However, the derivative of is , not x+ y.
If you are trying to use Picard's method to solve this equation, as Ackbeet suggested, then you are doing it wrong.
From dy/dx= x+y we get dy= (x+y)dx so that . Of course, since we don't know y but we can approximate it. If we are given y(0)= 1 (which you don't say but is implied by "y0= 1") as a first approximation, assume y(x)= 1 for all x. Then dy/dx= x+1 so that dy= (x+1)dx and [tex]y= (1/2)x^2+ x+ C[tex] and y(0)= C= 1 gives , as you have.
But then, and, again, gives which is NOT what you have. I expect you forgot the added x.
Now, and gives .
One more time: and gives
I think I would conjecture now that
and I would attempt to prove that by "induction"
When n= 0, that says that y0= 1 which is true.
Now assume that, for some k, so that
just as we wish.