Question Number 209246 by Erico last updated on 05/Jul/24
$$\mathrm{Find}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{x}} \frac{\mathrm{dt}}{\mathrm{t}+\mathrm{e}^{\mathrm{f}\left(\mathrm{t}\right)} } \\ $$
Answered by mr W last updated on 06/Jul/24
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{{t}+{e}^{{f}\left({t}\right)} } \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{{x}+{e}^{{f}\left({x}\right)} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}+{e}^{{y}} } \\ $$$$\frac{{dx}}{{dy}}={x}+{e}^{{y}} \\ $$$$\frac{{dx}}{{dy}}−{x}={e}^{{y}} \:\:\leftarrow\:{D}.{E}.\:{of}\:{type}\:{y}'+{p}\left({x}\right){y}={q}\left({x}\right) \\ $$$$\Rightarrow{x}=\frac{\int{e}^{−{y}} {e}^{{y}} {dy}+{C}}{{e}^{−{y}} }=\left({y}+{C}\right){e}^{{y}} \\ $$$$\Rightarrow{xe}^{{C}} =\left({y}+{C}\right){e}^{{y}+{C}} \\ $$$$\Rightarrow{y}+{C}={W}\left({xe}^{{C}} \right)\:\:\leftarrow\:{Lambert}\:{W}\:{function} \\ $$$${or} \\ $$$${y}={f}\left({x}\right)={W}\left({kx}\right)−\mathrm{ln}\:{k} \\ $$$${f}\left(\mathrm{0}\right)={W}\left(\mathrm{0}\right)−\mathrm{ln}\:{k}=\mathrm{0}\:\Rightarrow\mathrm{ln}\:{k}=\mathrm{0}\:\Rightarrow{k}=\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)={W}\left({x}\right) \\ $$