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-

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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