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the equation is x-absent

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

i can not see how you did that

that's not quite right since
y'' y+ (y')^2
and we want
y'' y- (y')^2

let\[y'=p\]\[y''=pp'\]

can you please retype your equation it's confusing me

can you please retype your equation it's confusing me

can you please retype your equation it's confusing me

can you please retype your equation it's confusing me

can you please retype your equation it's confusing me

can you please retype your equation it's confusing me

can you please retype your equation it's confusing me

ok

\[y=y(x)\]\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]or\[y{y''}+1=(y')^2\]

\[ypp'+1=p^2\]

\[p'+{1 \over py} = {p \over y}\]

*correction: A = +e^C

well we have y''=...
so integration twice will produce the two constants

right?

bleah ... that equation 2 posts above should
stuff . cosh^2 + 1 = stuff . sinh^2

i got lost