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Question Number 196427 by Erico last updated on 24/Aug/23
If f(x)=∫^( x) _( 0) (dt/(t+e^(−f(t)) )), determine f(x)
$$\mathrm{If}\:\:{f}\left({x}\right)=\underset{\:\mathrm{0}} {\int}^{\:{x}} \frac{{dt}}{{t}+{e}^{−{f}\left({t}\right)} },\:\mathrm{determine}\:{f}\left({x}\right) \\ $$
Answered by witcher3 last updated on 25/Aug/23
f(0)=0 f′(x)=(1/(x+e^(−f(x)) )) ⇔f′(x)(x+e^(−f(x)) )=1 f(x)=y ⇔(dy/dx)(x+e^(−y) )=1 ⇔dy(x+e^(−y) )−dx=0 ⇔dy(x+e^(−y) )f(y)−f(y)dx=0,let f bee such thatf(y)=−f′(y)⇒f(y)=e^(−y) ⇒(x+e^(−y) )e^(−y) dy−e^(−y) dx=0 exacte Differential Equation Ψ(x,y)solution ∂_y Ψ=(x+e^(−y) )e^(−y) ⇒Ψ(x,y)=−xe^(−y) −(e^(−2y) /2)+f(x) ∂_x Ψ=−e^(−y) ⇒Ψ=−xe^(−y) +f(y) ⇒f′(x)−e^(−y) =−e^(−y) ⇒f(x)=c xe^(−y) +f′(y)=xe^(−y) +e^(−2y) ⇒f(y)=−(e^(−2y) /2) Ψ(x,y)=−xe^(−y) −(e^(−2y) /2)+c=0 implicite solution ⇒f′(x)=0⇒f=c,−xe^(−y) + Ψ(x,y)=−xe^(−y) −(e^(−2y) /2)+c Ψ(x,f(x))=−xe^(−f(x)) −(e^(−2f(x)) /2)=C −xt−(t^2 /2)=c ⇒t^2 +2xt−2c=0 t=((−2x−(√(4x^2 +8c)))/2) −ln(−x+(√(x^2 +2c)))=ln((((√(x^2 +2c))+x)/(2c)))=f(x)
$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{e}^{−\mathrm{f}\left(\mathrm{x}\right)} } \\ $$$$\Leftrightarrow\mathrm{f}'\left(\mathrm{x}\right)\left(\mathrm{x}+\mathrm{e}^{−\mathrm{f}\left(\mathrm{x}\right)} \right)=\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$\Leftrightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}+\mathrm{e}^{−\mathrm{y}} \right)=\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{dy}\left(\mathrm{x}+\mathrm{e}^{−\mathrm{y}} \right)−\mathrm{dx}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{dy}\left(\mathrm{x}+\mathrm{e}^{−\mathrm{y}} \right)\mathrm{f}\left(\mathrm{y}\right)−\mathrm{f}\left(\mathrm{y}\right)\mathrm{dx}=\mathrm{0},\mathrm{let}\:\mathrm{f}\:\mathrm{bee} \\ $$$$\mathrm{such}\:\mathrm{thatf}\left(\mathrm{y}\right)=−\mathrm{f}'\left(\mathrm{y}\right)\Rightarrow\mathrm{f}\left(\mathrm{y}\right)=\mathrm{e}^{−\mathrm{y}} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{e}^{−\mathrm{y}} \right)\mathrm{e}^{−\mathrm{y}} \mathrm{dy}−\mathrm{e}^{−\mathrm{y}} \mathrm{dx}=\mathrm{0} \\ $$$$\mathrm{exacte}\:\mathrm{Differential}\:\mathrm{Equation} \\ $$$$\Psi\left(\mathrm{x},\mathrm{y}\right)\mathrm{solution} \\ $$$$\partial_{\mathrm{y}} \Psi=\left(\mathrm{x}+\mathrm{e}^{−\mathrm{y}} \right)\mathrm{e}^{−\mathrm{y}} \Rightarrow\Psi\left(\mathrm{x},\mathrm{y}\right)=−\mathrm{xe}^{−\mathrm{y}} −\frac{\mathrm{e}^{−\mathrm{2y}} }{\mathrm{2}}+\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\partial_{\mathrm{x}} \Psi=−\mathrm{e}^{−\mathrm{y}} \Rightarrow\Psi=−\mathrm{xe}^{−\mathrm{y}} +\mathrm{f}\left(\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{f}'\left(\mathrm{x}\right)−\mathrm{e}^{−\mathrm{y}} =−\mathrm{e}^{−\mathrm{y}} \Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{c} \\ $$$$\mathrm{xe}^{−\mathrm{y}} +\mathrm{f}'\left(\mathrm{y}\right)=\mathrm{xe}^{−\mathrm{y}} +\mathrm{e}^{−\mathrm{2y}} \Rightarrow\mathrm{f}\left(\mathrm{y}\right)=−\frac{\mathrm{e}^{−\mathrm{2y}} }{\mathrm{2}} \\ $$$$\Psi\left(\mathrm{x},\mathrm{y}\right)=−\mathrm{xe}^{−\mathrm{y}} −\frac{\mathrm{e}^{−\mathrm{2y}} }{\mathrm{2}}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{implicite}\:\mathrm{solution} \\ $$$$\Rightarrow\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{f}=\mathrm{c},−\mathrm{xe}^{−\mathrm{y}} + \\ $$$$\Psi\left(\mathrm{x},\mathrm{y}\right)=−\mathrm{xe}^{−\mathrm{y}} −\frac{\mathrm{e}^{−\mathrm{2y}} }{\mathrm{2}}+\mathrm{c} \\ $$$$\Psi\left(\mathrm{x},\mathrm{f}\left(\mathrm{x}\right)\right)=−\mathrm{xe}^{−\mathrm{f}\left(\mathrm{x}\right)} −\frac{\mathrm{e}^{−\mathrm{2f}\left(\mathrm{x}\right)} }{\mathrm{2}}=\mathrm{C} \\ $$$$−\mathrm{xt}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{c} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{2xt}−\mathrm{2c}=\mathrm{0} \\ $$$$\mathrm{t}=\frac{−\mathrm{2x}−\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{8c}}}{\mathrm{2}} \\ $$$$−\mathrm{ln}\left(−\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2c}}\right)=\mathrm{ln}\left(\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2c}}+\mathrm{x}}{\mathrm{2c}}\right)=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$ \\ $$
Answered by mr W last updated on 05/Jul/24
f(x)=∫_0 ^x (dt/(t+e^(−f(t)) )) f′(x)=(1/(x+e^(−f(x)) )) (dy/dx)=(1/(x+e^(−y) )) (dx/dy)=x+e^(−y) (dx/dy)−x=e^(−y) ⇒x=((∫e^(−y) e^(−y) dy+(C/2))/e^(−y) )=((−e^(−2y) +C)/(2e^(−y) ))=(1/2)(Ce^y −e^(−y) ) ⇒Ce^y −e^(−y) =2x ⇒Ce^(2y) −2xe^y −1=0 ⇒e^y =((x±(√(x^2 +C)))/C) ⇒y=ln ((x±(√(x^2 +C)))/C)
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{{t}+{e}^{−{f}\left({t}\right)} } \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{{x}+{e}^{−{f}\left({x}\right)} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}+{e}^{−{y}} } \\ $$$$\frac{{dx}}{{dy}}={x}+{e}^{−{y}} \\ $$$$\frac{{dx}}{{dy}}−{x}={e}^{−{y}} \\ $$$$\Rightarrow{x}=\frac{\int{e}^{−{y}} {e}^{−{y}} {dy}+\frac{{C}}{\mathrm{2}}}{{e}^{−{y}} }=\frac{−{e}^{−\mathrm{2}{y}} +{C}}{\mathrm{2}{e}^{−{y}} }=\frac{\mathrm{1}}{\mathrm{2}}\left({Ce}^{{y}} −{e}^{−{y}} \right) \\ $$$$\Rightarrow{Ce}^{{y}} −{e}^{−{y}} =\mathrm{2}{x} \\ $$$$\Rightarrow{Ce}^{\mathrm{2}{y}} −\mathrm{2}{xe}^{{y}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{e}^{{y}} =\frac{{x}\pm\sqrt{{x}^{\mathrm{2}} +{C}}}{{C}} \\ $$$$\Rightarrow{y}=\mathrm{ln}\:\frac{{x}\pm\sqrt{{x}^{\mathrm{2}} +{C}}}{{C}} \\ $$

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