Question

If the transformation $z = \log \tan \frac{x}{2}$ reduces the differential equation $\frac{d^2y}{dx^2} + \cot x \frac{dy}{dx} + 4y \csc^2 x = 0$ into the form $\frac{d^2y}{dz^2} + ky = 0$, then $k$ is equal to

• A. 4
• B. -4
• C. 2
• D. -2

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{d^2y}{dx^2} + \cot x \frac{dy}{dx} + 4y \csc^2 x = 0 \] and the transformation: \[ z = \log \tan \frac{x}{2} \]

Step 1: Finding expressions for derivatives of \( y \) in terms of \( z \)
First, we need to express the derivatives \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) in terms of \( \frac{dy}{dz} \) and \( \frac{d^2y}{dz^2} \). To do this, let's compute the derivatives of \( z \) with respect to \( x \). From the given transformation, we have: \[ z = \log \tan \frac{x}{2} \] Taking the derivative of \( z \) with respect to \( x \), we get: \[ \frac{dz}{dx} = \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} \] Now, using the chain rule, we can express \( \frac{dy}{dx} \) as: \[ \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = \frac{dy}{dz} \cdot \frac{1}{\sin x} \] Next, we find \( \frac{d^2y}{dx^2} \). Using the product rule and chain rule, we get: \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{dy}{dz} \cdot \frac{1}{\sin x} \right) \] \[ = \frac{d^2y}{dz^2} \cdot \left( \frac{1}{\sin x} \right)^2 - \frac{dy}{dz} \cdot \frac{\cos x}{\sin^2 x} \] 

Step 2: Substitute into the original equation
Now, we substitute the expressions for \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the given differential equation: \[ \frac{d^2y}{dx^2} + \cot x \frac{dy}{dx} + 4y \csc^2 x = 0 \] Substituting the derivatives, we get: \[ \frac{d^2y}{dz^2} \cdot \frac{1}{\sin^2 x} - \frac{dy}{dz} \cdot \frac{\cos x}{\sin^2 x} + \cot x \cdot \frac{dy}{dz} \cdot \frac{1}{\sin x} + 4y \csc^2 x = 0 \] Simplifying, we see that the terms involving \( \frac{dy}{dz} \) cancel out, leaving us with: \[ \frac{d^2y}{dz^2} + 4y = 0 \] 

Step 3: Conclusion
This is the required differential equation: \[ \frac{d^2y}{dz^2} + ky = 0 \] where \( k = -4 \).

Answer:

\[ \boxed{k = -4} \]