Question

If x=sinθ and y=sin kθ, then (1-x²)y₂-xy₁-αy=0, for α=

• A. k
• B. -k
• C. -k²
• D. k²

Answer 1

The Correct Option is C

Step 1: Define x and y. 

We are given \( x = \sin\theta \) and \( y = \sin(k\theta) \).

Step 2: Calculate y1 (dy/dx).

First, we find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \):

\[ \frac{dy}{d\theta} = k\cos(k\theta) \]

\[ \frac{dx}{d\theta} = \cos\theta \]

Now, we can find \( y_1 = \frac{dy}{dx} \) using the chain rule:

\[ y_1 = \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{k\cos(k\theta)}{\cos\theta} \]

Step 3: Calculate y2 (d^2y/dx^2).

We need to find \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(y_1) \). We use the chain rule again:

\[ y_2 = \frac{d}{dx}(y_1) = \frac{d}{d\theta}(y_1) \cdot \frac{d\theta}{dx} = \frac{\frac{d}{d\theta}(y_1)}{\frac{dx}{d\theta}} \]

First, let's find \( \frac{d}{d\theta}(y_1) \):

\[ \frac{d}{d\theta}\left( \frac{k\cos(k\theta)}{\cos\theta} \right) = k \frac{-k\sin(k\theta)\cos\theta - \cos(k\theta)(-\sin\theta)}{\cos^2\theta} = k \frac{\sin\theta\cos(k\theta) - k\sin(k\theta)\cos\theta}{\cos^2\theta} \]

Now, we can find \( y_2 \):

\[ y_2 = \frac{k \frac{\sin\theta\cos(k\theta) - k\sin(k\theta)\cos\theta}{\cos^2\theta}}{\cos\theta} = k \frac{\sin\theta\cos(k\theta) - k\sin(k\theta)\cos\theta}{\cos^3\theta} \]

Step 4: Substitute into the given equation.

We are given \( (1-x^2)y_2 - xy_1 - \alpha y = 0 \). Substitute the expressions for \( x, y, y_1, \) and \( y_2 \):

\[ (1-\sin^2\theta) \left( k \frac{\sin\theta\cos(k\theta) - k\sin(k\theta)\cos\theta}{\cos^3\theta} \right) - \sin\theta \left( \frac{k\cos(k\theta)}{\cos\theta} \right) - \alpha \sin(k\theta) = 0 \]

Since \( 1 - \sin^2\theta = \cos^2\theta \), we can simplify:

\[ \cos^2\theta \left( k \frac{\sin\theta\cos(k\theta) - k\sin(k\theta)\cos\theta}{\cos^3\theta} \right) - \sin\theta \left( \frac{k\cos(k\theta)}{\cos\theta} \right) - \alpha \sin(k\theta) = 0 \]

\[ k \frac{\sin\theta\cos(k\theta) - k\sin(k\theta)\cos\theta}{\cos\theta} - \frac{k\sin\theta\cos(k\theta)}{\cos\theta} - \alpha \sin(k\theta) = 0 \]

\[ \frac{k\sin\theta\cos(k\theta)}{\cos\theta} - \frac{k^2\sin(k\theta)\cos\theta}{\cos\theta} - \frac{k\sin\theta\cos(k\theta)}{\cos\theta} - \alpha \sin(k\theta) = 0 \]

\[ - \frac{k^2\sin(k\theta)\cos\theta}{\cos\theta} - \alpha \sin(k\theta) = 0 \]

\[ - k^2\sin(k\theta) - \alpha \sin(k\theta) = 0 \]

\[ (-k^2 - \alpha) \sin(k\theta) = 0 \]

Step 5: Solve for α.

For the equation to hold true for all \( \theta \), we must have:

\[ -k^2 - \alpha = 0 \]

\[ \alpha = -k^2 \]

Final Answer

Therefore, \( \alpha = -k^2 \).