Question Number 30420 by abdo imad last updated on 22/Feb/18
$${integrate}\:{y}^{''} =\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\left({y}^{'} \right)^{\mathrm{2}} }\:\:\:\:\:. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
$$\frac{{dy}'}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}'}{\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }}=\frac{{dx}}{\mathrm{2}} \\ $$$$\int\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$${t}={sinh}\left({u}\right)\Rightarrow{dt}={cosh}\left({u}\right){du}=\sqrt{\mathrm{1}+{sinh}^{\mathrm{2}} {u}}{du} \\ $$$${du}=\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$\int\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}=\int{du}={u}+\lambda={sinh}^{−\mathrm{1}} \left({t}\right)+\lambda \\ $$$${so}\:\int\frac{{dy}'}{\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }}={sinh}^{−\mathrm{1}} \left({y}'\right)=\frac{{x}}{\mathrm{2}}+{C}_{\mathrm{1}} \\ $$$${y}'={sinh}\left(\frac{{x}}{\mathrm{2}}+{C}_{\mathrm{1}} \right) \\ $$$${dy}={sinh}\left(\frac{{x}}{\mathrm{2}}+{C}_{\mathrm{1}} \right){dx} \\ $$$${y}=\mathrm{2}{cosh}\left(\frac{{x}}{\mathrm{2}}+{C}_{\mathrm{1}} \right)+{C}_{\mathrm{2}} \\ $$