f-x-x-2-2-f-x-0-f-x-

Question Number 197530 by Erico last updated on 20/Sep/23

f (x) - \frac{x^{2}}{2} f^{″} (x) = 0

f (x) = ?

Commented by AST last updated on 20/Sep/23

f (x) = x^{2} w o r k s .

Answered by witcher3 last updated on 20/Sep/23

f (x) = x^{a}

\Rightarrow x^{a} - \frac{1}{2} a (a - 1) x^{a} = 0

\Leftrightarrow (- \frac{a^{2}}{2} + \frac{a}{2} + 1) x^{a} = 0

\Rightarrow a^{2} - a - 2 = 0, a \in {- 1, 2}

f (x) = \frac{a}{x} + {bx}^{2}

\frac{1}{x}, x^{2}, are idependent 2 dimension space \Rightarrow

\forall f \in C_{2} R_{*} ∣ f - \frac{x^{2}}{2} f^{″} = 0 \exists (a, b) \in R^{2} such f = \frac{a}{x} + {bx}^{2}

Answered by sniper237 last updated on 21/Sep/23

y + {x y}^{'} - ({x y}^{'} + \frac{x^{2}}{2} y^{″}) = 0

\frac{d}{d x} (x y - \frac{x^{2}}{2} y^{'}) = 0

x y - \frac{x^{2}}{2} y^{'} = C

\overset{h o m o g e n s o l u t i o n}{\Rightarrow} y_{H} = {K x}^{2}

\overset{c t e v a r i a t %}{\Rightarrow} - \frac{x^{4}}{2} K^{'} (x) = C

\Rightarrow K (x) = \frac{A}{x^{3}} + B

\overset{g e n s o l u t %}{\Rightarrow} y = \frac{A}{x} + {B x}^{2}