let-a-b-gt-0-find-all-differentiable-function-f-0-0-such-that-f-a-x-bx-f-x-x-gt-0-

Question Number 204426 by universe last updated on 17/Feb/24

let a, b > 0 find all differentiable function

f : (0, \infty) \to (0, \infty) such that

f^{'} (\frac{a}{x}) = \frac{bx}{f (x)}, \forall x > 0

Answered by MrGaster last updated on 03/Feb/25

Let f (x) = \sqrt{{b x}^{2} + c}, c = \frac{a^{2} b}{4}

f^{'} (x) = \frac{b x}{\sqrt{{b x}^{2} + c}}

Substitute x = \frac{a}{y} into f^{'} (x) :

f^{'} (\frac{a}{y}) = \frac{b (\frac{a}{y})}{\sqrt{b {(\frac{a}{y})}^{2} + c}} = \frac{\frac{b a}{y}}{\sqrt{\frac{{b a}^{2}}{y^{2}} + c}}

\frac{\frac{b a}{y}}{\sqrt{\frac{{b a}^{2}}{y^{2}} + c}} = \frac{b y}{f (y)}

Substitute f (y) = \sqrt{{b y}^{2} + c} :

\frac{\frac{b y}{y}}{\sqrt{\frac{{b a}^{2}}{y^{2}} + c}} = \frac{b y}{f (y)}

Substitute f (y) = \sqrt{{b y}^{2} + c} :

\frac{b a}{\frac{y}{\sqrt{\frac{{b a}^{2}}{y^{2}} + c}}} = \frac{b y}{\sqrt{{b y}^{2} + c}}

\frac{b a}{y} \cdot \sqrt{{b y}^{2} + c} = b y \cdot \sqrt{\frac{{b a}^{2}}{y^{2}} + c}

{(\frac{b a}{y})}^{2} ({b y}^{2} + c) = b^{2} y^{2} (\frac{{b a}^{2}}{y^{2}} + c)

\frac{b^{2} a^{2}}{y^{2}} ({b y}^{2} + c) = b^{2} y^{2} (\frac{{b a}^{2}}{y^{2}} + c)

\frac{a^{2}}{y^{2}} ({b y}^{2} + c) = y^{2} (\frac{a^{2}}{y^{2}} + c)

a^{2} b + \frac{a^{2} c}{y^{2}} = a^{2} + {c y}^{2}

Subtitute c = \frac{a^{2} b}{4} :

a^{2} b + \frac{a^{2} \cdot \frac{a^{2} b}{4}}{y^{2}} = a^{2} + \frac{a^{2} b}{4} y^{2}

4 a^{2} {b y}^{2} + a^{4} b = 4 a^{2} y^{2} + a^{2} {b y}^{2}

a^{2} {b y}^{4} - 4 a^{2} {b y}^{2} + a^{4} b = 0

y^{4} - 4 y^{2} + a^{2} = 0

Let z = y^{2} :

z^{2} - 4 z + a^{2} = 0

z = 2 \pm \sqrt{4 - a^{2}}

y = \sqrt{2 \pm \sqrt{4 - a^{2}}}

so :

f (x) = \sqrt{{b x}^{2} + \frac{a^{2} b}{4}}

Facebook Tweet Pin