2.7 Q-5

Question Statement

Given that

$$x = \sin \theta, \quad y = \sin(m \theta),$$

show that

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

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Background and Explanation

This problem involves differentiating trigonometric functions. We are given expressions for $x$ and $y$ in terms of $\sin \theta$, and the goal is to differentiate them to obtain a differential equation in terms of $y_1$ (the first derivative of $y$ with respect to $x$) and $y_2$ (the second derivative of $y$ with respect to $x$). To solve this, we need to use the chain rule for derivatives and some trigonometric identities.

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Solution

Step 1: Express $y$ in terms of $x$

We are given:

$$x = \sin \theta \quad \Rightarrow \quad \theta = \sin^{-1}(x)$$

and

$$y = \sin(m \theta) \quad \Rightarrow \quad y = \sin(m \sin^{-1}(x)).$$

Step 2: Differentiate $y$ with respect to $x$

To find $y_1$, differentiate both sides of $y = \sin(m \sin^{-1}(x))$ with respect to $x$ using the chain rule:

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

The derivative is:

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

Step 3: Multiply both sides by $\sqrt{1 - x^2}$

This yields the expression for $y_1$ multiplied by $\sqrt{1 - x^2}$:

$$\left( \sqrt{1 - x^2} \right) y_1 = m \cos(m \sin^{-1}(x)).$$

Step 4: Square both sides

To get a more manageable expression, square both sides:

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

Step 5: Use the identity $\cos^2 \theta = 1 - \sin^2 \theta$

Now, apply the trigonometric identity $\cos^2 \theta = 1 - \sin^2 \theta$:

$$(1 - x^2) y_1^2 = m^2 \left( 1 - \sin^2(m \sin^{-1}(x)) \right).$$

Step 6: Substitute $y^2 = \sin^2(m \sin^{-1}(x))$

Since $y = \sin(m \sin^{-1}(x))$, we know that $y^2 = \sin^2(m \sin^{-1}(x))$, so we can substitute:

$$(1 - x^2) y_1^2 = m^2 (1 - y^2).$$

Step 7: Differentiate again with respect to $x$

Next, differentiate the entire equation with respect to $x$ to find $y_2$:

$$\frac{d}{dx} \left[ (1 - x^2) 2 y_1 y_2 - 2 x y_1^2 \right] = \frac{d}{dx} \left[ m^2 (-2y y_1) \right].$$

This simplifies to:

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

Step 8: Rearrange the terms

Now, rearrange the terms to isolate the desired equation:

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

Thus, we have proven that:

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

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Key Formulas or Methods Used

• Chain rule for derivatives: $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$
• Trigonometric identity: $\cos^2 \theta = 1 - \sin^2 \theta$

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Summary of Steps

1. Express $x = \sin \theta$ and $y = \sin(m \theta)$.
2. Differentiate $y = \sin(m \sin^{-1}(x))$ with respect to $x$ using the chain rule to get $y_1$.
3. Multiply both sides by $\sqrt{1 - x^2}$.
4. Square both sides to get $(1 - x^2) y_1^2 = m^2 (1 - y^2)$.
5. Differentiate again with respect to $x$ to get the second derivative $y_2$.
6. Rearrange the terms to show that $(1 - x^2) y_2 - x y_1 + m^2 y = 0$.

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