If 2 x-y1 - 5-y-1 - 5 then x2-1 d2 y-d x2-x d y-d x-A- 5y B- 25 yC- 25 y 2D- y -25

The correct option is B 25y Consider (y^1/5+y^ 1/5)^2=(y^1/5 y^ 1/5)^2+4 ∴ y^1/5 y^ 1/5=2√(x^2 1) ∴ y^1/5=x+√(x^2 1) ⇒ y=(x+√(x^2 1))^5 dy/dx=5(x+√(x^2 1))^4 ...

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Byju's Answer

Standard XII

Mathematics

Parametric Differentiation

If 2 x=y1 / 5...

Question

If $2 x = y^{1 / 5} + y^{- 1 / 5}$ then $(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} =$

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25y

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$25 y^{2}$

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y + 25

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Solution

The correct option is B
25y

Consider $(y^{1 / 5} + y^{- 1 / 5})^{2} = (y^{1 / 5} - y^{- 1 / 5})^{2} + 4$
$∴ y^{1 / 5} - y^{- 1 / 5} = 2 \sqrt{x^{2} - 1}$
$∴ y^{1 / 5} = x + \sqrt{x^{2} - 1} \Rightarrow y = (x + \sqrt{x^{2} - 1})^{5}$
$\frac{d y}{d x} = 5 (x + \sqrt{x^{2} - 1})^{4} (1 + \frac{2 x}{2 \sqrt{x^{2} - 1}})$
$(x^{2} - 1) {(\frac{d y}{d x})}^{2} = 25 y^{2}$
Again differentiate wrt.x

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Similar questions

If $2 x = y^{1 / 5} + y^{- 1 / 5}$ then $(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} =$

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2 x = y^{1 / 5} + y^{- 1 / 5}

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

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(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - n^{2} y = 0

Q. If

2 x = y^{\frac{1}{5}} + y^{\frac{- 1}{5}} and (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + λ x \frac{d y}{d x} + ky = 0, then λ + K

is equal to.

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If 2x=y1/5+y−1/5 then (x2−1)d2ydx2+xdydx=

The correct option is B 25y Consider (y1/5+y−1/5)2=(y1/5−y−1/5)2+4 ∴y1/5−y−1/5=2√x2−1 ∴y1/5=x+√x2−1 ⇒y=(x+√x2−1)5 dydx=5(x+√x2−1)4(1+2x2√x2−1) (x2−1)(dydx)2=25 y2 Again differentiate wrt.x

If $2 x = y^{1 / 5} + y^{- 1 / 5}$ then $(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} =$