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Question-202100




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=0  r^2 −1=0 ⇒r=±1  ⇒f(x)=C_1 e^x +C_2 e^(−x)   f(−x)=C_1 e^(−x) +C_2 e^x =f(x)=C_1 e^x +C_2 e^(−x)   ⇒C_1 =C_2 =C  ⇒f(x)=C(e^x +e^(−x) )  g(x)=f′(x)=C(e^x −e^(−x) )
$${f}'={g} \\ $$$${f}''={g}'={f} \\ $$$${f}''−{f}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{r}=\pm\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$${f}\left(−{x}\right)={C}_{\mathrm{1}} {e}^{−{x}} +{C}_{\mathrm{2}} {e}^{{x}} ={f}\left({x}\right)={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\Rightarrow{C}_{\mathrm{1}} ={C}_{\mathrm{2}} ={C} \\ $$$$\Rightarrow{f}\left({x}\right)={C}\left({e}^{{x}} +{e}^{−{x}} \right) \\ $$$${g}\left({x}\right)={f}'\left({x}\right)={C}\left({e}^{{x}} −{e}^{−{x}} \right) \\ $$

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