Find the special solution of y = y '= 0 when y' '+ (y)' ^ 2 = 1 and x = 0

There is no X-type
Let y '= P, Y "= PDP / dy
The original differential equation can be reduced to
pdp/dy+p^2=1
Separate variables
pdp/(p^2-1)=-dy
Two side integral
ln|p^2-1|=-2y+C
obtain
p^2=C'e^(-2y)+1
If x = 0, y = y '= 0, C' = - 1 can be obtained
Then p = ± √ [1-e ^ (- 2Y)]
That is dy / DX = ± √ [1-e ^ (- 2Y)]
Separate variables
dy/√[1-e^(-2y)]=±dx
A little bit
1/√[e^(2y)-1]d(e^y)=±dx
Two side integral
ln|e^y+√[e^(2y)-1]|=±x+C"
Under the initial condition x = 0, y = y '= 0, C "= 0 can be obtained
So the special solution of the equation is
ln|e^y+√[e^(2y)-1]|=±x
[the formula ∫ 1 / √ (x ^ 2-1) DX = ln| x + √ (x ^ 2-1) | + C is used]

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