if-xy-y-2-1-Find-d-2-y-dx-at-0-1-

Question Number 8115 by uchechukwu okorie favour last updated on 30/Sep/16

i f x y + y^{2} = 1 . F i n d \frac{d^{2} y}{d x} a t (0, 1)

Answered by prakash jain last updated on 30/Sep/16

x \frac{d y}{d x} + y + 2 y \frac{d y}{d x} = 0 \dots (i)

x \frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x} + \frac{d y}{d x} + 2 y \frac{d^{2} y}{{d x}^{2}} + 2 {(\frac{d y}{d x})}^{2} \dots (i i)

(x + 2 y) \frac{d^{2} y}{{d x}^{2}} = - 2 {(\frac{d y}{d x} + (\frac{d y}{d x}))}^{2}

f r o m (i) \frac{d y}{d x} = - \frac{y}{x + 2 y}

\frac{d^{2} y}{{d x}^{2}} = \frac{- 2 (- \frac{y}{x + 2 y} + {(- \frac{y}{x + 2 y})}^{2})}{(x + 2 y)}

x = 0, y = 1

\frac{d^{2} y}{{d x}^{2}} = \frac{- 2 (- \frac{1}{2} + \frac{1}{4})}{2} = - 1 (- \frac{1}{4}) = \frac{1}{4}

Answered by mrW1 last updated on 13/Feb/17

x = \frac{1 - y^{2}}{y} = \frac{1}{y} - y

\frac{d x}{d y} = - \frac{1}{y^{2}} - 1 = - \frac{1 + y^{2}}{y^{2}}

g = \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = - \frac{y^{2}}{1 + y^{2}} = - 1 + \frac{1}{1 + y^{2}}

\frac{d g}{d y} = - \frac{2 y}{{(1 + y^{2})}^{2}}

\frac{d^{2} y}{{d x}^{2}} = \frac{d g}{d x} = \frac{d g}{d y} \times \frac{d y}{d x} = \frac{2 y^{3}}{{(1 + y^{2})}^{3}} = 2 {(\frac{y}{1 + y^{2}})}^{3}

w i t h y = 1, \frac{d^{2} y}{{d x}^{2}} = \frac{1}{4}

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