Solve the following differential equation:
$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$

Solution

$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$

The equation can be expressed as

$\frac{dy}{dx} + \frac{y}{1 + x^{2}} = \frac{\tan^{- 1} x}{1 + x^{2}} . . . . . . . . . (i)$

This is a linear differential equation of the type $\frac{dy}{dx} + Py = Q$

$So, I . F = e^{\int \frac{1}{1 + x^{2}} dx} = e^{\tan^{- 1} x} Solution of (i) y e^{\tan^{- 1} x} = \int e^{\tan^{- 1} x} (\frac{\tan^{- 1} x}{1 + x^{2}}) dx . . . . . . . . (ii) For R . H . S ., let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} dx = dt$

Substituting in equation (ii)

${ye}^{\tan^{- 1} x} = \int e^{t} . tdt \Rightarrow y . e^{\tan^{- 1} x} = [{te}^{t} - e^{t}] + C \Rightarrow y . e^{\tan^{- 1} x} = e^{\tan^{- 1} x} (\tan^{- 1} x - 1) + C \Rightarrow y = (\tan^{- 1} x - 1) + {Ce}^{- \tan^{- 1} x}$

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Differential Equations

Solve the following differential equation:
$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$

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Mock Test Series

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Differential Equations

Solve the following differential equation: 1 + x2 dydx + y = tan-1 x

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Solve the following differential equation:
$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$