ODE5.4

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

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 $2 y'$ 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 $\frac{dy}{dx} = p, \frac{d^2y}{dx^2} = p\frac{dp}{dy}$ 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 $y' d x = d y$ 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.

Ordinary Differential Equations

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ODE5.4

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