Euler's method, Multiple choice does not match my answer.

Question

This is the original question.

Use Euler's method with h=0.2 to estimate y when x =1 if
$y' = (y^2-1) /2 $ and y(0) = 0
A. 7.690
B. 12.730
C. 13.504
D. 90.676

My answer follows.
n= 5, h= 0.2 a= x_0= 0, b=1, y0= 0
I'm using the formula
$y_{(n+1)} = y_n + h * y'$
and generated the following table.
Note this table shows the answer as $-0.47141$ because $y_{n+1}$ is on the previous line of the table below. However, this doesn't match any of the given answer choices. This is an employment test covering Advanced Placement Calculus BC, so this is all the context I have.

Where is the mistake?
Are there other variations called "Euler's method" which generate different answers?

x	h	y	y^2	y'	next y
0	0.2	0.00000	0.00000	-0.50000	-0.10000
0.2	0.2	-0.10000	0.01000	-0.49500	-0.19900
0.4	0.2	-0.19900	0.03960	-0.48020	-0.29504
0.6	0.2	-0.29504	0.08705	-0.45648	-0.38634
0.8	0.2	-0.38634	0.14925	-0.42537	* -0.47141 *
1	0.2	-0.47141	0.22223	-0.38889	-0.54919

Answer 1

There is something wrong with the problem as posed. The equilibria for $ \ y' = (y^2 - 1)/2 \ $ are $ \ y = \pm 1 \ \ , $ with $ \ y = -1 \ $ being the stable equilibrium (as the "flow" of your values for $ \ y \ $ suggests) and $ \ y = +1 \ $ is unstable. So it is only possible to reach large positive values of $ \ y \ $ for $ \ y(0) > 1 \ \ . $ My suspicion is that either the wrong initial value was given or the differential equation is not what was intended (the choices go with some other problem, or there is a significant typo, or the like).

Answer 2

You should get, for one row,

$$y(0.4) \approx y(0.2) + 0.2 \frac{y(0.2)^2-1}{2} \approx -0.1 + 0.2 \frac{(-0.1)^2 - 1}{2} \approx -0.199.$$

That said, the exact answer is $y(1) = \frac{1-e}{1+e} \approx -0.462$ so all of the large, positive multiple-choice answers are highly suspect, to say the least.