Euler Method - IVP

olski1

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.

CaptainBlack

MHF Hall of Fame
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