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

If f (x) = {\int_{0}}^{x} \frac{d t}{t + e^{- f (t)}}, determine f (x)

Answered by witcher3 last updated on 25/Aug/23

f (0) = 0

f^{'} (x) = \frac{1}{x + e^{- f (x)}}

\Leftrightarrow f^{'} (x) (x + e^{- f (x)}) = 1

f (x) = y

\Leftrightarrow \frac{dy}{dx} (x + e^{- y}) = 1

\Leftrightarrow dy (x + e^{- y}) - dx = 0

\Leftrightarrow dy (x + e^{- y}) f (y) - f (y) dx = 0, let f bee

such thatf (y) = - f^{'} (y) \Rightarrow f (y) = e^{- y}

\Rightarrow (x + e^{- y}) e^{- y} dy - e^{- y} dx = 0

exacte Differential Equation

Ψ (x, y) solution

\partial_{y} Ψ = (x + e^{- y}) e^{- y} \Rightarrow Ψ (x, y) = - {xe}^{- y} - \frac{e^{- 2 y}}{2} + f (x)

\partial_{x} Ψ = - e^{- y} \Rightarrow Ψ = - {xe}^{- y} + f (y)

\Rightarrow f^{'} (x) - e^{- y} = - e^{- y} \Rightarrow f (x) = c

{xe}^{- y} + f^{'} (y) = {xe}^{- y} + e^{- 2 y} \Rightarrow f (y) = - \frac{e^{- 2 y}}{2}

Ψ (x, y) = - {xe}^{- y} - \frac{e^{- 2 y}}{2} + c = 0

implicite solution

\Rightarrow f^{'} (x) = 0 \Rightarrow f = c, - {xe}^{- y} +

Ψ (x, y) = - {xe}^{- y} - \frac{e^{- 2 y}}{2} + c

Ψ (x, f (x)) = - {xe}^{- f (x)} - \frac{e^{- 2 f (x)}}{2} = C

- xt - \frac{t^{2}}{2} = c

\Rightarrow t^{2} + 2 xt - 2 c = 0

t = \frac{- 2 x - \sqrt{{4 x}^{2} + 8 c}}{2}

- \ln (- x + \sqrt{x^{2} + 2 c}) = \ln (\frac{\sqrt{x^{2} + 2 c} + x}{2 c}) = f (x)

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

f (x) = \int_{0}^{x} \frac{d t}{t + e^{- f (t)}}

f^{'} (x) = \frac{1}{x + e^{- f (x)}}

\frac{d y}{d x} = \frac{1}{x + e^{- y}}

\frac{d x}{d y} = x + e^{- y}

\frac{d x}{d y} - x = e^{- y}

\Rightarrow x = \frac{\int e^{- y} e^{- y} d y + \frac{C}{2}}{e^{- y}} = \frac{- e^{- 2 y} + C}{2 e^{- y}} = \frac{1}{2} ({C e}^{y} - e^{- y})

\Rightarrow {C e}^{y} - e^{- y} = 2 x

\Rightarrow {C e}^{2 y} - 2 {x e}^{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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