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Question Number 155101 by SANOGO last updated on 25/Sep/21

s o i t : y y^{'} + {x y}^{2} + x = 0, a v e c f (0) = 1

f^{″} (0) = ?

Commented by tabata last updated on 25/Sep/21

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

\frac{1}{2} l n ∣ y^{2} + 1 ∣ = - \frac{x^{2}}{2} + \frac{c}{2} \Rightarrow l n ∣ y^{2} + 1 ∣ = - x^{2} + c

y^{2} = k e^{- x^{2}} - 1 \Rightarrow y = \sqrt{k e^{- x^{2}} - 1}

f (0) = 1 \Rightarrow 1 = \sqrt{k - 1} \Rightarrow k = 2

∴ y = \sqrt{2 e^{- x^{2}} - 1}

y^{^{'}} = \frac{- 2 x e^{- x^{2}}}{\sqrt{2 e^{- x^{2}} - 1}} \Rightarrow y^{^{″}} = \frac{(\sqrt{2 e^{- x^{2}} - 1}) (4 x^{2} e^{- x^{2}} - 2 e^{- x^{2}}) - (2 x e^{- x^{2}}) (\frac{2 x e^{- x^{2}}}{\sqrt{2 e^{- x^{2}} - 1}})}{(2 e^{- x^{2}} - 1)}

f^{^{″}} (0) = \frac{(\sqrt{2 - 1}) (- 2)}{(2 - 1)} = - \frac{2}{1} = - 2

< M . T >

Commented by SANOGO last updated on 25/Sep/21

m e r c i b i e n

Commented by tabata last updated on 25/Sep/21

y o u a r e w e l c o m e

Answered by mr W last updated on 25/Sep/21

y (0) = 1

{y y}^{'} + {x y}^{2} + x = 0

1 \times y^{'} (0) + 0 \times 1^{2} + 0 = 0

\Rightarrow y^{'} (0) = 0

{y y}^{″} + {(y^{'})}^{2} + y^{2} + 2 {x y y}^{'} + 1 = 0

1 \times y^{″} (0) + {(0)}^{2} + 1^{2} + 2 \times 0 \times 1 \times 0 + 1 = 0

\Rightarrow y^{″} (0) = - 2

Commented by SANOGO last updated on 25/Sep/21

m e r c i b i e n

m e r c i b i e n

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