The solution of the differential equation $\left( {1 + {y^2}} \right) + \left( {x - {e^{{\mathop{\rm t}\nolimits} {{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}} = 0$ is

$x{e^{{{\tan }^{ - 1}}y}}{ = ^{{{\tan }^{ - 1}}y}} + c$

$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + c$

$\left( {x - 2} \right) = c{e^{ - {{\tan }^{ - 1}}y}}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of the differential equation

\left( {1 + {y^2}} \right) + \left( {x - {e^{{\mathop\mathrm{t}\nolimits} {{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}} = 0

View answer and explanation

Correct option: A

2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + c

Option A (

2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + c

) is correct because it is the solution obtained by solving the given linear differential equation using the integrating factor method.

Practice All Chapter MCQs →

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\lim \limits_{{x \rightarrow \infty}} y(t)