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Question Number 202100 by sonukgindia last updated on 20/Dec/23

Answered by mr W last updated on 21/Dec/23

$$f^{\prime }=g$$$$f^{″}=g^{\prime }=f$$$$f^{″}-f=0$$$$r^2-1=0\Rightarrow r=\pm 1$$$$\Rightarrow f\left(x\right)=C_1{e}^x+C_2{e}^{-x}$$$$f(-x)=C_1{e}^{-x}+C_2{e}^x=f\left(x\right)=C_1{e}^x+C_2{e}^{-x}$$$$\Rightarrow C_1=C_2=C$$$$\Rightarrow f\left(x\right)=C(e^x+e^{-x})$$$$g\left(x\right)=f^{\prime }\left(x\right)=C(e^x-e^{-x})$$

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