Find f ?

Find $f$ for which we have : $f ' \left(1\right) = f \left(1\right) = 1$ $f \left(x\right) > 0$ ${x}^{3} f ' ' \left(x\right) - x f ' \left(x\right) + 2 f \left(x\right) = 0$ $x > 0$

Jan 10, 2018

$f \left(x\right) = {e}^{\frac{x - 1}{x}}$ $x > 0$

Explanation:

We have

${x}^{3} f ' ' \left(x\right) - x f ' \left(x\right) + 2 f \left(x\right) = 0$

$x > 0$ , ${x}^{3} > 0$

so

$f ' ' \left(x\right) - \frac{1}{x} ^ 2 f ' \left(x\right) + \frac{2 f \left(x\right)}{x} ^ 3 = 0$ $\iff$

$f ' ' \left(x\right) = \frac{1}{x} ^ 2 \cdot f ' \left(x\right) - \frac{2}{x} ^ 3 \cdot f \left(x\right)$ $\iff$

$f ' ' \left(x\right) = \frac{1}{x} ^ 2 \cdot f ' \left(x\right) + \left(\frac{1}{x} ^ 2\right) ' f \left(x\right)$ $\iff$

$\left(f ' \left(x\right)\right) ' = \left(\frac{1}{x} ^ 2 \cdot f \left(x\right)\right) '$

$f ' \left(x\right) = \frac{1}{x} ^ 2 \cdot f \left(x\right) + {c}_{1}$

For $x = 1$

$f ' \left(1\right) = f \left(1\right) + {c}_{1} \iff$ ${c}_{1} = 0$

Therefore, $f ' \left(x\right) = f \frac{x}{x} ^ 2$

$f ' \left(x\right) - \frac{1}{x} ^ 2 \cdot f \left(x\right) = 0$

$f ' \left(x\right) + \left(\frac{1}{x}\right) ' f \left(x\right) = 0$

${e}^{\frac{1}{x}} f ' \left(x\right) + {e}^{\frac{1}{x}} \left(\frac{1}{x}\right) ' f \left(x\right) = 0$

$\left({e}^{\frac{1}{x}} f \left(x\right)\right) ' = 0$

${e}^{\frac{1}{x}} f \left(x\right) = {c}_{2}$

For $x = 1$ $\to$ ${e}^{1} f \left(1\right) = {c}_{2}$ $\iff {c}_{2} = e$

As a result ,

${e}^{\frac{1}{x}} f \left(x\right) = e$

$f \left(x\right) = e \cdot {e}^{- \frac{1}{x}}$

$f \left(x\right) = {e}^{\frac{x - 1}{x}}$ , $x > 0$