Question Number 202100 by sonukgindia last updated on 20/Dec/23
Answered by mr W last updated on 21/Dec/23
$${f}'={g} \\ $$$${f}''={g}'={f} \\ $$$${f}''−{f}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{r}=\pm\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$${f}\left(−{x}\right)={C}_{\mathrm{1}} {e}^{−{x}} +{C}_{\mathrm{2}} {e}^{{x}} ={f}\left({x}\right)={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\Rightarrow{C}_{\mathrm{1}} ={C}_{\mathrm{2}} ={C} \\ $$$$\Rightarrow{f}\left({x}\right)={C}\left({e}^{{x}} +{e}^{−{x}} \right) \\ $$$${g}\left({x}\right)={f}'\left({x}\right)={C}\left({e}^{{x}} −{e}^{−{x}} \right) \\ $$