solve-the-differention-equation-x-p-3-p-2-since-p-y-

Question Number 143165 by mohammad17 last updated on 11/Jun/21

s o l v e t h e d i f f e r e n t i o n e q u a t i o n

x = p^{3} - p + 2 s i n c e : p = y^{'}

Commented by mohammad17 last updated on 11/Jun/21

? ? ? ? ? ? ?

Commented by mr W last updated on 11/Jun/21

y o u {c a n}^{'} t e x p e c t a n a n s w e r t o a

q u e s t i o n w h i c h i s n o q u e s t i o n .

x = p^{3} - p + 2 i s n e i t h e r a d i f f e r e n t i a l

e q u a t i o n n o r a d i f f e r e n t i o n e q u a t i o n .

i t i s e v e n n o t a n e q u a t i o n a t a l l .

Commented by mohammad17 last updated on 11/Jun/21

s i r i t s d i f f e r e n t i a l e q u a t i o n s u c h t h a t

p = y^{^{'}}

Commented by mr W last updated on 11/Jun/21

b u t y o u {d i d n}^{'} t g i v e t h i s b e f o r e .

b e s i d e s, w h a t i s m e a n t w i t h y^{'} ?

i s i t y^{'} = \frac{d y}{d x} o r y^{'} = \frac{d y}{d p} ?

i n m a t h e m s t i c s t h e t h i n g s m u s t b e

e x a c t a n d c l e a r!

Answered by Cwesi last updated on 10/Jun/21

\frac{d x}{d p} = 3 p^{2} - 1

Answered by mr W last updated on 11/Jun/21

y^{'} = \frac{d y}{d x} = p

\frac{d y}{d p} \times \frac{d p}{d x} = p

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

y = \frac{3 p^{4}}{4} - \frac{p^{2}}{2} + \frac{C_{1}}{4}

3 p^{4} - 2 p^{2} + C_{1} - 4 y = 0

\Rightarrow p^{2} = \frac{1 \pm \sqrt{1 + 3 (4 y - C_{1})}}{3} = \frac{1 \pm \sqrt{12 y + C}}{3}

\Rightarrow p = \pm \sqrt{\frac{1 \pm \sqrt{12 y + C}}{3}}

\Rightarrow x = p (p^{2} - 1) + 2

\Rightarrow x = \pm \sqrt{\frac{1 \pm \sqrt{12 y + C}}{3}} (\frac{- 2 \pm \sqrt{12 y + C}}{3}) + 2

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