\(\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