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Question Number 144608 by ArielVyny last updated on 26/Jun/21

find all aplication f in R→R f∈C^2 ∀x∈R. f′′(x)+f(−x)=x

$${find}\:{all}\:{aplication}\:{f}\:{in}\:\mathbb{R}\rightarrow\mathbb{R}\:\:{f}\in{C}^{\mathrm{2}} \\ $$$$\forall{x}\in\mathbb{R}.\:\:{f}''\left({x}\right)+{f}\left(−{x}\right)={x} \\ $$

Answered by Olaf_Thorendsen last updated on 27/Jun/21

f′′(x)+f(−x) = x (1) Let u(x) = f(x)+x (1) : u′′(x)+u(−x)+x = x u′′(x)+u(−x) = 0 u′′(−x)+u(x) = 0 −u′′′(−x)+u′(x) = 0 u^((4)) (−x)+u′′(x) = 0 u^((4)) (−x)−u(−x) = 0 u^((4)) (x)−u(x) = 0 r^4 −1 = 0 r = 1, i, −1, −i u(x) = C_1 e^x +C_2 e^(−x) +C_3 e^(ix) +C_4 e^(−ix) or : u(x) = acoshx+bsinhx+ccosx+dsinx f(x) = u(x)−x f(x) = acoshx+bsinhx+ccosx+dsinx−x f′′(x) = acoshx+bsinhx−ccosx−dsinx f(−x) = acoshx−bsinhx+ccosx−dsinx+x With (1) necessarily a = 0, d = 0 Finally, the genaral solution is : f(x) = αsinhx+βcosx−x

$${f}''\left({x}\right)+{f}\left(−{x}\right)\:=\:{x}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\left({x}\right)\:=\:{f}\left({x}\right)+{x} \\ $$$$\left(\mathrm{1}\right)\::\:{u}''\left({x}\right)+{u}\left(−{x}\right)+{x}\:=\:{x} \\ $$$${u}''\left({x}\right)+{u}\left(−{x}\right)\:=\:\mathrm{0} \\ $$$${u}''\left(−{x}\right)+{u}\left({x}\right)\:=\:\mathrm{0} \\ $$$$−{u}'''\left(−{x}\right)+{u}'\left({x}\right)\:=\:\mathrm{0} \\ $$$${u}^{\left(\mathrm{4}\right)} \left(−{x}\right)+{u}''\left({x}\right)\:=\:\mathrm{0} \\ $$$${u}^{\left(\mathrm{4}\right)} \left(−{x}\right)−{u}\left(−{x}\right)\:=\:\mathrm{0} \\ $$$${u}^{\left(\mathrm{4}\right)} \left({x}\right)−{u}\left({x}\right)\:=\:\mathrm{0} \\ $$$${r}^{\mathrm{4}} −\mathrm{1}\:=\:\mathrm{0} \\ $$$${r}\:=\:\mathrm{1},\:{i},\:−\mathrm{1},\:−{i} \\ $$$${u}\left({x}\right)\:=\:\mathrm{C}_{\mathrm{1}} {e}^{{x}} +\mathrm{C}_{\mathrm{2}} {e}^{−{x}} +\mathrm{C}_{\mathrm{3}} {e}^{{ix}} +\mathrm{C}_{\mathrm{4}} {e}^{−{ix}} \\ $$$$\mathrm{or}\:: \\ $$$${u}\left({x}\right)\:=\:{a}\mathrm{cosh}{x}+{b}\mathrm{sinh}{x}+{c}\mathrm{cos}{x}+{d}\mathrm{sin}{x} \\ $$$$ \\ $$$${f}\left({x}\right)\:=\:{u}\left({x}\right)−{x} \\ $$$${f}\left({x}\right)\:=\:{a}\mathrm{cosh}{x}+{b}\mathrm{sinh}{x}+{c}\mathrm{cos}{x}+{d}\mathrm{sin}{x}−{x} \\ $$$$ \\ $$$${f}''\left({x}\right)\:=\:{a}\mathrm{cosh}{x}+{b}\mathrm{sinh}{x}−{c}\mathrm{cos}{x}−{d}\mathrm{sin}{x} \\ $$$${f}\left(−{x}\right)\:=\:{a}\mathrm{cosh}{x}−{b}\mathrm{sinh}{x}+{c}\mathrm{cos}{x}−{d}\mathrm{sin}{x}+{x} \\ $$$$\mathrm{With}\:\left(\mathrm{1}\right)\:\mathrm{necessarily}\:{a}\:=\:\mathrm{0},\:{d}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Finally},\:\mathrm{the}\:\mathrm{genaral}\:\mathrm{solution}\:\mathrm{is}\:: \\ $$$${f}\left({x}\right)\:=\:\alpha\mathrm{sinh}{x}+\beta\mathrm{cos}{x}−{x} \\ $$

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