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Question Number 182039 by liuxinnan last updated on 03/Dec/22

if f′′′(x)=f(x) find what is f(x)

$${if}\:{f}'''\left({x}\right)={f}\left({x}\right) \\ $$$${find}\:{what}\:{is}\:{f}\left({x}\right)\: \\ $$

Answered by FelipeLz last updated on 03/Dec/22

f′′′(x)−f(x) = 0 f(x) = e^(rx) → r^3 e^(rx) −e^(rx) = 0 e^(rx) (r^3 −1) = 0 e^(rx) ≠ 0 ∀x → r^3 −1 = 0 (r−1)(r^2 +r+1) = 0 r_1 = 1, r_2 = −(1/2)(1−(√3)i), r_3 = −(1/2)(1+(√3)i) f(x) = c_1 e^x +c_2 (1/( (√e^x )))cos(((√3)/2)x)+c_3 (1/( (√e^x )))sin(((√3)/2)x)

$${f}'''\left({x}\right)−{f}\left({x}\right)\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)\:=\:{e}^{{rx}} \:\rightarrow\:{r}^{\mathrm{3}} {e}^{{rx}} −{e}^{{rx}} \:=\:\mathrm{0} \\ $$$${e}^{{rx}} \left({r}^{\mathrm{3}} −\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${e}^{{rx}} \:\neq\:\mathrm{0}\:\forall{x}\:\rightarrow\:{r}^{\mathrm{3}} −\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${r}_{\mathrm{1}} \:=\:\mathrm{1},\:{r}_{\mathrm{2}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\sqrt{\mathrm{3}}{i}\right),\:{r}_{\mathrm{3}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}{i}\right) \\ $$$${f}\left({x}\right)\:=\:{c}_{\mathrm{1}} {e}^{{x}} +{c}_{\mathrm{2}} \frac{\mathrm{1}}{\:\sqrt{{e}^{{x}} }}\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)+{c}_{\mathrm{3}} \frac{\mathrm{1}}{\:\sqrt{{e}^{{x}} }}\mathrm{sin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right) \\ $$

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