Question Number 126791 by BHOOPENDRA last updated on 24/Dec/20
$${yy}''−\left({y}'\right)^{\mathrm{2}} ={e}^{{ax}\:} \:{find}\:{genral}\:{solution}? \\ $$
Commented by BHOOPENDRA last updated on 24/Dec/20
$${help}\:{me}\:{out}\:{this}? \\ $$
Answered by Olaf last updated on 26/Dec/20
$${yy}''−{y}'^{\mathrm{2}} \:=\:{e}^{{ax}} \:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{y}\:=\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ax}} {u}\:\left(\mathrm{2}\right) \\ $$$${y}'\:=\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ax}} \left({u}'+\frac{\mathrm{1}}{\mathrm{2}}{au}\right) \\ $$$${y}''\:=\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ax}} \left({u}''+{au}'+\frac{\mathrm{1}}{\mathrm{4}}{a}^{\mathrm{2}} {u}\right) \\ $$$$\left(\mathrm{1}\right)\::\:\left({u}''+{au}'+\frac{\mathrm{1}}{\mathrm{4}}{a}^{\mathrm{2}} {u}\right){u}−\left({u}'+\frac{\mathrm{1}}{\mathrm{2}}{au}\right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${u}''{u}−{u}'^{\mathrm{2}} \:=\:\mathrm{1}\:\left(\mathrm{3}\right) \\ $$$${u}'''{u}+{u}''{u}'−\mathrm{2}{u}'{u}''\:=\:\mathrm{0} \\ $$$${u}'''{u}−{u}'{u}''\:=\:\mathrm{0}\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{4}\right)\boldsymbol{\div}{uu}''\::\:\frac{{u}'''}{{u}''}−\frac{{u}'}{{u}}\:=\:\mathrm{0} \\ $$$$\mathrm{ln}\mid{u}''\mid−\mathrm{ln}\mid{u}\mid\:=\:\mathrm{C}_{\mathrm{1}} \\ $$$${u}''\:=\:\mathrm{C}_{\mathrm{2}} {u} \\ $$$${u}\:=\:\alpha{e}^{\sqrt{\mathrm{C}_{\mathrm{2}} }{x}} +\beta{e}^{−\sqrt{\mathrm{C}_{\mathrm{2}} }{x}} \:=\:\alpha{e}^{\mathrm{C}_{\mathrm{3}} {x}} +\beta{e}^{−\mathrm{C}_{\mathrm{3}} {x}} \\ $$$${u}'\:=\:\alpha\mathrm{C}_{\mathrm{3}} {e}^{\mathrm{C}_{\mathrm{3}} {x}} −\beta\mathrm{C}_{\mathrm{3}} {e}^{−\mathrm{C}_{\mathrm{3}} {x}} \\ $$$${u}''\:=\:\alpha\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} {e}^{\mathrm{C}_{\mathrm{3}} {x}} +\beta\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} {e}^{−\mathrm{C}_{\mathrm{3}} {x}} \\ $$$$\left(\mathrm{3}\right)\::\:\left(\alpha\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} {e}^{\mathrm{C}_{\mathrm{3}} {x}} +\beta\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} {e}^{−\mathrm{C}_{\mathrm{3}} {x}} \right)\left(\alpha{e}^{\mathrm{C}_{\mathrm{3}} {x}} +\beta{e}^{−\mathrm{C}_{\mathrm{3}} {x}} \right)−\left(\alpha{e}^{\mathrm{C}_{\mathrm{3}} {x}} −\beta{e}^{−\mathrm{C}_{\mathrm{3}} {x}} \right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\alpha^{\mathrm{2}} \mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} −\alpha^{\mathrm{2}} \:=\:\mathrm{0}\:\left(\mathrm{5}\right) \\ $$$$\beta^{\mathrm{2}} \mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} −\beta^{\mathrm{2}} \:=\:\mathrm{0}\:\left(\mathrm{6}\right) \\ $$$$\mathrm{4}\alpha\beta\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:=\:\mathrm{1}\:\left(\mathrm{7}\right) \\ $$$$\left(\mathrm{5}\right)\:\mathrm{and}\:\left(\mathrm{6}\right)\::\:\mathrm{C}_{\mathrm{3}} \:=\:\pm\mathrm{1} \\ $$$$\left(\mathrm{7}\right)\::\:\beta\:=\:\frac{\mathrm{1}}{\mathrm{4}\alpha} \\ $$$${u}\:=\:\alpha{e}^{{x}} +\frac{\mathrm{1}}{\mathrm{4}\alpha}{e}^{−{x}} \:\mathrm{or}\:\alpha{e}^{−{x}} +\frac{\mathrm{1}}{\mathrm{4}\alpha}{e}^{{x}} \\ $$$$\left(\mathrm{2}\right)\::\:{y}\:=\:{e}^{\frac{\mathrm{1}}{\mathrm{4}}{ax}} \left(\alpha{e}^{{x}} +\frac{\mathrm{1}}{\mathrm{4}\alpha}{e}^{−{x}} \right)\:\mathrm{or}\:{e}^{\frac{\mathrm{1}}{\mathrm{4}}{ax}} \left(\alpha{e}^{−{x}} +\frac{\mathrm{1}}{\mathrm{4}\alpha}{e}^{{x}} \right) \\ $$
Commented by BHOOPENDRA last updated on 26/Dec/20
$${thanks}\:{alot}\:{sir} \\ $$