(d^2 y/dx^2 )+6y((dy/dx))^2 =0 Please solve the differential eq.


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Question Number 55597 by ajfour last updated on 27/Feb/19

$\frac{d^{2} y}{{d x}^{2}} + 6 y {(\frac{d y}{d x})}^{2} = 0$ $P l e a s e s o l v e t h e d i f f e r e n t i a l e q .$
	Answered by mr W last updated on 27/Feb/19

	$u = \frac{d y}{d x}$ $\frac{d^{2} y}{{d x}^{2}} = \frac{d u}{d x} = \frac{d u}{d y} \times \frac{d y}{d x} = u \frac{d u}{d y}$ $\Rightarrow u \frac{d u}{d y} + 6 {y u}^{2} = 0$ $\Rightarrow \frac{d u}{d y} + 6 y u = 0$ $\Rightarrow \frac{d u}{u} = - 6 y d y$ $\Rightarrow \int \frac{d u}{u} = - 6 \int y d y$ $\Rightarrow \ln u = - 3 y^{2} + C$ $\Rightarrow u = \frac{d y}{d x} = {c e}^{- 3 y^{2}}$ $\Rightarrow e^{3 y^{2}} d y = c d x$ $\Rightarrow \int e^{3 y^{2}} d y = c \int d x$ $\Rightarrow \frac{\sqrt{π}}{2 \sqrt{3}} \times \frac{2}{\sqrt{π}} \int e^{{(\sqrt{3} y)}^{2}} d (\sqrt{3} y) + d = c x$ $\Rightarrow \frac{\sqrt{π} e r f i (\sqrt{3} y)}{2 \sqrt{3}} + d = c x$ $\Rightarrow x = c_{1} e r f i (\sqrt{3} y) + c_{2}$ $w i t h e r f i () = i m a g i n a r y e r r o r f u n c t i o n$
	Commented by mr W last updated on 27/Feb/19

	Commented by mr W last updated on 27/Feb/19

	$a n y o t h e r s o l u t i o n s i r ?$
	Commented by ajfour last updated on 27/Feb/19

	$T h a n k y o u S i r, g o e s b e y o n d m y$ $e x p e c t a t i o n .$ $i f y = x^{3} + p x + q$ $a n d i f i s w a p x w i t h y,$ $x = y^{3} + p y + q$ $T h e n 1 = (3 y^{2} + p) \frac{d y}{d x} . . . (i)$ $0 = 6 y \frac{d y}{d x} + (3 y^{2} + p) \frac{d^{2} y}{{d x}^{2}}$ $u s i n g (i)$ $6 y \frac{d y}{d x} + \frac{(\frac{d^{2} y}{{d x}^{2}})}{(\frac{d y}{d x})} = 0$ $\Rightarrow \frac{d^{2} y}{{d x}^{2}} + 6 y {(\frac{d y}{d x})}^{2} = 0$ $I t s t h i s w a y t h a t i g o t t h e q u e s t i o n$ $S i r .$
	Commented by ajfour last updated on 27/Feb/19

	$k n o w i n g t h e n t h a t o t h e r s o l u t i o n$ $i s y^{3} + p y + q = x$ $o n l y w e n e e d t o w r i t e y i n t e r m s$ $o f x .$ $l e t y = u + v$ ${(u + v)}^{3} + p (u + v) + q = x$ $u^{3} + v^{3} + (u + v) [3 u v + p] = x - q$ $l e t u v = - \frac{p}{3} \Rightarrow {(u v)}^{3} = - \frac{p^{3}}{27}$ $\Rightarrow u^{3} + v^{3} = x - q$ $\Rightarrow u^{3}, v^{3} a r e r o o t s o f$ $z^{2} - (x - q) z - \frac{p^{3}}{27} = 0$ $u^{3}, v^{3} = \frac{x - q}{2} \pm \sqrt{{(\frac{x - q}{2})}^{2} + \frac{p^{3}}{27}}$ $h e n c e o u r y (x) c a n e v e n b e$ $y = {[\frac{x - q}{2} + \sqrt{{(\frac{x - q}{2})}^{2} + \frac{p^{3}}{27}}]}^{1 / 3}$ $+ {[\frac{x - q}{2} - \sqrt{{(\frac{x - q}{2})}^{2} + \frac{p^{3}}{27}}]}^{1 / 3}$ $p, q b e i n g t h e a r b i t r a r y c o n s t a n t s .$ $(\underset{⌣}{∵}) C o n g r a t u l a t i o n s S i r,$ $i t s y o u w h o f o u n d t h e n e w c u b i c$ $f o r m u l a .$
	Commented by mr W last updated on 27/Feb/19

	$i n e e d s o m e t i m e t o c a t c h y o u f u l l y .$
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