If log e y = 3 sin - 1 x , then ( 1 - x 2 ) y ' ' - x y ' at x = 1 2 is equal to [2024]

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Solution

Q.
$If \log_{e} y = 3 \sin^{- 1} x, then (1 - x^{2}) y'' - x y' at x = \frac{1}{2} is equal to$ [2024]

1 $9 e^{\frac{π}{6}}$

2 $9 e^{\frac{π}{2}}$

3 $3 e^{\frac{π}{6}}$

4 $3 e^{\frac{π}{2}}$

Ans.
(2)

$\log_{e} y = 3 \sin^{- 1} x$

Differentiating w.r.t. $' x'$ we get

$\frac{1}{y} y' = \frac{3}{\sqrt{1 - x^{2}}}$

$y' = \frac{3 y}{\sqrt{1 - x^{2}}}$ ...(i)

Again differentiating w.r.t. $' x'$ we get

$y'' = 3 [\frac{\sqrt{1 - x^{2}} y' - y \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x)}{(1 - x^{2})}]$

$\Rightarrow (1 - x^{2}) y'' = 3 [\sqrt{1 - x^{2}} \cdot \frac{3 y}{\sqrt{1 - x^{2}}} + \frac{x y}{\sqrt{1 - x^{2}}}]$ [Using (i)]

$= 3 [3 y + \frac{x y}{\sqrt{1 - x^{2}}}]$

Now, $y' (\frac{1}{2}) = \frac{3 e^{3 \sin^{- 1} (\frac{1}{2})}}{\sqrt{1 - \frac{1}{4}}} = \frac{3 e^{π / 2}}{\frac{\sqrt{3}}{2}} = 2 \sqrt{3} e^{π / 2}$

$(1 - x^{2}) y'' (x) at (x = \frac{1}{2}) = 3 [3 e^{π / 2} + \frac{\frac{1}{2} e^{π / 2}}{\frac{\sqrt{3}}{2}}]$

$= 3 [3 e^{π / 2} + \frac{1}{\sqrt{3}} e^{π / 2}] = 3 e^{π / 2} [3 + \frac{1}{\sqrt{3}}]$

So, $(1 - x^{2}) y'' - x y'$ at $x = \frac{1}{2}$ is given by

$3 e^{π / 2} [3 + \frac{1}{\sqrt{3}}] - \frac{1}{2} (2 \sqrt{3} e^{π / 2}) = 9 e^{π / 2}$

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