Question

If y = cot x show that \[\frac{d^2 y}{d x^2} + 2y\frac{dy}{dx} = 0\] ?

Answer 1

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.