It

**isn't** equal to

. Where did you get that? You must be looking at an example that has nothing to do with this given problem. f(t, Yn(t)) is simply f applied to t and whatever you have just calculated Yn(t) to be!

In this case

. Since we are given that y(0)= 1, a reasonable starting approximation for y is the constant y= 1; Then we have

and, integrating

. In order to have y(0)= 1, we must have y(0)= (1/3)0+ C= 1 so C= 1 and

. Replacing the y in the differential equation with that,

and integrating,

. Again, y(0)= 1 gives C= 1. Take

and integrate again.