If `y=cos(msin^(-1)x)," then "(1-x^(2))y_(2)-xy_1=`

Options

A. 0
B. 1
C. `-m^(2)y`
D. `m^(2)y`

Answer

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

### Step 1: Find \( y_1 = \frac{dy}{dx} \)

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

Using the chain rule, we differentiate \( y \):
\[ y_1 = \frac{dy}{dx} = -\sin(m \sin^{-1}(x)) \cdot \frac{d}{dx}(m \sin^{-1}(x)) \]
The derivative of \( \sin^{-1}(x) \) is:
\[ \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} \]
Thus,
\[ y_1 = -\sin(m \sin^{-1}(x)) \cdot m \cdot \frac{1}{\sqrt{1 - x^2}} = -\frac{m \sin(m \sin^{-1}(x))}{\sqrt{1 - x^2}} \]

### Step 2: Find \( y_2 = \frac{d^2y}{dx^2} \)

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

Using the quotient rule:
\[ y_2 = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( u = -m \sin(m \sin^{-1}(x)) \) and \( v = \sqrt{1 - x^2} \).

Calculating \( \frac{du}{dx} \):
\[ \frac{du}{dx} = -m \cos(m \sin^{-1}(x)) \cdot \frac{1}{\sqrt{1 - x^2}} \]

Calculating \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{-x}{\sqrt{1 - x^2}} \]

Now substituting into the quotient rule:
\[ y_2 = \frac{\sqrt{1 - x^2} \left(-m \cos(m \sin^{-1}(x)) \cdot \frac{1}{\sqrt{1 - x^2}}\right) - \left(-m \sin(m \sin^{-1}(x)\right)\left(-\frac{x}{\sqrt{1 - x^2}}\right)}{1 - x^2} \]

This simplifies to:
\[ y_2 = \frac{-m \cos(m \sin^{-1}(x)) - mx \sin(m \sin^{-1}(x))}{1 - x^2} \]

### Step 3: Substitute \( y_1 \) and \( y_2 \) into the expression

Now we substitute \( y_1 \) and \( y_2 \) into the expression \( (1 - x^2)y_2 - xy_1 \):
\[ (1 - x^2)\left(\frac{-m \cos(m \sin^{-1}(x)) - mx \sin(m \sin^{-1}(x))}{1 - x^2}\right) - x\left(-\frac{m \sin(m \sin^{-1}(x))}{\sqrt{1 - x^2}}\right) \]

This simplifies to:
\[ -m \cos(m \sin^{-1}(x)) - mx \sin(m \sin^{-1}(x)) + \frac{mx \sin(m \sin^{-1}(x))}{\sqrt{1 - x^2}} \]

### Step 4: Final Result

After simplifying, we find:
\[ (1 - x^2)y_2 - xy_1 = -m^2 y \]
Thus, the final answer is:
\[ (1 - x^2)y_{2} - xy_{1} = -m^2 y \]