The solution of $\left( {x + y + 1} \right)\frac{{dy}}{{dx}} = 1$ is

$y = \left( {x + 2} \right) + c{e^x}$

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

$x = {\left( {y + 2} \right)^2} + c{e^y}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

\left( {x + y + 1} \right)\frac{{dy}}{{dx}} = 1

View answer and explanation

Correct option: C

x = - \left( {y + 2} \right) + c{e^y}

Option C,

x = - \left( {y + 2} \right) + c{e^y}

, is the correct solution because it satisfies the given differential equation after solving it as a linear differential equation in terms of x.

Practice All Chapter MCQs →

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