Differential equation

Let $f:(0, +\infty) \rightarrow \mathbb{R}$ be a twice differentiable function such that

$x^2 f''(x) + x f'(x) = x - 2 \quad \text{for all} \;\; x>0$

If $f(1) =0$ , $f'(1)=1$ find an explicit formula of $f$ .

Solution

We have successively

$\begin{align*} x^2 f''(x) + x f'(x) = x - 2 &\Rightarrow x f''(x) + f'(x) = 1 - \frac{2}{x} \\ &\Rightarrow \left ( x f'(x) \right )' = \left ( x - 2 \ln x \right )' \\ &\Rightarrow x f'(x) = x - 2 \ln x + c_1 \\ &\!\!\!\!\!\!\!\!\!\!\!\overset{x=1 \Rightarrow c_1=0}{=\! =\! =\!=\! =\! =\!\Rightarrow } x f'(x) = x - 2 \ln x \\ &\Rightarrow f'(x) = 1 - \frac{2 \ln x}{x} \\ &\Rightarrow \left ( f(x) \right ) ' = \left ( x - \ln^2 x \right )' \\ &\Rightarrow f(x) = x - \ln^2 x + c \\ &\!\!\!\!\!\!\!\!\!\!\!\overset{x=1 \Rightarrow c=-1}{=\! =\! =\!=\! =\! =\!\Rightarrow } f(x) = x - \ln^2 x - 1 \end{align*}$

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