\[ y'(x)+y(x)^2-1=0 \] ✓ Mathematica : cpu = 0.0730206 (sec), leaf count = 34
\[\left \{\left \{y(x)\to \frac {e^{2 x}-e^{2 c_1}}{e^{2 x}+e^{2 c_1}}\right \}\right \}\] ✓ Maple : cpu = 0.047 (sec), leaf count = 8
\[ \left \{ y \left ( x \right ) =\tanh \left ( x+{\it \_C1} \right ) \right \} \]
\begin {align} \frac {dy}{dx}+y^{2}\left ( x\right ) -1 & =0\nonumber \\ \frac {dy}{dx} & =1-y^{2}\left ( x\right ) \tag {1} \end {align}
This is separable. Hence
\begin {align*} \frac {dy}{dx}\frac {1}{1-y^{2}\left ( x\right ) } & =1\\ \frac {dy}{1-y^{2}\left ( x\right ) } & =dx \end {align*}
Integrating\[ \int \frac {dy}{1-y^{2}\left ( x\right ) }=x+C \] Using \(\int \frac {1}{a+by^{2}}dy=\frac {\sqrt {-\frac {a}{b}}\tanh ^{-1}\left ( \frac {y}{\sqrt {\frac {-a}{b}}}\right ) }{a}\) and since \(a=1,b=-1\), then \(\int \frac {dy}{1-y^{2}\left ( x\right ) }=\tanh ^{-1}\left ( y\right ) \) and the above becomes
\[ \tanh ^{-1}\left ( y\right ) =x+C \]
Therefore
\begin {equation} y=\tanh \left ( x+C\right ) \tag {2} \end {equation}
In terms of exponential, since \(\tanh \left ( u\right ) =\frac {e^{u}-e^{-u}}{e^{u}+e^{-u}}\) then (2) can also be written as
\[ y=\frac {e^{x+C}-e^{-\left ( x+C\right ) }}{e^{x+C}+e^{-\left ( x+C\right ) }}=\frac {e^{x}e^{C}-e^{-x}e^{-C}}{e^{x}e^{C}+e^{-x}e^{-C}}\]
Multiplying numerator and denominator by \(e^{-C}e^{x}\)
\[ y=\frac {e^{2x}-e^{-2C}}{e^{2x}+e^{-2C}}\]
To get same answer as Mathematica, since \(C\) is constant, let \(C_{1}=-C\), then
\[ y=\frac {e^{2x}-e^{2C_{1}}}{e^{2x}+e^{2C_{1}}}\]