show-that-x-x-2-x-1-is-an-explicit-solution-to-linear-equation-d-2-y-dx-2-2y-x-2-0-

Question Number 98367 by bobhans last updated on 13/Jun/20

show that φ (x) = x^{2} - x^{- 1} is an explicit

solution to linear equation \frac{d^{2} y}{{dx}^{2}} - \frac{2 y}{x^{2}} = 0

Commented by john santu last updated on 13/Jun/20

the function φ (x) = x^{2} - x^{- 1}, φ^{^{'}} (x) = 2 x + x^{- 2} and

φ^{″} (x) = 2 - {2 x}^{- 3} are defined for all x \neq 0

substitution of φ (x) for y in equation

gives (2 - {2 x}^{- 3}) - \frac{2}{x^{2}} (x^{2} - x^{- 1}) =

(2 - {2 x}^{- 3}) - (2 - {2 x}^{- 3}) = 0 . since this

is valid for any x \neq 0, the function

φ (x) = x^{2} - x^{- 1} is an explicit solution

for \frac{d^{2} y}{{dx}^{2}} - \frac{2 y}{x^{2}} = 0 on (- \infty, 0) \cup (0, \infty)

Commented by bobhans last updated on 13/Jun/20

thank you both

Answered by abdomathmax last updated on 13/Jun/20

φ (x) = x^{2} - \frac{1}{x} \Rightarrow φ^{^{'}} (x) = 2 x + \frac{1}{x^{2}} and

φ^{^{″}} (x) = 2 - \frac{2 x}{x^{4}} = 2 - \frac{2}{x^{3}} \Rightarrow φ^{^{″}} (x) - \frac{2}{x^{2}} φ (x)

= 2 - \frac{2}{x^{3}} - \frac{2}{x^{2}} (x^{2} - \frac{1}{x})

= 2 - \frac{2}{x^{3}} - 2 + \frac{2}{x^{3}} = 0 so φ is solution of ? (de)

y^{^{″}} - \frac{2 y}{x^{2}} = 0