The solution of $\frac{{dy}}{{dx}} = 1 + y + {y^2} + x + xy + x{y^2}$ is

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

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

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

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The solution of

\frac{{dy}}{{dx}} = 1 + y + {y^2} + x + xy + x{y^2}

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it is the solution obtained by separating variables and integrating the given differential equation.

Practice All Chapter MCQs →

y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

{x^2} + {y^2} = 1

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

y=y_{1}(x)

y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

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