Question

If $$ y = \sin^{-1} x, $$ then $$ (1 - x^2)y_2 - xy_1 = 0. $$ 

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 2y \)

Hint:

For inverse trigonometric functions, differentiate carefully and use algebraic simplifications.

Answer 1

The Correct Option is A

Step 1: Differentiating \[ y_1 = \frac{1}{\sqrt{1-x^2}}. \] Differentiating again: \[ y_2 = \frac{x}{(1-x^2)^{3/2}}. \] Step 2: Substituting into the given equation \[ (1 - x^2)y_2 - xy_1 = (1-x^2) \cdot \frac{x}{(1-x^2)^{3/2}} - x \cdot \frac{1}{\sqrt{1-x^2}}. \] Simplifying, both terms cancel: \[ 0. \]

Answer 2

Given:
\[ y = \sin^{-1} x \] and the differential equation:
\[ (1 - x^2) y'' - x y' = ? \]

Step 1: Find the first derivative:
\[ y' = \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} \]

Step 2: Find the second derivative:
\[ y'' = \frac{d}{dx} \left( \frac{1}{\sqrt{1 - x^2}} \right ) = \frac{d}{dx} (1 - x^2)^{-\frac{1}{2}} = -\frac{1}{2} (1 - x^2)^{-\frac{3}{2}} \times (-2x) = \frac{x}{(1 - x^2)^{3/2}} \]

Step 3: Substitute into the equation:
\[ (1 - x^2) y'' - x y' = (1 - x^2) \times \frac{x}{(1 - x^2)^{3/2}} - x \times \frac{1}{\sqrt{1 - x^2}} = \frac{x (1 - x^2)}{(1 - x^2)^{3/2}} - \frac{x}{(1 - x^2)^{1/2}} = 0 \]

Therefore,
\[ \boxed{0} \]