Question
Solve the differential equation $ (\tan^{–1} \text{x – y) dx = (1 + x}^{2}) \text{ dy}.$

Answer

Given differential equation can be written as
$(1 + \text{x}^{2}) \frac{\text{dy}}{\text{dx}} + \text{y} = \tan^{-1} \text{x} \Rightarrow \frac{\text{dy}}{\text{dx}} + \frac{1} {1 + \text{x}^{2}} \text{y} = \frac{\tan^{-1} \text{x}}{\text{1 + x}^{2}}$
Integrating factor $ = \text{e}^{\tan^{-1}} \text{x}.$
$\therefore \text{Solution is y . e}^{\tan^{-1}}\text{x} = \int \tan^{-1} \text{x. e}^{\tan^{-1} \text{x}} \frac{1}{1 + \text{x}^{2}} \text{dx}$
$\Rightarrow \text{y. e}^{\tan^{-1}} \text{x} = \text{e}^{\tan^{-1}} \text{x} . (\tan^{-1}\text{x} - 1) + \text{c}$
$\text{ or } \text{ y} = (\tan^{-1} \text{x - 1)} + \text{c . e}^{-\tan^{-1_{\text{x}}}}$

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