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Question
If y = cot x show that \[\frac{d^2 y}{d x^2} + 2y\frac{dy}{dx} = 0\] ?
Solution
Here,
\[y = \cot x\]
\[\text { Differentiating w . r . t . x, we get } \]
\[\frac{d y}{d x} = - {cosec}^2 x\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = - 2 \ cosec \ x \times \left( - cosec \ x \ cot \ x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 2 \ {cosec}^2 \ x \ cot x\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - 2y\frac{dy}{dx}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} + 2y\frac{dy}{dx} = 0\]
Hence proved.
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