If `y=e^(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 given \( y = e^{m \sin^{-1}(x)} \), we need to find \( (1 + x^2)y'' - xy' \).

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

We start by differentiating \( y \) with respect to \( x \):

\[ y' = \frac{dy}{dx} = \frac{d}{dx} \left( e^{m \sin^{-1}(x)} \right) \]

Using the chain rule, we have:

\[ y' = e^{m \sin^{-1}(x)} \cdot \frac{d}{dx}(m \sin^{-1}(x)) \]

The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \). Thus:

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

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

Next, we differentiate \( y' \) to find \( y'' \):

\[ y'' = \frac{d}{dx} \left( e^{m \sin^{-1}(x)} \cdot \frac{m}{\sqrt{1 - x^2}} \right) \]

Using the product rule:

\[ y'' = \left( \frac{d}{dx} \left( e^{m \sin^{-1}(x)} \right) \cdot \frac{m}{\sqrt{1 - x^2}} \right) + \left( e^{m \sin^{-1}(x)} \cdot \frac{d}{dx} \left( \frac{m}{\sqrt{1 - x^2}} \right) \right) \]

We already found \( \frac{d}{dx} \left( e^{m \sin^{-1}(x)} \right) = e^{m \sin^{-1}(x)} \cdot \frac{m}{\sqrt{1 - x^2}} \). Now we need to differentiate \( \frac{m}{\sqrt{1 - x^2}} \):

\[ \frac{d}{dx} \left( \frac{m}{\sqrt{1 - x^2}} \right) = m \cdot \frac{d}{dx} \left( (1 - x^2)^{-1/2} \right) = m \cdot \left( -\frac{1}{2} (1 - x^2)^{-3/2} \cdot (-2x) \right) = \frac{mx}{(1 - x^2)^{3/2}} \]

Now substituting back into the expression for \( y'' \):

\[ y'' = e^{m \sin^{-1}(x)} \cdot \frac{m}{\sqrt{1 - x^2}} \cdot \frac{m}{\sqrt{1 - x^2}} + e^{m \sin^{-1}(x)} \cdot \frac{mx}{(1 - x^2)^{3/2}} \]

This simplifies to:

\[ y'' = e^{m \sin^{-1}(x)} \left( \frac{m^2}{1 - x^2} + \frac{mx}{(1 - x^2)^{3/2}} \right) \]

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

Now we substitute \( y, y', y'' \) into the expression:

\[ (1 + x^2)y'' - xy' = (1 + x^2)e^{m \sin^{-1}(x)} \left( \frac{m^2}{1 - x^2} + \frac{mx}{(1 - x^2)^{3/2}} \right) - x \left( e^{m \sin^{-1}(x)} \cdot \frac{m}{\sqrt{1 - x^2}} \right) \]

### Step 4: Simplify the expression

Distributing \( (1 + x^2) \):

\[ = e^{m \sin^{-1}(x)} \left( (1 + x^2) \left( \frac{m^2}{1 - x^2} + \frac{mx}{(1 - x^2)^{3/2}} \right) - x \cdot \frac{m}{\sqrt{1 - x^2}} \right) \]

After simplification, we find that the expression reduces to:

\[ = m^2 e^{m \sin^{-1}(x)} \]

### Final Result

Thus, we have:

\[ (1 + x^2)y'' - xy' = m^2 y \]