Question

If log y = tan⁻¹ x, show that (1 + x²)y₂ + (2x − 1) y₁ = 0 ?

Answer 1

Here,

\[\log y = \tan^{- 1} x\]

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

\[\frac{1}{y} \times y_1 = \frac{1}{1 + x^2}\]

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

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

\[ \Rightarrow \left( 1 + x^2 \right) y_2 + 2x y_1 - y_1 = 0\]

\[ \Rightarrow \left( 1 + x^2 \right) y_2 + \left( 2x - 1 \right) y_1 = 0\]

Hence proved.