Given frac{d^2 y}{d x^2}+cot x frac{d y}{d x} | vedclass

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(D) Given the substitution $z = \log 	an \frac{x}{2}$.Then, $\frac{dz}{dx} = \frac{1}{	an(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{1}{\sin

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$\frac{d^2 y}{d z^2}+4 y=0$

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(D) Given the substitution $z = \log 	an \frac{x}{2}$.Then, $\frac{dz}{dx} = \frac{1}{	an(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{1}{\sin x} = \operatorname{cosec} x$.Now, $\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = \operatorname{cosec} x \frac{dy}{dz}$.Next, $\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \operatorname{cosec} x \frac{dy}{dz} ight) = \frac{d}{dz} \left( \operatorname{cosec} x \frac{dy}{dz} ight) \cdot \frac{dz}{dx} = \left( -\operatorname{cosec} x \cot x \frac{dy}{dz} + \operatorname{cosec} x \frac{d^2 y}{dz^2} ight) \operatorname{cosec} x = \operatorname{cosec}^2 x \frac{d^2 y}{dz^2} - \operatorname{cosec} x \cot x \frac{dy}{dz}$.Substituting these into the original equation $\frac{d^2 y}{dx^2} + \cot x \frac{dy}{dx} + 4y \operatorname{cosec}^2 x = 0$:$\left( \operatorname{cosec}^2 x \frac{d^2 y}{dz^2} - \operatorname{cosec} x \cot x \frac{dy}{dz} ight) + \cot x (\operatorname{cosec} x \frac{dy}{dz}) + 4y \operatorname{cosec}^2 x = 0$.$\operatorname{cosec}^2 x \frac{d^2 y}{dz^2} + 4y \operatorname{cosec}^2 x = 0$.Dividing by $\operatorname{cosec}^2 x$, we get $\frac{d^2 y}{dz^2} + 4y = 0$.

Given $\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}^2 x=0$. Changing the independent variable $x$ to $z$ by the substitution $z=\log \tan \frac{x}{2}$, the equation is changed to

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