Question

If y = (cot⁻¹ x)², prove that y₂(x² + 1)² + 2x (x² + 1) y₁ = 2 ?

Answer 1

Here,

\[y = \left( \cot^{- 1} x \right)^2 \]

\[\text { Differentiating w . r . t . x, we get }\]

\[ y_1 = 2co t^{- 1} x \times \frac{- 1}{1 + x^2} = \frac{- 2co t^{- 1} x}{1 + x^2}\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[ y_2 = \frac{2 + 4x \cot^{- 1} x}{\left( 1 + x^2 \right)^2}\]

\[ \Rightarrow y_2 = \frac{2}{\left( 1 + x^2 \right)^2} + \frac{2x \times 2 \cot^{- 1} x}{\left( 1 + x^2 \right)\left( 1 + x^2 \right)}\]

\[ \Rightarrow y_2 = \frac{2}{\left( 1 + x^2 \right)^2} - \frac{2x y_1}{\left( 1 + x^2 \right)}\]

\[ \Rightarrow \left( 1 + x^2 \right)^2 y_2 = 2 - 2x y_1 \left( 1 + x^2 \right)\]

\[ \Rightarrow \left( 1 + x^2 \right)^2 y_2 + 2x y_1 \left( 1 + x^2 \right) = 2\]

Hence proved.