If y - e m sin -1 x- then 1- x 2 d 2 y - dx 2- x dy - dx - ky -0- where k is equal toA- 2 B- - m 2C- -1 D- m 2

The correct option is D m^2 nbsp;y=e^msin^ 1x, dydx=m e^msin^ 1x√(1 x^2)=my√(1 x^2) ddx·dydx=m√(1 x^2)·me^msin^ 1x√(1 x^2)+e^msin^ 1xx√(1 x^2)1 x^2 (1 x^2)d ...

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Byju's Answer

Standard XII

Mathematics

Integrating Factor

If y = e m si...

Question

If $y = e^{m {sin}^{- 1} x}$ , then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to

m^{2}

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- 1

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- m^{2}

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Solution

The correct option is A $m^{2}$
$y = e^{m {sin}^{- 1} x}, \frac{d y}{d x} = \frac{m e^{m {sin}^{- 1} x}}{\sqrt{1 - x^{2}}} = \frac{m y}{\sqrt{1 - x^{2}}} \frac{d}{d x} \cdot \frac{d y}{d x} = m \frac{\sqrt{1 - x^{2}} \cdot \frac{m e^{m {sin}^{- 1} x}}{\sqrt{1 - x^{2}}} + e^{m {sin}^{- 1} x} \frac{x}{\sqrt{1 - x^{2}}}}{1 - x^{2}} (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = m^{2} y + x \frac{d y}{d x} (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$
Comparing this equation with the given equation we have
$k = m^{2}$

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Similar questions

Q. If

y = e^{m {sin}^{- 1} x}

, then

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

, where

k

is equal to

Q. If

y = e^{m s i n^{- 1} x}

, then show that

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

y = e^{m {sin}^{- 1} x}

, then

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

Q. If

y = e^{m {sin}^{- 1} x}, - 1 \leq x \leq 1

, show that

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

Q. If

y = {sin}^{- 1} x

, then prove that

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

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If y=emsin−1x, then (1−x2)d2ydx2−xdydx−ky=0, where k is equal to

The correct option is A m2 y=emsin−1x,dydx=memsin−1x√1−x2=my√1−x2ddx⋅dydx=m√1−x2⋅memsin−1x√1−x2+emsin−1xx√1−x21−x2(1−x2)d2ydx2=m2y+xdydx(1−x2)d2ydx2−xdydx−m2y=0 Comparing this equation with the given equation we have k=m2

If $y = e^{m {sin}^{- 1} x}$ , then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to