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Question: If $\log_e y = 3 \sin^{-1} x$, then $(1-x^2)y'' - xy'$ at $x=\frac{1}{2}$ is equal to:...

If log⁡ey=3sin⁡−1x\log_e y = 3 \sin^{-1} x, then (1−x2)y′′−xy′(1-x^2)y'' - xy' at x=12x=\frac{1}{2} is equal to:

A

9eπ/69e^{\pi/6}

A

9eπ/29e^{\pi/2}

A

3eπ/63e^{\pi/6}

A

3eπ/23e^{\pi/2}

Answer

9eπ/29e^{\pi/2}

Explanation

Solution

Given log⁡ey=3sin⁡−1x\log_e y = 3 \sin^{-1} x, we can write y=e3sin⁡−1xy = e^{3 \sin^{-1} x}.

Differentiating with respect to xx: 1yy′=3⋅11−x2\frac{1}{y} y' = 3 \cdot \frac{1}{\sqrt{1-x^2}} 1−x2y′=3y\sqrt{1-x^2} y' = 3y (1)

Differentiating equation (1) with respect to xx: −x1−x2y′+1−x2y′′=3y′\frac{-x}{\sqrt{1-x^2}} y' + \sqrt{1-x^2} y'' = 3y'

Multiply by 1−x2\sqrt{1-x^2}: −xy′+(1−x2)y′′=3(1−x2)y′1−x21−x2-x y' + (1-x^2) y'' = 3(1-x^2) \frac{y'}{\sqrt{1-x^2}} \sqrt{1-x^2} (1−x2)y′′−xy′=31−x2y′(1-x^2) y'' - xy' = 3\sqrt{1-x^2} y'

Substitute 1−x2y′=3y\sqrt{1-x^2} y' = 3y from equation (1): (1−x2)y′′−xy′=3(3y)=9y(1-x^2) y'' - xy' = 3(3y) = 9y

Now, evaluate at x=12x = \frac{1}{2}: At x=12x = \frac{1}{2}, sin⁡−1x=sin⁡−112=π6\sin^{-1} x = \sin^{-1} \frac{1}{2} = \frac{\pi}{6}. So, y=e3sin⁡−1x=e3(π6)=eπ2y = e^{3 \sin^{-1} x} = e^{3(\frac{\pi}{6})} = e^{\frac{\pi}{2}}.

Therefore, (1−x2)y′′−xy′(1-x^2) y'' - xy' at x=12x = \frac{1}{2} is 9y=9eπ29y = 9e^{\frac{\pi}{2}}.