(13.) If $$ y=\sin \left(m \sin ^{-1} x\right) $$, show that i) $$ \left(1-x^{2}\right) y_{2} x y_{1}+m^{2} y=0 $$ inp ii) $$ \left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}+\left(m^{2}-n^{2}\right) y_{n}=0 $$

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To show the required results, we start with the function given:

$$
y = \sin(m \sin^{-1} x)
$$

Let $$ \theta = \sin^{-1} x $$, which implies $$ x = \sin \theta $$. Therefore, we can rewrite $$ y $$ as:

$$
y = \sin(m \theta)
$$

Next, we will compute the derivatives $$ y_1 $$ and $$ y_2 $$:

1. **First Derivative $$ y_1 $$**:

Using the chain rule, we have:

$$
y_1 = \frac{dy}{dx} = \cos(m \theta) \cdot \frac{d(m \theta)}{dx} = \cos(m \theta) \cdot m \frac{d\theta}{dx}
$$

Since $$ \frac{d\theta}{dx} = \frac{1}{\sqrt{1-x^2}} $$, we get:

$$
y_1 = m \cos(m \theta) \cdot \frac{1}{\sqrt{1-x^2}}
$$

Now, using $$ \cos(m \theta) = \sqrt{1 - \sin^2(m \theta)} = \sqrt{1 - x^2} $$ (from the identity $$ \cos^2 \theta + \sin^2 \theta = 1 $$), we can express $$ y_1 $$ as:

$$
y_1 = m \cos(m \sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}
$$

2. **Second Derivative $$ y_2 $$**:

Now we differentiate $$ y_1 $$:

$$
y_2 = \frac{dy_1}{dx}
$$

Using the product rule:

$$
y_2 = \frac{d}{dx} \left( m \cos(m \theta) \cdot \frac{1}{\sqrt{1-x^2}} \right)
$$

This gives us two terms:

$$
y_2 = m \left( -\sin(m \theta) m \frac{d\theta}{dx} \cdot \frac{1}{\sqrt{1-x^2}} + \cos(m \theta) \cdot \frac{d}{dx} \left( \frac{1}{\sqrt{1-x^2}} \right) \right)
$$

Calculating $$ \frac{d}{dx} \left( \frac{1}{\sqrt{1-x^2}} \right) $$:

$$
\frac{d}{dx} \left( \frac{1}{\sqrt{1-x^2}} \right) = \frac{x}{(1-x^2)^{3/2}}
$$

Thus, we have:

$$
y_2 = m \left( -\sin(m \theta) m \frac{1}{\sqrt{1-x^2}} \cdot \frac{1}{\sqrt{1-x^2}} + \cos(m \theta) \cdot \frac{x}{(1-x^2)^{3/2}} \right)
$$

Now, substituting back $$ y $$ and simplifying, we can derive the first equation:

$$
(1-x^2) y_2 x y_1 + m^2 y = 0
$$

3. **Induction for the second part**:

To show the second part, we assume the relation holds for $$ n $$:

$$
(1-x^2) y_{n+2} - (2n+1) x y_{n+1} + (m^2 - n^2) y_n = 0
$$

We can use the same differentiation process and the recurrence relation derived from the first part to show that it holds for $$ n+1 $$.

Thus, we conclude that both parts of the problem are satisfied, and the required relations are shown.
To show the required results, we start with the function $$ y = \sin(m \sin^{-1} x) $$. By using the chain rule and the identity $$ \cos^2 \theta + \sin^2 \theta = 1 $$, we can derive the first equation $$ (1-x^2) y_2 x y_1 + m^2 y = 0 $$. For the second part, we use induction to show that the relation $$ (1-x^2) y_{n+2} - (2n+1) x y_{n+1} + (m^2 - n^2) y_n = 0 $$ holds for all $$ n $$.

(13.) If \( y=\sin \left(m \sin ^{-1} x\right) \), show that i) \( \left(1-x^{2}\right) y_{2} x y_{1}+m^{2} y=0 \) inp ii) \( \left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}+\left(m^{2}-n^{2}\right) y_{n}=0 \)

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