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Mathematics Question on Differential equations

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

A

9eπ/69e^{\pi/6}

B

3eπ/63e^{\pi/6}

C

3eπ/23e^{\pi/2}

D

9eπ/29e^{\pi/2}

Answer

9eπ/29e^{\pi/2}

Explanation

Solution

We are given the equation log⁡ey=3sin⁡−1x\log_e y = 3 \sin^{-1} x. We need to find (1−x2)y′′−xy′(1 - x^2) y'' - xy' at x=12x = \frac{1}{2}.

Step 1: Differentiate yy

Start by differentiating y=e3sin⁡−1xy = e^{3 \sin^{-1} x}:

1y⋅y′=3⋅11−x2  ⟹  y′=3y1−x2\frac{1}{y} \cdot y' = 3 \cdot \frac{1}{\sqrt{1 - x^2}} \implies y' = \frac{3y}{\sqrt{1 - x^2}}

At x=12x = \frac{1}{2}, we get:

y′=3eπ63=23eπ63y' = \frac{3e^{\frac{\pi}{6}}}{\sqrt{3}} = \frac{2\sqrt{3}e^{\frac{\pi}{6}}}{3}

Step 2: Differentiate y′y' to find y′′y''

Now differentiate y′y' to get y′′y'':

y′′=ddx(3y1−x2)y'' = \frac{d}{dx} \left( \frac{3y}{\sqrt{1 - x^2}} \right)

Using the quotient rule, we get:

y′′=3(1−x2⋅y′−y⋅(−x1−x2)(1−x2))y'' = 3 \left( \frac{\sqrt{1 - x^2} \cdot y' - y \cdot \left( -\frac{x}{\sqrt{1 - x^2}} \right)}{(1 - x^2)} \right)

Substitute the expression for y′y':

Step 3: Substitute x=12x = \frac{1}{2}

At x=12x = \frac{1}{2}, we substitute values for yy and y′y':

(1−x2)y′′=3(3eπ6+eπ63)=3eπ6(3+13)(1 - x^2) y'' = 3 \left( 3e^{\frac{\pi}{6}} + \frac{e^{\frac{\pi}{6}}}{\sqrt{3}} \right) = 3e^{\frac{\pi}{6}} \left( 3 + \frac{1}{\sqrt{3}} \right)

Now, calculate (1−x2)y′′−xy′(1 - x^2) y'' - xy':

(1−x2)y′′−xy′=3eπ6(3+13)−12×23eπ63(1 - x^2) y'' - xy' = 3e^{\frac{\pi}{6}} \left( 3 + \frac{1}{\sqrt{3}} \right) - \frac{1}{2} \times \frac{2\sqrt{3}e^{\frac{\pi}{6}}}{3}

After simplifying:

(1−x2)y′′−xy′=9eπ2(1 - x^2) y'' - xy' = 9e^{\frac{\pi}{2}}

Thus, the correct answer is:

9eπ2\boxed{9e^{\frac{\pi}{2}}}