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Question Number 196427 by Erico last updated on 24/Aug/23

$$\mathrm{If}f\left(x\right)={\underset{0}{\int }}^x\frac{dt}{t+e^{-f\left(t\right)}},\mathrm{determine}f\left(x\right)$$

Answered by witcher3 last updated on 25/Aug/23

$$\mathrm{f}\left(0\right)=0$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}+{\mathrm{e}}^{-\mathrm{f}\left(\mathrm{x}\right)}}$$$$\iff {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)(\mathrm{x}+{\mathrm{e}}^{-\mathrm{f}\left(\mathrm{x}\right)})=1$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y}$$$$\iff \frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{x}+{\mathrm{e}}^{-\mathrm{y}})=1$$$$\iff \mathrm{dy}(\mathrm{x}+{\mathrm{e}}^{-\mathrm{y}})-\mathrm{dx}=0$$$$\iff \mathrm{dy}(\mathrm{x}+{\mathrm{e}}^{-\mathrm{y}})\mathrm{f}\left(\mathrm{y}\right)-\mathrm{f}\left(\mathrm{y}\right)\mathrm{dx}=0,\mathrm{let}\mathrm{f}\mathrm{bee}$$$$\mathrm{such}\mathrm{thatf}\left(\mathrm{y}\right)=-{\mathrm{f}}^{\prime }\left(\mathrm{y}\right)\Rightarrow \mathrm{f}\left(\mathrm{y}\right)={\mathrm{e}}^{-\mathrm{y}}$$$$\Rightarrow (\mathrm{x}+{\mathrm{e}}^{-\mathrm{y}}){\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}-{\mathrm{e}}^{-\mathrm{y}}\mathrm{dx}=0$$$$\mathrm{exacte}\mathrm{Differential}\mathrm{Equation}$$$$\mathrm{\Psi }(\mathrm{x},\mathrm{y})\mathrm{solution}$$$${\partial }_{\mathrm{y}}\mathrm{\Psi }=(\mathrm{x}+{\mathrm{e}}^{-\mathrm{y}}){\mathrm{e}}^{-\mathrm{y}}\Rightarrow \mathrm{\Psi }(\mathrm{x},\mathrm{y})=-{\mathrm{xe}}^{-\mathrm{y}}-\frac{{\mathrm{e}}^{-2\mathrm{y}}}{2}+\mathrm{f}\left(\mathrm{x}\right)$$$${\partial }_{\mathrm{x}}\mathrm{\Psi }=-{\mathrm{e}}^{-\mathrm{y}}\Rightarrow \mathrm{\Psi }=-{\mathrm{xe}}^{-\mathrm{y}}+\mathrm{f}\left(\mathrm{y}\right)$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)-{\mathrm{e}}^{-\mathrm{y}}=-{\mathrm{e}}^{-\mathrm{y}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{c}$$$${\mathrm{xe}}^{-\mathrm{y}}+{\mathrm{f}}^{\prime }\left(\mathrm{y}\right)={\mathrm{xe}}^{-\mathrm{y}}+{\mathrm{e}}^{-2\mathrm{y}}\Rightarrow \mathrm{f}\left(\mathrm{y}\right)=-\frac{{\mathrm{e}}^{-2\mathrm{y}}}{2}$$$$\mathrm{\Psi }(\mathrm{x},\mathrm{y})=-{\mathrm{xe}}^{-\mathrm{y}}-\frac{{\mathrm{e}}^{-2\mathrm{y}}}{2}+\mathrm{c}=0$$$$\mathrm{implicite}\mathrm{solution}$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=0\Rightarrow \mathrm{f}=\mathrm{c},-{\mathrm{xe}}^{-\mathrm{y}}+$$$$\mathrm{\Psi }(\mathrm{x},\mathrm{y})=-{\mathrm{xe}}^{-\mathrm{y}}-\frac{{\mathrm{e}}^{-2\mathrm{y}}}{2}+\mathrm{c}$$$$\mathrm{\Psi }(\mathrm{x},\mathrm{f}\left(\mathrm{x}\right))=-{\mathrm{xe}}^{-\mathrm{f}\left(\mathrm{x}\right)}-\frac{{\mathrm{e}}^{-2\mathrm{f}\left(\mathrm{x}\right)}}{2}=\mathrm{C}$$$$-\mathrm{xt}-\frac{{\mathrm{t}}^2}{2}=\mathrm{c}$$$$\Rightarrow {\mathrm{t}}^2+2\mathrm{xt}-2\mathrm{c}=0$$$$\mathrm{t}=\frac{-2\mathrm{x}-\sqrt{{4\mathrm{x}}^2+8\mathrm{c}}}{2}$$$$-\ln (-\mathrm{x}+\sqrt{{\mathrm{x}}^2+2\mathrm{c}})=\ln \left(\frac{\sqrt{{\mathrm{x}}^2+2\mathrm{c}}+\mathrm{x}}{2\mathrm{c}}\right)=\mathrm{f}\left(\mathrm{x}\right)$$$$$$

Answered by mr W last updated on 05/Jul/24

$$f\left(x\right)={\int }_0^x\frac{dt}{t+e^{-f\left(t\right)}}$$$$f^{\prime }\left(x\right)=\frac{1}{x+e^{-f\left(x\right)}}$$$$\frac{dy}{dx}=\frac{1}{x+e^{-y}}$$$$\frac{dx}{dy}=x+e^{-y}$$$$\frac{dx}{dy}-x=e^{-y}$$$$\Rightarrow x=\frac{\int e^{-y}{e}^{-y}dy+\frac{C}{2}}{e^{-y}}=\frac{-e^{-2y}+C}{2e^{-y}}=\frac{1}{2}({Ce}^y-e^{-y})$$$$\Rightarrow {Ce}^y-e^{-y}=2x$$$$\Rightarrow {Ce}^{2y}-2{xe}^y-1=0$$$$\Rightarrow e^y=\frac{x\pm \sqrt{x^2+C}}{C}$$$$\Rightarrow y=\ln \frac{x\pm \sqrt{x^2+C}}{C}$$

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