JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Medium

1 of 1

Application +4 | -1

The solution of the differential equation

\left( {1 + {y^2}} \right){\tan ^{ - 1}}xdx + y\left( {1 + {x^2}} \right)dy = 0

View answer and explanation

Correct option: B

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

Option B (

\log (1 + {y^2}) + {({{\tan }^{ - 1}}x)^2} = c

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

More Questions from DIFFERENTIAL EQUATIONS

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)