If the transformation z = log tan frac{x}{2} | vedclass

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(B) Given the transformation $z = \log 	an \frac{x}{2}$.First, find $\frac{d z}{d x}$:$\frac{d z}{d x} = \frac{1}{	an \frac{x}{2}} \cdot \sec^2 \fra

Vedclass · Accepted Answer

$4$

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(B) Given the transformation $z = \log 	an \frac{x}{2}$.First, find $\frac{d z}{d x}$:$\frac{d z}{d x} = \frac{1}{	an \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.Now, express $\frac{d y}{d z}$ in terms of $x$:$\frac{d y}{d z} = \frac{d y}{d x} \cdot \frac{d x}{d z} = \frac{d y}{d x} \cdot \sin x$.Next, find $\frac{d^2 y}{d z^2}$:$\frac{d^2 y}{d z^2} = \frac{d}{d z} \left( \sin x \frac{d y}{d x} ight) = \frac{d}{d x} \left( \sin x \frac{d y}{d x} ight) \cdot \frac{d x}{d z} = \left( \cos x \frac{d y}{d x} + \sin x \frac{d^2 y}{d x^2} ight) \cdot \sin x = \sin x \cos x \frac{d y}{d x} + \sin^2 x \frac{d^2 y}{d x^2}$.Substitute this into the target equation $\frac{d^2 y}{d z^2} + k y = 0$:$\sin^2 x \frac{d^2 y}{d x^2} + \sin x \cos x \frac{d y}{d x} + k y = 0$.Divide by $\sin^2 x$:$\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + k \operatorname{cosec}^2 x \cdot y = 0$.Comparing this with the given equation $\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + 4 \operatorname{cosec}^2 x \cdot y = 0$, we get $k = 4$.

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

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