Question Number 88003 by Ar Brandon last updated on 07/Apr/20
![Determine all functions f[0,1]→Ω such that ∀x∈[0,1] f ′(x)+f(x)=f(0)+f(1)](Q88003.png)
$${Determine}\:{all}\:{functions}\:{f}\left[\mathrm{0},\mathrm{1}\right]\rightarrow\Omega \\ $$$${such}\:{that}\:\forall{x}\in\left[\mathrm{0},\mathrm{1}\right]\:{f}\:'\left({x}\right)+{f}\left({x}\right)={f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right) \\ $$
Commented by mr W last updated on 07/Apr/20
$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)={k} \\ $$$${f}'\left({x}\right)+{f}\left({x}\right)={k} \\ $$$$\frac{{dy}}{{y}−{k}}=−{dx} \\ $$$$\mathrm{ln}\:\mid{y}−{k}\mid=−{x}+{C} \\ $$$${y}={k}+{ce}^{−{x}} \\ $$$${f}\left(\mathrm{0}\right)={k}+{c} \\ $$$${f}\left(\mathrm{1}\right)={k}+\frac{{c}}{{e}} \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)={k}+{c}+{k}+\frac{{c}}{{e}}={k} \\ $$$${k}=−{c}\left(\mathrm{1}+\frac{\mathrm{1}}{{e}}\right) \\ $$$$\Rightarrow{y}={f}\left({x}\right)={c}\left({e}^{−{x}} −\mathrm{1}−\frac{\mathrm{1}}{{e}}\right) \\ $$
Commented by Ar Brandon last updated on 07/Apr/20
$${Thanks}\: \\ $$