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

f i n d a l l a p l i c a t i o n f i n R \to R f \in C^{2}

\forall x \in R . f^{″} (x) + f (- x) = 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^{i x} + C_{4} e^{- i x}

or :

u (x) = a \cosh x + b \sinh x + c \cos x + d \sin x

f (x) = u (x) - x

f (x) = a \cosh x + b \sinh x + c \cos x + d \sin x - x

f^{″} (x) = a \cosh x + b \sinh x - c \cos x - d \sin x

f (- x) = a \cosh x - b \sinh x + c \cos x - d \sin x + x

With (1) necessarily a = 0, d = 0

Finally, the genaral solution is :

f (x) = α \sinh x + β \cos x - x