by **Norwin** » Thu Sep 28, 2017 2:22 pm

I doubt very so much that a closed type resolution exists. It might, but i'd be amazed. However it lends itself fantastically for numerical fixing. 1) y' is constantly +ve so the operate is certainly not decreasing 2) as x --> ±?, y' --> zero So y(x) may just seem similar to the erf(x) perform. *** good truely I feel the reply is y(x) = 0 identically. If y'(zero) = zero, then y certainly not will get round to relocating from y(zero) = zero. **** For other preliminary conditions, y(x) will resemble the error function, but it'll be contained between consecutive values n?. That is for the reason that each time y = n?, y' = zero, and as a result y=n? will constantly be an aymptote to the operate.