solve the differention equation x=p^3 −p+2 since:p=y′


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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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