Question

y'' + (y')^2 + 1 = 0

Answer 1

are you looking for the double derivative of y?

Answer 2

I want to know what y is.

Answer 3

second order diff equations are out of my reach

Answer 4

Oh, alright, how about integrals, because I got it so far as to have an integral?

Answer 5

v=y'
v'=y''

v'+v^2+1=0 which is saying v'=-(v^2+1)

dv/dx=-(v^2+1)

dv/(v^2+1)=-dx

Answer 6

\[\text{ArcTan}[v]\]=-x

Answer 7

no y?stick issue

Answer 8

i believe it has something to do with coefficients and "e"

Answer 9

...but i could be seeing it wrong :)