given y'(t) = g (y(t),t), y(0)=y₀ what is y(b)?

Simplest:

y(t) y(t_a) + y'(t_a) (t-t_a)

but we know y'(t_a) = g(y(t_a),ta)

y(t) y(t_a) + g(y(t_a),t_a) (t-t_a)

Forward Euler Method:
y(t_n+1) = y(t_n) + g(y(t_n),t_n) (t_n+1-t_n)

Keeping the next term in the Taylor series:

y(t) y(t_a) + y'(t_a) (t-t_a) + (1/2)y''(t_a) (t-t_a)²

we know y'(t_a) = g(y(t_a),t_a)

and we can approximate the next derivative numerically:

y"(t_a) = g'(y(t_a),t_a) [ g(y(t),t) - g(y(t_a),t_a) ]/(t-t_a)

y(t) y(t_a) + g(y(t_a),t_a) (t-t_a) + (1/2){ [ g(y(t),t) - g(y(t_a),t_a) ]/(t-t_a) }(t-t_a)²

which simplifies to:

y(t) y(t_a) + (1/2) [ g(y(t),t) + g(y(t_a),t_a) ] (t - t_a)

Adams-Moulton (AM2) algorithm:
y(t_n+1) = y(t_n) + (1/2) [ g(y(t_n+1),t_n+1) + g(y(t_n),t_n) ] (t_n+1 - t_n)

(unfortunately this implicit equation usually has to be solved with a Root - Finder)