The solution of $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy - 4{x^2} = 0$ is

$3x\left( {1 + {y^2}} \right) = 4{y^3} + c$

$3x\left( {1 - {y^2}} \right) = 4{y^3} + c$

$3y\left( {1 + {y^2}} \right) = 4{x^3} + c$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The solution of

\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy - 4{x^2} = 0

View answer and explanation

Correct option: B

3y\left( {1 + {x^2}} \right) = 4{x^3} + c

Option B (

3y\left( {1 + {x^2}} \right) = 4{x^3} + c

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

Practice All Chapter MCQs →

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