Question

Să se arate ca funcția urmatoare verifica identitatea scrisă in dreptul ei :
f(x)=e la puterea radical din x + e la puterea minus radical din x , x>0
4x f''(x) + 2f'(x)-f(x)=0 , x>0

Answer 1

$f(x)=e^{\sqrt x}+e^{-\sqrt x};\\\\f^{'}(x)=\left(e^{\sqrt x}\right)^{'}+\left(e^{-\sqrt x}\right)^{'}=e^{\sqrt x}\cdot\left(\sqrt x\right)^{'}+e^{-\sqrt x}\cdot\left(-\sqrt x\right)^{'}=\\\\=e^{\sqrt x}\cdot\dfrac{1}{2\sqrt x}+e^{-\sqrt x}\cdot\dfrac{-1}{2\sqrt x}=\dfrac1{2\sqrt x}\left(e^{\sqrt x}-e^{-\sqrt x}\right);\\\\2f^{'}(x)-f(x)=\dfrac2{2\sqrt x}\left(e^{\sqrt x}-e^{-\sqrt x}\right)-e^{\sqrt x}-e^{-\sqrt x}=\\\\=\dfrac{1}{\sqrt x}\cdot\left(e^{\sqrt x}-e^{-\sqrt x}-\sqrt x\cdot e^{\sqrt x}-\sqrt x\cdot e^{-\sqrt x}\right)\;(1);\\\\f^{''}(x)=\left[\dfrac1{2\sqrt x}\left(e^{\sqrt x}-e^{-\sqrt x}\right)\right]^{'}=\dfrac{\dfrac1{2\sqrt x}\left(e^{\sqrt x}+e^{-\sqrt x}\right)\cdot 2\sqrt x-\left(e^{\sqrt x}-e^{-\sqrt x}\right)\cdot 2\cdot\dfrac1{2\sqrt x}}{4x}=\\\\=\dfrac{\dfrac{1}{\sqrt x}\cdot\left(\sqrt x\cdot e^{\sqrt x}+\sqrt x\cdot e^{-\sqrt x}-e^{\sqrt x}+e^{-\sqrt x}\right)}{4x}\;(2).$

Dacă înmulţeşti relaţia (2) cu 4x şi o aduni la relaţia (1), toţi termenii se reduc, deci vei obţine 0.

Green eyes.