We want to solve
y′′=1+(y′)2
The usual thing is to replace y′=v and therefore we get
v′=1+v2.
This ODE can be solved by the usual methods, just use what you know. For example separation of variables i.e. solve for v
∫1+v2dv=∫dx
arctan(v)=x+C.
We get v(x)=tan(x+C) where C is some constant. Now resubstitute i.e.
y′=v=tan(x+C)
Just integrate both sides with respect to x i.e.
y(x)=∫tan(x+C) dx
This can be solved easily, for example use substitution s=cos(x+C)
∫tan(x+C) dx=−∫sds=−log(s)=−log(cos(x+C))
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