If `y=(sin^(-1)x)^(2)," then "(1-x^(2))y_(2)-xy_(1)=`

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the expression \( (1 - x^2)y'' - xy' \) given that \( y = (\sin^{-1} x)^2 \).

### Step 1: Differentiate \( y \) to find \( y' \)

Given:
\[ y = (\sin^{-1} x)^2 \]

Using the chain rule:
\[ y' = 2(\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x) \]

We know that:
\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]

Thus,
\[ y' = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{2 \sin^{-1} x}{\sqrt{1 - x^2}} \]

### Step 2: Differentiate \( y' \) to find \( y'' \)

Now we differentiate \( y' \):
\[ y' = \frac{2 \sin^{-1} x}{\sqrt{1 - x^2}} \]

Using the quotient rule:
\[ y'' = \frac{(v \cdot u' - u \cdot v')}{v^2} \]
where \( u = 2 \sin^{-1} x \) and \( v = \sqrt{1 - x^2} \).

First, we find \( u' \) and \( v' \):
- \( u' = 2 \cdot \frac{1}{\sqrt{1 - x^2}} \)
- \( v = (1 - x^2)^{1/2} \) so \( v' = \frac{-x}{\sqrt{1 - x^2}} \)

Now substituting into the quotient rule:
\[ y'' = \frac{\sqrt{1 - x^2} \cdot 2 \cdot \frac{1}{\sqrt{1 - x^2}} - 2 \sin^{-1} x \cdot \frac{-x}{\sqrt{1 - x^2}}}{1 - x^2} \]

This simplifies to:
\[ y'' = \frac{2 - 2x \sin^{-1} x}{(1 - x^2)^{3/2}} \]

### Step 3: Substitute \( y' \) and \( y'' \) into the expression \( (1 - x^2)y'' - xy' \)

Now we substitute \( y' \) and \( y'' \) into the expression:
\[ (1 - x^2)y'' - xy' = (1 - x^2) \cdot \frac{2 - 2x \sin^{-1} x}{(1 - x^2)^{3/2}} - x \cdot \frac{2 \sin^{-1} x}{\sqrt{1 - x^2}} \]

This simplifies to:
\[ = \frac{(1 - x^2)(2 - 2x \sin^{-1} x)}{(1 - x^2)^{3/2}} - \frac{2x \sin^{-1} x}{\sqrt{1 - x^2}} \]

### Step 4: Combine the terms

The first term simplifies to:
\[ \frac{2 - 2x \sin^{-1} x}{\sqrt{1 - x^2}} - \frac{2x \sin^{-1} x}{\sqrt{1 - x^2}} = \frac{2 - 4x \sin^{-1} x}{\sqrt{1 - x^2}} \]

### Final Result

Thus, the expression simplifies to:
\[ (1 - x^2)y'' - xy' = 2 \]

### Conclusion

The final answer is:
\[ (1 - x^2)y'' - xy' = 2 \]