I think I'm on the right track

since we can also say Euler's Method is yk=y0+f(y)t

therefore, yk=y0+(2-y)t

So to get y's I plug for each t...(1,.5,.25), 4 times..(y1,y2,y3,y4)

and for the x's can I use xn+1=xn+h or xn+1=xn+t? therefore I compute the same way as I did for y

so my table of the approximate values would look like this

for t=1.0

y t x

yk=y0+(2-y)t xn+1=xn+t

y1=1+(2-1)1=2 1.0 x1=0+1=1

y2=2+(2-2)1=2 1.0 x2=1+1=2

y3=2+(2-2)1=2 1.0 x3=2+1=3

y4=2+(2-2)1=2 1.0 x4=3+1=4

Then I would do the same for t=0.5 and t=0.25. Am I on the right track?

The way I would graph, would be x-vs-y

Now I don't know what the question is trying to ask with "what predictions can I make about the actual solutions to the initial-value problem"?

Do I have to differentiate? If yes then what?