

\frac{dy}{dx} = (x^2y-2e^x)
$

Will the formula using euler method look like:

$

y_{n+1} = y_{n} + h({x_{n}}^2y_{n}-2e^{x_{n}})

$

, with h = 0.25

I was just unsure if this is how you do it for exponentials.

• May 15th 2010, 02:14 AM
CaptainBlack
Quote:

Originally Posted by olski1
Just to clarify that I am on the right track, for the following problem:

$

\frac{dy}{dx} = (x^2y-2e^x)
$

Will the formula using euler method look like:

$

y_{n+1} = y_{n} + h({x_{n}}^2y_{n}-2e^{x_{n}})

$

, with h = 0.25

That is correct