#### 2.12   ODE No. 12

$y'(x)+y(x)^2-1=0$ Mathematica : cpu = 0.0718015 (sec), leaf count = 34

$\left \{\left \{y(x)\to \frac {e^{2 x}-e^{2 c_1}}{e^{2 c_1}+e^{2 x}}\right \}\right \}$ Maple : cpu = 0.033 (sec), leaf count = 8

$\left \{ y \left ( x \right ) =\tanh \left ( x+{\it \_C1} \right ) \right \}$

Hand solution

\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

$y=\frac {e^{2x}-e^{2C_{1}}}{e^{2x}+e^{2C_{1}}}$