prove-that-y-c-2-cx-its-solution-of-the-differention-equation-4xy-2xy-y-0-m-o-

Question Number 115829 by mohammad17 last updated on 28/Sep/20

p r o v e t h a t {(y - c)}^{2} = c x i t s s o l u t i o n o f t h e d i f f e r e n t i o n e q u a t i o n 4 {x y}^{^{″}} + 2 {x y}^{^{'}} - y = 0

(m . o)

Answered by $@y@m last updated on 29/Sep/20

y - c = \sqrt{c x}

y = \sqrt{c} . \sqrt{x} + c \dots (1)

y^{'} = \sqrt{c} . \frac{1}{2 \sqrt{x}} \dots (2)

y^{″} = \frac{\sqrt{c}}{2} . \frac{- 1}{2 x \sqrt{x}}

y^{″} = - \frac{\sqrt{c}}{4} . \frac{1}{x \sqrt{x}} \dots (3)

N o w,

4 {x y}^{^{″}} + 2 {x y}^{^{'}} - y = 4 x \times (- \frac{\sqrt{c}}{4} . \frac{1}{x \sqrt{x}}) + 2 x \times \frac{\sqrt{c}}{2 \sqrt{x}} - \sqrt{c x} - c

= \frac{- \sqrt{c}}{x \sqrt{x}} + \sqrt{c x} - . \sqrt{c x} - c

= \frac{- \sqrt{c}}{x \sqrt{x}} - c

\neq 0

P l . c h e c k q u e s t i o n

O R, m y s o l u t i o n .

:)

Commented by mohammad17 last updated on 29/Sep/20

t h a n k y o u s i r