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Question
If \[y = e^{\tan^{- 1} x}\] prove that (1 + x2)y2 + (2x − 1)y1 = 0 ?
Solution
Here,
\[y = e^{\tan^{- 1} x }\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = e^{\tan^{- 1} x } \times \frac{1}{1 + x^2}\]
\[\text { Differentiating again w . r . t . x, we get}\]
\[\frac{d^2 y}{d x^2} = e^{\tan^{- 1} x} \frac{1}{\left( 1 + x^2 \right)^2} + e^{\tan^{- 1} x} \frac{- 2x}{\left( 1 + x^2 \right)^2}\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} = \frac{e^{\tan^{- 1} x}}{\left( 1 + x^2 \right)} - \frac{2x e^{\tan^{- 1} x}}{\left( 1 + x^2 \right)}\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} = \frac{d y}{d x} - 2x\frac{d y}{d x}\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{d y}{d x} = 0\]
Hence proved
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