Euler Method - IVP

Apr 2010
65
2
Just to clarify that I am on the right track, for the following problem:


\(\displaystyle

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

Will the formula using euler method look like:

\(\displaystyle

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.

thanks in advance

 

CaptainBlack

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Just to clarify that I am on the right track, for the following problem:


\(\displaystyle

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

Will the formula using euler method look like:

\(\displaystyle

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

\)

, with h = 0.25



That is correct