#### 2.60   ODE No. 60

$y'(x)-\frac {\sqrt {y(x)^2-1}}{\sqrt {x^2-1}}=0$ Mathematica : cpu = 0.11609 (sec), leaf count = 173

$\left \{\left \{y(x)\to -\frac {1}{2} e^{-c_1} \sqrt {2 e^{4 c_1} x^2+2 e^{4 c_1} \sqrt {(x-1) (x+1)} x+2 e^{2 c_1}-e^{4 c_1}+2 x^2-2 \sqrt {(x-1) (x+1)} x-1}\right \},\left \{y(x)\to \frac {1}{2} e^{-c_1} \sqrt {2 e^{4 c_1} x^2+2 e^{4 c_1} \sqrt {(x-1) (x+1)} x+2 e^{2 c_1}-e^{4 c_1}+2 x^2-2 \sqrt {(x-1) (x+1)} x-1}\right \}\right \}$ Maple : cpu = 0.013 (sec), leaf count = 29

$\left \{ \ln \left ( x+\sqrt {{x}^{2}-1} \right ) -\ln \left ( y \left ( x \right ) +\sqrt { \left ( y \left ( x \right ) \right ) ^{2}-1} \right ) +{\it \_C1}=0 \right \}$

Hand solution

\begin {equation} y^{\prime }=\pm \frac {\sqrt {y^{2}-1}}{\sqrt {x^{2}-1}}\tag {1} \end {equation}

Separable. For the positive case

\begin {align*} \frac {dy}{dx}\frac {1}{\sqrt {y^{2}-1}} & =\frac {1}{\sqrt {x^{2}-1}}\\ \frac {dy}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}} & =\frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}} \end {align*}

Integrating

$\int \frac {dy}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}}=\int \frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}}+C$

But $$\int \frac {dy}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}}=\tanh ^{-1}\frac {y}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}}=\ln \left ( y+\sqrt {y^{2}-1}\right )$$, hence

$\ln \left ( y+\sqrt {y^{2}-1}\right ) =\ln \left ( x+\sqrt {x^{2}-1}\right ) +C$

For the negative case

\begin {align*} \frac {dy}{dx}\frac {1}{\sqrt {y^{2}-1}} & =-\frac {1}{\sqrt {x^{2}-1}}\\ \frac {dy}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}} & =-\frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}} \end {align*}

Integrating

$\int \frac {dy}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}}=-\int \frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}}+C$

But $$\int \frac {dy}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}}=\tanh ^{-1}\frac {y}{\left ( y^{2}-1\right ) ^{\frac {1}{2}}}=\ln \left ( y+\sqrt {y^{2}-1}\right )$$, hence

$\ln \left ( y+\sqrt {y^{2}-1}\right ) =-\ln \left ( x+\sqrt {x^{2}-1}\right ) +C$

Therefore

$\ln \left ( y+\sqrt {y^{2}-1}\right ) =\pm \ln \left ( x+\sqrt {x^{2}-1}\right ) +C$