Show that x^2 + 2xy + 3y^2 = 1, is a solution to the differential equation (x + 3y)^2 (d^2 y/dx^2 ) + 2(x^2 + 2xy + 2y^3 ) = 0


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Question Number 173736 by Tawa11 last updated on 17/Jul/22

$Show that x^{2} + 2 xy + {3 y}^{2} = 1, is a solution to the$ $differential equation {(x + 3 y)}^{2} \frac{d^{2} y}{{dx}^{2}} + 2 (x^{2} + 2 xy + {2 y}^{3}) = 0$
	Answered by mindispower last updated on 17/Jul/22

	$\frac{d}{d x} (x^{2} + 2 x y + 3 y^{2}) = 0$ $\Leftrightarrow (2 x + 2 y + 2 x \frac{d y}{d x} + \frac{6 d y}{d x} y) = 0$ $\frac{d}{d x} (2 x + 2 y + 2 x \frac{d y}{d x} + \frac{6 d y}{d x} y) = 0$ $⇎ 2 + \frac{2 d y}{d x} + 2 \frac{d y}{d x} + 2 x \frac{d^{2} y}{{d x}^{2}} + \frac{6 d^{2} y}{{d x}^{2}} y + 6 {(\frac{d y}{d x})}^{2} = 0$ $\frac{d y}{d x} = \frac{- 2 x - 2 y}{2 x + 6 y}$ $\frac{d^{2} y}{{d x}^{2}} = \frac{1}{2 x + 6 y} (- 2 + \frac{8 x + 8 y}{2 x + 6 y} - 6 {(\frac{2 x + 2 y}{2 x + 6 y})}^{2})$ ${(x + 3 y)}^{3} \frac{d^{2} y}{{d x}^{2}} = \frac{1}{2} {(x + 3 y)}^{2} (- 2 + \frac{4 x + 4 y}{x + 3 y} - 6 {(\frac{x + y}{x + 3 y})}^{2_{)}}$ $= - 3 {(x + y)}^{2} - {(x + 3 y)}^{2} + (2 x + 2 y) (x + 3 y)$ $- 2 x^{2} - 6 y^{2} - 4 x y$ $\Rightarrow {(x + 3 y)}^{2} \frac{d^{2} y}{{d x}^{2}} + 2 (x^{2} + 2 x y + 3 y^{2}) = 0$ $M a y b e e i h a v e m i s t a k s o r r y s i r$
	Commented by Tawa11 last updated on 17/Jul/22

	$God bless you sir .$
	Commented by Tawa11 last updated on 17/Jul/22

	$In conclusion, that mean x^{2} + 2 xy + {3 y}^{2} = 1 is not a solution .$
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