Question Number 7245 by 2402@gmail.com last updated on 18/Aug/16
$${Q}\mathrm{1}.{if}\:{y}^{\mathrm{1}/{n}} +{y}^{−\mathrm{1}/{n}} \:=\mathrm{2}\:{find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:. \\ $$$$ \\ $$$$ \\ $$
Commented by Yozzia last updated on 19/Aug/16
$${y}^{{a}} +{y}^{−{a}} ={c}\:\:\:\left({y}\neq\mathrm{0},\:{c}\neq\mathrm{0},\:{a}>\mathrm{0}\right)\: \\ $$$${Implicit}\:{differentiation}\:{gives} \\ $$$${ay}^{{a}−\mathrm{1}} {y}'−{ay}^{−{a}−\mathrm{1}} {y}'=\mathrm{0} \\ $$$${y}'\left({ay}^{{a}−\mathrm{1}} −{ay}^{−{a}−\mathrm{1}} \right)=\mathrm{0}\:\:\:\left(\ast\right) \\ $$$${Further}\:{implicit}\:{differentiation}\:{gives} \\ $$$${y}''\left({ay}^{{a}−\mathrm{1}} −{ay}^{−{a}−\mathrm{1}} \right)+{y}'\left({a}\left({a}−\mathrm{1}\right){y}^{{a}−\mathrm{2}} {y}'+{a}\left({a}+\mathrm{1}\right){y}^{−{a}−\mathrm{2}} {y}'\right)=\mathrm{0} \\ $$$${y}''=−\frac{{a}\left({y}'\right)^{\mathrm{2}} {y}^{−\mathrm{2}} \left(\left({a}−\mathrm{1}\right){y}^{{a}} +\left({a}+\mathrm{1}\right){y}^{−{a}} \right)}{{ay}^{−\mathrm{1}} \left({y}^{{a}} −{y}^{−{a}} \right)} \\ $$$${y}''=−\frac{\left({y}'\right)^{\mathrm{2}} {y}^{−\mathrm{1}} \left({a}\left({y}^{{a}} +{y}^{−{a}} \right)−{y}^{{a}} +{y}^{−{a}} \right)}{{y}^{{a}} −{y}^{−{a}} } \\ $$$${y}''=−\frac{\left({y}'\right)^{\mathrm{2}} {y}^{−\mathrm{1}} \left({ac}−{y}^{{a}} +{y}^{−{a}} \right)}{{y}^{{a}} −{y}^{−{a}} } \\ $$$${If}\:{y}'=\mathrm{0}\:{from}\:\left(\ast\right),\:\Rightarrow{y}''=\mathrm{0}. \\ $$$${If}\:{y}'\neq\mathrm{0}\Rightarrow{y}^{{a}} ={y}^{−{a}} \Rightarrow\:{y}''\:{is}\:{undefined}. \\ $$