# ODE5.4

${\frac {1}{y}}={\frac {y''}{1+y'^{2}}}$

Multiply both sides by $2y'\,$:

${\frac {2y'}{y}}={\frac {2y'y''}{1+y'^{2}}}$

Observe that the numerator of the term on the right side is the derivative (with respect to $x$) of the denominator of that term (indeed, that is why the factor $2y'$ was chosen -- a sort of integrating factor). Integrate both sides with respect to $x$; the result is

$2\ln |y|=\ln(1+y'^{2})+a\,$

(Alternatively, since the differential equation does not explicitly involve $x$, the substitution ${\frac {dy}{dx}}=p,{\frac {d^{2}y}{dx^{2}}}=p{\frac {dp}{dy}}$ converts the equation to a first-order equation in $p$ and $y$ which is separable. Integrating and back-substituting $dy/dx\,$ for $p\,$ yields the same result as given above, as of course it must.)

Exponentiating both sides, we get$y^{2}=c^{2}(1+y'^{2})\,$

where $c^{2}=e^{a}\,$ (we are free to write $c^{2}$ since $e^{a}>0$ for all real $a$).

Solving the above for $y'$,

$y'=\pm {\frac {{\sqrt {y^{2}-c^{2}}}}{c}}$

Divide both sides by ${\sqrt {y^{2}-c^{2}}}$ and integrate:

$\int {\frac {dy}{{\sqrt {y^{2}-c^{2}}}}}=\int \pm {\frac {dx}{c}}$ (where we have used $y'dx=dy$ in the integral on the left)

Use the trig substitution $y=c\,\sec(\theta )$, $dy=c\,\sec(\theta )\tan(\theta )\,d\theta$ to get

$\int \sec(\theta )\,d\theta =\pm \int {\frac {dx}{c}}\qquad$ Integrating,

$\ln {|\sec(\theta )+\tan(\theta )|}=\pm x/c+d\,$ or after back-substitution,

$\ln {|y/c+{\sqrt {y^{2}/c^{2}-1}}|}=\pm x/c+d\qquad$ Exponentiating both sides and multiplying both sides by $c$,

$y+{\sqrt {y^{2}-1}}=ce^{{\pm (x/c)+d}}\qquad$ Solving this for $y$,

$y={\frac {c\left(e^{{2(\pm (x/c)+d)}}+1\right)}{2e^{{\pm (x/c)+d}}}}=c\,{\frac {e^{{\pm (x/c)+d}}+e^{{\mp (x/c)-d}}}{2}}$

Either case may be written in the form $y=c\cosh(x/c\,+\,b)\,$ where $b$ and $c$ are arbitrary constants, as may be verified by substituting this solution into the original differential equation.