`Y_1 = (dy)/(dx) and Y_2 =(d^2 y)/( dx^2)` . If ` Y= sin ( m sin^(-1)"" x)`, then `(1 -x^2) y_2 - xy_1 +m^2 y =`

Options

A. 1
B. `-1`
C. `0`
D. `2`

Answer

To solve the problem, we need to find the expression \( (1 - x^2) y_2 - x y_1 + m^2 y \) where \( y = \sin(m \sin^{-1}(x)) \), \( y_1 = \frac{dy}{dx} \), and \( y_2 = \frac{d^2y}{dx^2} \).

### Step 1: Find \( y_1 \)

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

Using the chain rule:
\[ y_1 = \frac{dy}{dx} = \cos(m \sin^{-1}(x)) \cdot \frac{d}{dx}(m \sin^{-1}(x)) \]

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

Thus,
\[ \frac{d}{dx}(m \sin^{-1}(x)) = m \cdot \frac{1}{\sqrt{1 - x^2}} \]

So,
\[ y_1 = \cos(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \]

### Step 2: Find \( y_2 \)

Now we differentiate \( y_1 \) to find \( y_2 \):
\[ y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}\left( \cos(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \right) \]

Using the product rule:
\[ y_2 = \left( \frac{d}{dx} \left( \cos(m \sin^{-1}(x)) \right) \cdot \frac{m}{\sqrt{1 - x^2}} \right) + \left( \cos(m \sin^{-1}(x)) \cdot \frac{d}{dx} \left( \frac{m}{\sqrt{1 - x^2}} \right) \right) \]

For the first part, we differentiate \( \cos(m \sin^{-1}(x)) \):
\[ \frac{d}{dx} \left( \cos(m \sin^{-1}(x)) \right) = -\sin(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \]

For the second part, we differentiate \( \frac{m}{\sqrt{1 - x^2}} \):
\[ \frac{d}{dx} \left( \frac{m}{\sqrt{1 - x^2}} \right) = m \cdot \frac{x}{(1 - x^2)^{3/2}} \]

Putting it all together:
\[ y_2 = \left( -\sin(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \cdot \frac{m}{\sqrt{1 - x^2}} \right) + \left( \cos(m \sin^{-1}(x)) \cdot \frac{mx}{(1 - x^2)^{3/2}} \right) \]

### Step 3: Substitute into the expression

Now we substitute \( y_1 \) and \( y_2 \) into the expression \( (1 - x^2) y_2 - x y_1 + m^2 y \):
\[ (1 - x^2) y_2 - x y_1 + m^2 y \]

Substituting \( y_1 \) and \( y_2 \):
1. Calculate \( (1 - x^2) y_2 \)
2. Calculate \( -x y_1 \)
3. Add \( m^2 y \)

After substituting and simplifying, we will find that the entire expression equals zero:
\[ (1 - x^2) y_2 - x y_1 + m^2 y = 0 \]

### Final Result
Thus, the final answer is:
\[ (1 - x^2) y_2 - x y_1 + m^2 y = 0 \]