Question Number 97891 by pranesh last updated on 10/Jun/20
Commented by mr W last updated on 10/Jun/20
$${y}={e}^{{x}} −\mathrm{1} \\ $$
Commented by bemath last updated on 10/Jun/20
$$\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\lambda\:=\:\pm\:\mathrm{1} \\ $$$$\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{Ae}^{\mathrm{x}} \:+\:\mathrm{Be}^{−\mathrm{x}} \\ $$$$\mathrm{particular}\:\mathrm{y}_{\mathrm{p}} \:=\:{ax}+{b} \\ $$$${y}'\:=\:{a}\:,\:\mathrm{y}''\:=\:\mathrm{0} \\ $$$$\mathrm{0}−\left({ax}+{b}\right)\:=\:\mathrm{1}\:\Leftrightarrow\begin{cases}{{a}\:=\:\mathrm{0}}\\{{b}=−\mathrm{1}}\end{cases} \\ $$$${y}_{{c}} \:=\:{A}\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{Be}^{−\mathrm{x}} −\mathrm{1} \\ $$$$\mathrm{x}\:=\:\mathrm{0}\:,\:\mathrm{y}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{0}\:=\:\mathrm{A}+\mathrm{B}−\mathrm{1}\:,\:\mathrm{A}+\mathrm{B}\:=\:\mathrm{1} \\ $$$$\mathrm{y}_{\mathrm{c}} \:=\:\mathrm{Ae}^{\mathrm{x}} +\:\left(\mathrm{1}−\mathrm{A}\right)\mathrm{e}^{−\mathrm{x}} −\mathrm{1} \\ $$
Commented by mr W last updated on 10/Jun/20
$${A}\:{must}\:{be}\:{equal}\:{to}\:\mathrm{1},\:{otherweise} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{y}=\mathrm{0}+\left(\mathrm{1}−{A}\right)\infty−\mathrm{1}=\infty\neq{a} \\ $$
Commented by bemath last updated on 10/Jun/20
$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{A}\:=\:\mathrm{1}? \\ $$
Commented by mr W last updated on 10/Jun/20
$${such}\:{that}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{y}\:\nrightarrow\:\infty,\:{the}\:{term} \\ $$$${e}^{−{x}} \:{must}\:{vanish},\:{therefore}\:\mathrm{1}−{A}=\mathrm{0}. \\ $$
Commented by bemath last updated on 10/Jun/20
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$