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Differential equation (II)

Let f:(0, +\infty) \rightarrow \mathbb{R} be a differentiable function such that f(1) = \frac{1}{2} and

    \[f'(x) + e^{f(x)} = x + \frac{1}{x} \quad \text{for all} \;\; x >0\]

Find an explicit formula for f.

Solution

Let us consider the function g(x) = f(x) - \ln x - \frac{x^2}{2} which is clearly differentiable. Hence,

    \begin{align*} f'(x) + e^{f(x)} = x + \frac{1}{x} &\Leftrightarrow \left ( g(x) + \ln x + \frac{x^2}{2} \right )' + e^{g(x) + \ln x + \frac{x^2}{2}} = x + \frac{1}{x} \\ &\Leftrightarrow g'(x) + \frac{1}{x} + x + x e^{g(x)} e^{x^2/2}= x + \frac{1}{x} \\ &\Leftrightarrow g'(x) + x e^{g(x)} e^{x^2/2} =0 \\ &\Leftrightarrow g'(x) = - x e^{g(x)} e^{x^2/2} \\ &\Leftrightarrow e^{-g(x)} g'(x) = -x e^{x^2/2} \\ &\Leftrightarrow -e^{-g(x)} g'(x) = x e^{x^2/2} \\ &\Leftrightarrow \left ( e^{-g(x)} \right ) ' = \left ( e^{x^2/2} \right )' \\ &\Rightarrow e^{-g(x)} = e^{x^2/2} + c \\ &\!\!\!\!\!\!\!\!\!\! \overset{x=1 \Rightarrow c =1-\sqrt{e} }{=\! =\! =\! =\! =\! =\! =\! \Rightarrow} e^{-g(x)} = e^{x^2/2} + 1 - \sqrt{e} \\ &\Rightarrow g(x) = - \ln \left ( e^{x^2/2} + 1 - \sqrt{e} \right ) \end{align*}

Thus,

    \[f(x) = \ln x + \frac{x^2}{2} - \ln \left ( e^{x^2/2} + 1 - \sqrt{e} \right )\]

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