Question Number 158687 by mahdipoor last updated on 07/Nov/21
$$\forall{x}\in{R}\:\:{f}\left({x}\right)={f}\:'\:\left({x}\right)\:\:\:\:\:\:{and}\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${prove}\:\:{f}\left({a}+{b}\right)={f}\left({a}\right)×{f}\left({b}\right) \\ $$
Answered by mr W last updated on 07/Nov/21
$${y}'=\frac{{dy}}{{dx}}={y} \\ $$$$\frac{{dy}}{{y}}={dx} \\ $$$$\int\frac{{dy}}{{y}}=\int{dx} \\ $$$$\mathrm{ln}\:{y}−\mathrm{ln}\:{C}={x} \\ $$$$\mathrm{ln}\:\frac{{y}}{{C}}={x} \\ $$$$\Rightarrow{y}={Ce}^{{x}} \\ $$$${y}\left(\mathrm{0}\right)={C}=\mathrm{1} \\ $$$$\Rightarrow{y}={f}\left({x}\right)={e}^{{x}} \\ $$$${f}\left({a}+{b}\right)={e}^{{a}+{b}} ={e}^{{a}} {e}^{{b}} ={f}\left({a}\right){f}\left({b}\right) \\ $$
Commented by mahdipoor last updated on 07/Nov/21
$${can}\:{you}\:{prove}\:{without}\:{use}\:{e}^{{x}} \:? \\ $$
Commented by mr W last updated on 07/Nov/21
$${f}\left({x}\right)={e}^{{x}} \:{is}\:{the}\:{only}\:{solution}\:{satifying} \\ $$$${f}'\left({x}\right)={f}\left({x}\right)\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1}. \\ $$