Solve the differential equation (1 + x2) dy dx + y = e^ ^ -1 x. - Vidyadip

Question

Solve the differential equation (1 + x²) $\frac{\text{dy}}{\text{dx}} + \text{y} = \text{e}^{\tan^{-1}\text{x}.}$

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Answer

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

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