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Differential equation (I)

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

    \[f'(x) \sqrt{x^2+1} = f(x) \quad \text{for all} \;\; x \in \mathbb{R}\]

Find an explicit formula for f.

Solution

We have successively

    \begin{align*} f'(x)\sqrt{x^2+1}=f(x) &\Leftrightarrow f'(x)\sqrt{x^2+1}-f(x)=0 \\ &\Leftrightarrow f'(x)\sqrt{x^2+1}-xf'(x)+\frac{xf(x)}{\sqrt{x^2+1}}-f(x)=0 \\ &\Leftrightarrow f'(x)\left ( \sqrt{x^2+1}-x \right )+f(x)\left ( \frac{x}{\sqrt{x^2+1}}-1 \right )=0 \\ &\Rightarrow \left [ f(x)\left ( \sqrt{x^2+1}-x \right ) \right ]'=(c)'\\ &\!\!\!\!\!\!\!\!\!\!\!\!\overset{f(0)=1\Rightarrow c=1}{=\!\!=\!=\!=\!=\!=\!=\!=\! \Rightarrow}f(x)=x+\sqrt{x^2+1}, \; x \in \mathbb{R} \end{align*}

which satisfies the given conditions.

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